cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A080390 Least x such that gcd(A001405(x+1), A001405(x)) = n.

Original entry on oeis.org

1, 5, 14, 27, 34, 11, 62, 183, 98, 39, 142, 107, 220, 69, 44, 495, 322, 143, 436, 139, 482, 637, 574, 119, 674, 233, 782, 55, 898, 29, 1146, 4063, 230, 441, 174, 467, 1516, 493, 662, 239, 1762, 377, 2020, 175, 764, 781, 2302, 2543, 2596, 949, 968, 207, 3126, 1241
Offset: 1

Views

Author

Labos Elemer, Mar 12 2003

Keywords

Crossrefs

Programs

  • Mathematica
    f[x_] := GCD[Binomial[x+1, Floor[x+1/2]], Binomial[x, Floor[x/2]]] t=Table[0, {257}]; Do[s=f[n]; If[s<258&&t[[s]]==0, t[[s]]=n], {n, 1, 10000}]; t

Formula

a(n) = Min{x; A057977(x)=n}.

A080391 Number of integers among the following n+1 quotients: A001405(n)/A001405(k), k=0,..,n. Number of values k for which the n-th central binomial coefficient[cbc] is divisible by the k-th cbc-s k=0,..n.

Original entry on oeis.org

1, 2, 3, 3, 5, 4, 5, 3, 6, 6, 7, 6, 7, 6, 7, 4, 8, 5, 7, 4, 5, 6, 7, 4, 5, 8, 9, 8, 9, 8, 9, 4, 8, 7, 9, 10, 13, 12, 13, 13, 18, 9, 11, 8, 9, 8, 9, 7, 9, 8, 9, 8, 9, 4, 5, 8, 9, 12, 13, 4, 5, 4, 5, 4, 9, 9, 12, 9, 11, 8, 9, 4, 5, 4, 5, 10, 11, 8, 9, 7, 9, 14, 15, 14, 15, 14, 15, 12, 13, 12, 13, 12, 13
Offset: 0

Views

Author

Labos Elemer, Mar 13 2003

Keywords

Examples

			n=11: the 12 relevant C(11,5)/C(k,floor(k/2)) quotients are {462,462,231,154,77,231/5,231/10,66/5,33/5,11/3,11/6,1}; of which a(12)=6 are integers.
		

Crossrefs

Cf. A001405.

Programs

  • Mathematica
    Table[Count[Table[IntegerQ[Binomial[n, Floor[n/2]]/ Binomial[i, Floor[i/2]]], {i, 0, n}], True], {n, 1, 256}]

A000079 Powers of 2: a(n) = 2^n.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592
Offset: 0

Views

Author

Keywords

Comments

2^0 = 1 is the only odd power of 2.
Number of subsets of an n-set.
There are 2^(n-1) compositions (ordered partitions) of n (see for example Riordan). This is the unlabeled analog of the preferential labelings sequence A000670.
This is also the number of weakly unimodal permutations of 1..n + 1, that is, permutations with exactly one local maximum. E.g., a(4) = 16: 12345, 12354, 12453, 12543, 13452, 13542, 14532 and 15432 and their reversals. - Jon Perry, Jul 27 2003 [Proof: see next line! See also A087783.]
Proof: n must appear somewhere and there are 2^(n-1) possible choices for the subset that precedes it. These must appear in increasing order and the rest must follow n in decreasing order. QED. - N. J. A. Sloane, Oct 26 2003
a(n+1) is the smallest number that is not the sum of any number of (distinct) earlier terms.
Same as Pisot sequences E(1, 2), L(1, 2), P(1, 2), T(1, 2). See A008776 for definitions of Pisot sequences.
With initial 1 omitted, same as Pisot sequences E(2, 4), L(2, 4), P(2, 4), T(2, 4). - David W. Wilson
Not the sum of two or more consecutive numbers. - Lekraj Beedassy, May 14 2004
Least deficient or near-perfect numbers (i.e., n such that sigma(n) = A000203(n) = 2n - 1). - Lekraj Beedassy, Jun 03 2004. [Comment from Max Alekseyev, Jan 26 2005: All the powers of 2 are least deficient numbers but it is not known if there exists a least deficient number that is not a power of 2.]
Almost-perfect numbers referred to as least deficient or slightly defective (Singh 1997) numbers. Does "near-perfect numbers" refer to both almost-perfect numbers (sigma(n) = 2n - 1) and quasi-perfect numbers (sigma(n) = 2n + 1)? There are no known quasi-perfect or least abundant or slightly excessive (Singh 1997) numbers.
The sum of the numbers in the n-th row of Pascal's triangle; the sum of the coefficients of x in the expansion of (x+1)^n.
The Collatz conjecture (the hailstone sequence will eventually reach the number 1, regardless of which positive integer is chosen initially) may be restated as (the hailstone sequence will eventually reach a power of 2, regardless of which positive integer is chosen initially).
The only hailstone sequence which doesn't rebound (except "on the ground"). - Alexandre Wajnberg, Jan 29 2005
With p(n) as the number of integer partitions of n, p(i) is the number of parts of the i-th partition of n, d(i) is the number of different parts of the i-th partition of n, m(i,j) is the multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i = 1..p(n)} (p(i)! / (Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
The number of binary relations on an n-element set that are both symmetric and antisymmetric. Also the number of binary relations on an n-element set that are symmetric, antisymmetric and transitive.
The first differences are the sequence itself. - Alexandre Wajnberg and Eric Angelini, Sep 07 2005
a(n) is the largest number with shortest addition chain involving n additions. - David W. Wilson, Apr 23 2006
Beginning with a(1) = 0, numbers not equal to the sum of previous distinct natural numbers. - Giovanni Teofilatto, Aug 06 2006
For n >= 1, a(n) is equal to the number of functions f:{1, 2, ..., n} -> {1, 2} such that for a fixed x in {1, 2, ..., n} and a fixed y in {1, 2} we have f(x) != y. - Aleksandar M. Janjic and Milan Janjic, Mar 27 2007
Let P(A) be the power set of an n-element set A. Then a(n) is the number of pairs of elements {x,y} of P(A) for which x = y. - Ross La Haye, Jan 09 2008
a(n) is the number of permutations on [n+1] such that every initial segment is an interval of integers. Example: a(3) counts 1234, 2134, 2314, 2341, 3214, 3241, 3421, 4321. The map "p -> ascents of p" is a bijection from these permutations to subsets of [n]. An ascent of a permutation p is a position i such that p(i) < p(i+1). The permutations shown map to 123, 23, 13, 12, 3, 2, 1 and the empty set respectively. - David Callan, Jul 25 2008
2^(n-1) is the largest number having n divisors (in the sense of A077569); A005179(n) is the smallest. - T. D. Noe, Sep 02 2008
a(n) appears to match the number of divisors of the modified primorials (excluding 2, 3 and 5). Very limited range examined, PARI example shown. - Bill McEachen, Oct 29 2008
Successive k such that phi(k)/k = 1/2, where phi is Euler's totient function. - Artur Jasinski, Nov 07 2008
A classical transform consists (for general a(n)) in swapping a(2n) and a(2n+1); examples for Jacobsthal A001045 and successive differences: A092808, A094359, A140505. a(n) = A000079 leads to 2, 1, 8, 4, 32, 16, ... = A135520. - Paul Curtz, Jan 05 2009
This is also the (L)-sieve transform of {2, 4, 6, 8, ..., 2n, ...} = A005843. (See A152009 for the definition of the (L)-sieve transform.) - John W. Layman, Jan 23 2009
a(n) = a(n-1)-th even natural number (A005843) for n > 1. - Jaroslav Krizek, Apr 25 2009
For n >= 0, a(n) is the number of leaves in a complete binary tree of height n. For n > 0, a(n) is the number of nodes in an n-cube. - K.V.Iyer, May 04 2009
Permutations of n+1 elements where no element is more than one position right of its original place. For example, there are 4 such permutations of three elements: 123, 132, 213, and 312. The 8 such permutations of four elements are 1234, 1243, 1324, 1423, 2134, 2143, 3124, and 4123. - Joerg Arndt, Jun 24 2009
Catalan transform of A099087. - R. J. Mathar, Jun 29 2009
a(n) written in base 2: 1,10,100,1000,10000,..., i.e., (n+1) times 1, n times 0 (A011557(n)). - Jaroslav Krizek, Aug 02 2009
Or, phi(n) is equal to the number of perfect partitions of n. - Juri-Stepan Gerasimov, Oct 10 2009
These are the 2-smooth numbers, positive integers with no prime factors greater than 2. - Michael B. Porter, Oct 04 2009
A064614(a(n)) = A000244(n) and A064614(m) < A000244(n) for m < a(n). - Reinhard Zumkeller, Feb 08 2010
a(n) is the largest number m such that the number of steps of iterations of {r - (largest divisor d < r)} needed to reach 1 starting at r = m is equal to n. Example (a(5) = 32): 32 - 16 = 16; 16 - 8 = 8; 8 - 4 = 4; 4 - 2 = 2; 2 - 1 = 1; number 32 has 5 steps and is the largest such number. See A105017, A064097, A175125. - Jaroslav Krizek, Feb 15 2010
a(n) is the smallest proper multiple of a(n-1). - Dominick Cancilla, Aug 09 2010
The powers-of-2 triangle T(n, k), n >= 0 and 0 <= k <= n, begins with: {1}; {2, 4}; {8, 16, 32}; {64, 128, 256, 512}; ... . The first left hand diagonal T(n, 0) = A006125(n + 1), the first right hand diagonal T(n, n) = A036442(n + 1) and the center diagonal T(2*n, n) = A053765(n + 1). Some triangle sums, see A180662, are: Row1(n) = A122743(n), Row2(n) = A181174(n), Fi1(n) = A181175(n), Fi2(2*n) = A181175(2*n) and Fi2(2*n + 1) = 2*A181175(2*n + 1). - Johannes W. Meijer, Oct 10 2010
Records in the number of prime factors. - Juri-Stepan Gerasimov, Mar 12 2011
Row sums of A152538. - Gary W. Adamson, Dec 10 2008
A078719(a(n)) = 1; A006667(a(n)) = 0. - Reinhard Zumkeller, Oct 08 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=1, a(n) equals the number of 2-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Equals A001405 convolved with its right-shifted variant: (1 + 2x + 4x^2 + ...) = (1 + x + 2x^2 + 3x^3 + 6x^4 + 10x^5 + ...) * (1 + x + x^2 + 2x^3 + 3x^4 + 6x^5 + ...). - Gary W. Adamson, Nov 23 2011
The number of odd-sized subsets of an n+1-set. For example, there are 2^3 odd-sized subsets of {1, 2, 3, 4}, namely {1}, {2}, {3}, {4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. Also, note that 2^n = Sum_{k=1..floor((n+1)/2)} C(n+1, 2k-1). - Dennis P. Walsh, Dec 15 2011
a(n) is the number of 1's in any row of Pascal's triangle (mod 2) whose row number has exactly n 1's in its binary expansion (see A007318 and A047999). (The result of putting together A001316 and A000120.) - Marcus Jaiclin, Jan 31 2012
A204455(k) = 1 if and only if k is in this sequence. - Wolfdieter Lang, Feb 04 2012
For n>=1 apparently the number of distinct finite languages over a unary alphabet, whose minimum regular expression has alphabetic width n (verified up to n=17), see the Gruber/Lee/Shallit link. - Hermann Gruber, May 09 2012
First differences of A000225. - Omar E. Pol, Feb 19 2013
This is the lexicographically earliest sequence which contains no arithmetic progression of length 3. - Daniel E. Frohardt, Apr 03 2013
a(n-2) is the number of bipartitions of {1..n} (i.e., set partitions into two parts) such that 1 and 2 are not in the same subset. - Jon Perry, May 19 2013
Numbers n such that the n-th cyclotomic polynomial has a root mod 2; numbers n such that the n-th cyclotomic polynomial has an even number of odd coefficients. - Eric M. Schmidt, Jul 31 2013
More is known now about non-power-of-2 "Almost Perfect Numbers" as described in Dagal. - Jonathan Vos Post, Sep 01 2013
Number of symmetric Ferrers diagrams that fit into an n X n box. - Graham H. Hawkes, Oct 18 2013
Numbers n such that sigma(2n) = 2n + sigma(n). - Jahangeer Kholdi, Nov 23 2013
a(1), ..., a(floor(n/2)) are all values of permanent on set of square (0,1)-matrices of order n>=2 with row and column sums 2. - Vladimir Shevelev, Nov 26 2013
Numbers whose base-2 expansion has exactly one bit set to 1, and thus has base-2 sum of digits equal to one. - Stanislav Sykora, Nov 29 2013
A072219(a(n)) = 1. - Reinhard Zumkeller, Feb 20 2014
a(n) is the largest number k such that (k^n-2)/(k-2) is an integer (for n > 1); (k^a(n)+1)/(k+1) is never an integer (for k > 1 and n > 0). - Derek Orr, May 22 2014
If x = A083420(n), y = a(n+1) and z = A087289(n), then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
The mini-sequence b(n) = least number k > 0 such that 2^k ends in n identical digits is given by {1, 18, 39}. The repeating digits are {2, 4, 8} respectively. Note that these are consecutive powers of 2 (2^1, 2^2, 2^3), and these are the only powers of 2 (2^k, k > 0) that are only one digit. Further, this sequence is finite. The number of n-digit endings for a power of 2 with n or more digits id 4*5^(n-1). Thus, for b(4) to exist, one only needs to check exponents up to 4*5^3 = 500. Since b(4) does not exist, it is clear that no other number will exist. - Derek Orr, Jun 14 2014
The least number k > 0 such that 2^k ends in n consecutive decreasing digits is a 3-number sequence given by {1, 5, 25}. The consecutive decreasing digits are {2, 32, 432}. There are 100 different 3-digit endings for 2^k. There are no k-values such that 2^k ends in '987', '876', '765', '654', '543', '321', or '210'. The k-values for which 2^k ends in '432' are given by 25 mod 100. For k = 25 + 100*x, the digit immediately before the run of '432' is {4, 6, 8, 0, 2, 4, 6, 8, 0, 2, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus, we see the digit before '432' will never be a 5. So, this sequence is complete. - Derek Orr, Jul 03 2014
a(n) is the number of permutations of length n avoiding both 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Numbers n such that sigma(n) = sigma(2n) - phi(4n). - Farideh Firoozbakht, Aug 14 2014
This is a B_2 sequence: for i < j, differences a(j) - a(i) are all distinct. Here 2*a(n) < a(n+1) + 1, so a(n) - a(0) < a(n+1) - a(n). - Thomas Ordowski, Sep 23 2014
a(n) counts n-walks (closed) on the graph G(1-vertex; 1-loop, 1-loop). - David Neil McGrath, Dec 11 2014
a(n-1) counts walks (closed) on the graph G(1-vertex; 1-loop, 2-loop, 3-loop, 4-loop, ...). - David Neil McGrath, Jan 01 2015
b(0) = 4; b(n+1) is the smallest number not in the sequence such that b(n+1) - Prod_{i=0..n} b(i) divides b(n+1) - Sum_{i=0..n} b(i). Then b(n) = a(n) for n > 2. - Derek Orr, Jan 15 2015
a(n) counts the permutations of length n+2 whose first element is 2 such that the permutation has exactly one descent. - Ran Pan, Apr 17 2015
a(0)-a(30) appear, with a(26)-a(30) in error, in tablet M 08613 (see CDLI link) from the Old Babylonian period (c. 1900-1600 BC). - Charles R Greathouse IV, Sep 03 2015
Subsequence of A028982 (the squares or twice squares sequence). - Timothy L. Tiffin, Jul 18 2016
A000120(a(n)) = 1. A000265(a(n)) = 1. A000593(a(n)) = 1. - Juri-Stepan Gerasimov, Aug 16 2016
Number of monotone maps f : [0..n] -> [0..n] which are order-increasing (i <= f(i)) and idempotent (f(f(i)) = f(i)). In other words, monads on the n-th ordinal (seen as a posetal category). Any monad f determines a subset of [0..n] that contains n, by considering its set of monad algebras = fixed points { i | f(i) = i }. Conversely, any subset S of [0..n] containing n determines a monad on [0..n], by the function i |-> min { j | i <= j, j in S }. - Noam Zeilberger, Dec 11 2016
Consider n points lying on a circle. Then for n>=2 a(n-2) gives the number of ways to connect two adjacent points with nonintersecting chords. - Anton Zakharov, Dec 31 2016
Satisfies Benford's law [Diaconis, 1977; Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
Also the number of independent vertex sets and vertex covers in the n-empty graph. - Eric W. Weisstein, Sep 21 2017
Also the number of maximum cliques in the n-halved cube graph for n > 4. - Eric W. Weisstein, Dec 04 2017
Number of pairs of compositions of n corresponding to a seaweed algebra of index n-1. - Nick Mayers, Jun 25 2018
The multiplicative group of integers modulo a(n) is cyclic if and only if n = 0, 1, 2. For n >= 3, it is a product of two cyclic groups. - Jianing Song, Jun 27 2018
k^n is the determinant of n X n matrix M_(i, j) = binomial(k + i + j - 2, j) - binomial(i+j-2, j), in this case k=2. - Tony Foster III, May 12 2019
Solutions to the equation Phi(2n + 2*Phi(2n)) = 2n. - M. Farrokhi D. G., Jan 03 2020
a(n-1) is the number of subsets of {1,2,...,n} which have an element that is the size of the set. For example, for n = 4, a(3) = 8 and the subsets are {1}, {1,2}, {2,3}, {2,4}, {1,2,3}, {1,3,4}, {2,3,4}, {1,2,3,4}. - Enrique Navarrete, Nov 21 2020
a(n) is the number of self-inverse (n+1)-order permutations with 231-avoiding. E.g., a(3) = 8: [1234, 1243, 1324, 1432, 2134, 2143, 3214, 4321]. - Yuchun Ji, Feb 26 2021
For any fixed k > 0, a(n) is the number of ways to tile a strip of length n+1 with tiles of length 1, 2, ... k, where the tile of length k can be black or white, with the restriction that the first tile cannot be black. - Greg Dresden and Bora Bursalı, Aug 31 2023

Examples

			There are 2^3 = 8 subsets of a 3-element set {1,2,3}, namely { -, 1, 2, 3, 12, 13, 23, 123 }.
		

References

  • Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 1016.
  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education Journal, Vol. 31, No. 1, pp. 24-28, Winter 1997.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 73, 84.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.5 Logarithms and §8.1 Terminology, pp. 150, 264.
  • Paul J. Nahin, An Imaginary Tale: The Story of sqrt(-1), Princeton University Press, Princeton, NJ. 1998, pp. 69-70.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 273.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 124.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • V. E. Tarakanov, Combinatorial problems on binary matrices, Combin. Analysis, MSU, 5 (1980), 4-15. (Russian)
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

This is the Hankel transform (see A001906 for the definition) of A000984, A002426, A026375, A026387, A026569, A026585, A026671 and A032351. - John W. Layman, Jul 31 2000
Euler transform of A001037, A209406 (multisets), inverse binomial transform of A000244, binomial transform of A000012.
Complement of A057716.
Boustrophedon transforms: A000734, A000752.
Range of values of A006519, A007875, A011782, A030001, A034444, A037445, A053644, and A054243.
Cf. A018900, A014311, A014312, A014313, A023688, A023689, A023690, A023691 (sum of 2, ..., 9 distinct powers of 2).
Cf. A090129.
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).

Programs

  • Haskell
    a000079 = (2 ^)
    a000079_list = iterate (* 2) 1
    -- Reinhard Zumkeller, Jan 22 2014, Mar 05 2012, Dec 29 2011
    
  • Magma
    [2^n: n in [0..40]]; // Vincenzo Librandi, Feb 17 2014
    
  • Magma
    [n le 2 select n else 5*Self(n-1)-6*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Feb 17 2014
    
  • Maple
    A000079 := n->2^n; [ seq(2^n,n=0..50) ];
    isA000079 := proc(n)
        local fs;
        fs := numtheory[factorset](n) ;
        if n = 1 then
            true ;
        elif nops(fs) <> 1 then
            false;
        elif op(1,fs) = 2 then
            true;
        else
            false ;
        end if;
    end proc: # R. J. Mathar, Jan 09 2017
  • Mathematica
    Table[2^n, {n, 0, 50}]
    2^Range[0, 50] (* Wesley Ivan Hurt, Jun 14 2014 *)
    LinearRecurrence[{2}, {2}, {0, 20}] (* Eric W. Weisstein, Sep 21 2017 *)
    CoefficientList[Series[1/(1 - 2 x), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
    NestList[2# &, 1, 40] (* Harvey P. Dale, Oct 07 2019 *)
  • Maxima
    A000079(n):=2^n$ makelist(A000079(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    A000079(n)=2^n \\ Edited by M. F. Hasler, Aug 27 2014
    
  • PARI
    unimodal(n)=local(x,d,um,umc); umc=0; for (c=0,n!-1, x=numtoperm(n,c); d=0; um=1; for (j=2,n,if (x[j]x[j-1] && d==1,um=0); if (um==0,break)); if (um==1,print(x)); umc+=um); umc
    
  • Python
    def a(n): return 1<Michael S. Branicky, Jul 28 2022
    
  • Python
    def is_powerof2(n) -> bool: return n and (n & (n - 1)) == 0  # Peter Luschny, Apr 10 2025
  • Scala
    (List.fill(20)(2: BigInt)).scanLeft(1: BigInt)( * ) // Alonso del Arte, Jan 16 2020
    
  • Scheme
    (define (A000079 n) (expt 2 n)) ;; Antti Karttunen, Mar 21 2017
    

Formula

a(n) = 2^n.
a(0) = 1; a(n) = 2*a(n-1).
G.f.: 1/(1 - 2*x).
E.g.f.: exp(2*x).
a(n)= Sum_{k = 0..n} binomial(n, k).
a(n) is the number of occurrences of n in A000523. a(n) = A001045(n) + A001045(n+1). a(n) = 1 + Sum_{k = 0..(n - 1)} a(k). The Hankel transform of this sequence gives A000007 = [1, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Feb 25 2004
n such that phi(n) = n/2, for n > 1, where phi is Euler's totient (A000010). - Lekraj Beedassy, Sep 07 2004
a(n + 1) = a(n) XOR 3*a(n) where XOR is the binary exclusive OR operator. - Philippe Deléham, Jun 19 2005
a(n) = StirlingS2(n + 1, 2) + 1. - Ross La Haye, Jan 09 2008
a(n+2) = 6a(n+1) - 8a(n), n = 1, 2, 3, ... with a(1) = 1, a(2) = 2. - Yosu Yurramendi, Aug 06 2008
a(n) = ka(n-1) + (4 - 2k)a(n-2) for any integer k and n > 1, with a(0) = 1, a(1) = 2. - Jaume Oliver Lafont, Dec 05 2008
a(n) = Sum_{l_1 = 0..n + 1} Sum_{l_2 = 0..n}...Sum_{l_i = 0..n - i}...Sum_{l_n = 0..1} delta(l_1, l_2, ..., l_i, ..., l_n) where delta(l_1, l_2, ..., l_i, ..., l_n) = 0 if any l_i <= l_(i+1) and l_(i+1) != 0 and delta(l_1, l_2, ..., l_i, ..., l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(0) = 1, a(1) = 2; a(n) = a(n-1)^2/a(n-2), n >= 2. - Jaume Oliver Lafont, Sep 22 2009
a(n) = A173786(n, n)/2 = A173787(n + 1, n). - Reinhard Zumkeller, Feb 28 2010
If p[i] = i - 1 and if A is the Hessenberg matrix of order n defined by: A[i, j] = p[j - i + 1], (i <= j), A[i, j] = -1, (i = j + 1), and A[i, j] = 0 otherwise. Then, for n >= 1, a(n-1) = det A. - Milan Janjic, May 02 2010
If p[i] = Fibonacci(i-2) and if A is the Hessenberg matrix of order n defined by: A[i, j] = p[j - i + 1], (i <= j), A[i, j] = -1, (i = j + 1), and A[i, j] = 0 otherwise. Then, for n >= 2, a(n-2) = det A. - Milan Janjic, May 08 2010
The sum of reciprocals, 1/1 + 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + ... = 2. - Mohammad K. Azarian, Dec 29 2010
a(n) = 2*A001045(n) + A078008(n) = 3*A001045(n) + (-1)^n. - Paul Barry, Feb 20 2003
a(n) = A118654(n, 2).
a(n) = A140740(n+1, 1).
a(n) = A131577(n) + A011782(n) = A024495(n) + A131708(n) + A024493(n) = A000749(n) + A038503(n) + A038504(n) + A038505(n) = A139761(n) + A139748(n) + A139714(n) + A133476(n) + A139398(n). - Paul Curtz, Jul 25 2011
a(n) = row sums of A007318. - Susanne Wienand, Oct 21 2011
a(n) = Hypergeometric([-n], [], -1). - Peter Luschny, Nov 01 2011
G.f.: A(x) = B(x)/x, B(x) satisfies B(B(x)) = x/(1 - x)^2. - Vladimir Kruchinin, Nov 10 2011
a(n) = Sum_{k = 0..n} A201730(n, k)*(-1)^k. - Philippe Deléham, Dec 06 2011
2^n = Sum_{k = 1..floor((n+1)/2)} C(n+1, 2k-1). - Dennis P. Walsh, Dec 15 2011
A209229(a(n)) = 1. - Reinhard Zumkeller, Mar 07 2012
A001227(a(n)) = 1. - Reinhard Zumkeller, May 01 2012
Sum_{n >= 1} mobius(n)/a(n) = 0.1020113348178103647430363939318... - R. J. Mathar, Aug 12 2012
E.g.f.: 1 + 2*x/(U(0) - x) where U(k) = 6*k + 1 + x^2/(6*k+3 + x^2/(6*k + 5 + x^2/U(k+1) )); (continued fraction, 3-step). - Sergei N. Gladkovskii, Dec 04 2012
a(n) = det(|s(i+2,j)|, 1 <= i,j <= n), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 04 2013
a(n) = det(|ps(i+1,j)|, 1 <= i,j <= n), where ps(n,k) are Legendre-Stirling numbers of the first kind (A129467). - Mircea Merca, Apr 06 2013
G.f.: W(0), where W(k) = 1 + 2*x*(k+1)/(1 - 2*x*(k+1)/( 2*x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
a(n-1) = Sum_{t_1 + 2*t_2 + ... + n*t_n = n} multinomial(t_1 + t_2 + ... + t_n; t_1, t_2, ..., t_n). - Mircea Merca, Dec 06 2013
Construct the power matrix T(n,j) = [A^*j]*[S^*(j-1)] where A(n)=(1,1,1,...) and S(n)=(0,1,0,0,...) (where * is convolution operation). Then a(n-1) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Jan 01 2015
a(n) = A000005(A002110(n)). - Ivan N. Ianakiev, May 23 2016
From Ilya Gutkovskiy, Jul 18 2016: (Start)
Exponential convolution of A000012 with themselves.
a(n) = Sum_{k=0..n} A011782(k).
Sum_{n>=0} a(n)/n! = exp(2) = A072334.
Sum_{n>=0} (-1)^n*a(n)/n! = exp(-2) = A092553. (End)
G.f.: (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = A090129(x) = (1 + 2x + 2x^2 + 4x^3 + 8x^4 + ...). - Gary W. Adamson, Sep 13 2016
a(n) = A000045(n + 1) + A000045(n) + Sum_{k = 0..n - 2} A000045(k + 1)*2^(n - 2 - k). - Melvin Peralta, Dec 22 2017
a(n) = 7*A077020(n)^2 + A077021(n)^2, n>=3. - Ralf Steiner, Aug 08 2021
a(n)= n + 1 + Sum_{k=3..n+1} (2*k-5)*J(n+2-k), where Jacobsthal number J(n) = A001045(n). - Michael A. Allen, Jan 12 2022
Integral_{x=0..Pi} cos(x)^n*cos(n*x) dx = Pi/a(n) (see Nahin, pp. 69-70). - Stefano Spezia, May 17 2023

Extensions

Clarified a comment T. D. Noe, Aug 30 2009
Edited by Daniel Forgues, May 12 2010
Incorrect comment deleted by Matthew Vandermast, May 17 2014
Comment corrected to match offset by Geoffrey Critzer, Nov 28 2014

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

Views

Author

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

Keywords

Comments

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

Examples

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
  • Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
  • William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
  • David Hök, Parvisa mönster i permutationer [Swedish], 2007.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
  • Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
  • Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
  • A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
  • John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
  • Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
  • Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Axiom
    -- (start)
    )set expose add constructor OutputForm
    pascal(0,n) == 1
    pascal(n,n) == 1
    pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
    pascalRow(n) == [pascal(i,n) for i in 0..n]
    displayRow(n) == output center blankSeparate pascalRow(n)
    for i in 0..20 repeat displayRow i -- (end)
    
  • GAP
    Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018
  • Haskell
    a007318 n k = a007318_tabl !! n !! k
    a007318_row n = a007318_tabl !! n
    a007318_list = concat a007318_tabl
    a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
    -- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
    -- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010
    
  • Magma
    /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
    
  • Maple
    A007318 := (n,k)->binomial(n,k);
  • Mathematica
    Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
    Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
  • Maxima
    create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011
    
  • Python
    # See Hobson link. Further programs:
    from math import prod,factorial
    def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
    
  • Python
    from math import comb, isqrt
    def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024
    
  • Sage
    def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014
    

Formula

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Extensions

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A000110 Bell or exponential numbers: number of ways to partition a set of n labeled elements.

Original entry on oeis.org

1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159, 5832742205057, 51724158235372, 474869816156751, 4506715738447323, 44152005855084346, 445958869294805289, 4638590332229999353, 49631246523618756274
Offset: 0

Keywords

Comments

The leading diagonal of its difference table is the sequence shifted, see Bernstein and Sloane (1995). - N. J. A. Sloane, Jul 04 2015
Also the number of equivalence relations that can be defined on a set of n elements. - Federico Arboleda (federico.arboleda(AT)gmail.com), Mar 09 2005
a(n) = number of nonisomorphic colorings of a map consisting of a row of n+1 adjacent regions. Adjacent regions cannot have the same color. - David W. Wilson, Feb 22 2005
If an integer is squarefree and has n distinct prime factors then a(n) is the number of ways of writing it as a product of its divisors. - Amarnath Murthy, Apr 23 2001
Consider rooted trees of height at most 2. Letting each tree 'grow' into the next generation of n means we produce a new tree for every node which is either the root or at height 1, which gives the Bell numbers. - Jon Perry, Jul 23 2003
Begin with [1,1] and follow the rule that [1,k] -> [1,k+1] and [1,k] k times, e.g., [1,3] is transformed to [1,4], [1,3], [1,3], [1,3]. Then a(n) is the sum of all components: [1,1] = 2; [1,2], [1,1] = 5; [1,3], [1,2], [1,2], [1,2], [1,1] = 15; etc. - Jon Perry, Mar 05 2004
Number of distinct rhyme schemes for a poem of n lines: a rhyme scheme is a string of letters (e.g., 'abba') such that the leftmost letter is always 'a' and no letter may be greater than one more than the greatest letter to its left. Thus 'aac' is not valid since 'c' is more than one greater than 'a'. For example, a(3)=5 because there are 5 rhyme schemes: aaa, aab, aba, abb, abc; also see example by Neven Juric. - Bill Blewett, Mar 23 2004
In other words, number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= 1 + max(prefix) for k >= 1, see example (cf. A080337 and A189845). - Joerg Arndt, Apr 30 2011
Number of partitions of {1, ..., n+1} into subsets of nonconsecutive integers, including the partition 1|2|...|n+1. E.g., a(3)=5: there are 5 partitions of {1,2,3,4} into subsets of nonconsecutive integers, namely, 13|24, 13|2|4, 14|2|3, 1|24|3, 1|2|3|4. - Augustine O. Munagi, Mar 20 2005
Triangle (addition) scheme to produce terms, derived from the recurrence, from Oscar Arevalo (loarevalo(AT)sbcglobal.net), May 11 2005:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
... [This is Aitken's array A011971]
With P(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..P(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (1/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
a(n+1) is the number of binary relations on an n-element set that are both symmetric and transitive. - Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005
If the rule from Jon Perry, Mar 05 2004, is used, then a(n-1) = [number of components used to form a(n)] / 2. - Daniel Kuan (dkcm(AT)yahoo.com), Feb 19 2006
a(n) is the number of functions f from {1,...,n} to {1,...,n,n+1} that satisfy the following two conditions for all x in the domain: (1) f(x) > x; (2) f(x)=n+1 or f(f(x))=n+1. E.g., a(3)=5 because there are exactly five functions that satisfy the two conditions: f1={(1,4),(2,4),(3,4)}, f2={(1,4),(2,3),(3,4)}, f3={(1,3),(2,4),(3,4)}, f4={(1,2),(2,4),(3,4)} and f5={(1,3),(2,3),(3,4)}. - Dennis P. Walsh, Feb 20 2006
Number of asynchronic siteswap patterns of length n which have no zero-throws (i.e., contain no 0's) and whose number of orbits (in the sense given by Allen Knutson) is equal to the number of balls. E.g., for n=4, the condition is satisfied by the following 15 siteswaps: 4444, 4413, 4242, 4134, 4112, 3441, 2424, 1344, 2411, 1313, 1241, 2222, 3131, 1124, 1111. Also number of ways to choose n permutations from identity and cyclic permutations (1 2), (1 2 3), ..., (1 2 3 ... n) so that their composition is identity. For n=3 we get the following five: id o id o id, id o (1 2) o (1 2), (1 2) o id o (1 2), (1 2) o (1 2) o id, (1 2 3) o (1 2 3) o (1 2 3). (To see the bijection, look at Ehrenborg and Readdy paper.) - Antti Karttunen, May 01 2006
a(n) is the number of permutations on [n] in which a 3-2-1 (scattered) pattern occurs only as part of a 3-2-4-1 pattern. Example: a(3) = 5 counts all permutations on [3] except 321. See "Eigensequence for Composition" reference a(n) = number of permutation tableaux of size n (A000142) whose first row contains no 0's. Example: a(3)=5 counts {{}, {}, {}}, {{1}, {}}, {{1}, {0}}, {{1}, {1}}, {{1, 1}}. - David Callan, Oct 07 2006
From Gottfried Helms, Mar 30 2007: (Start)
This sequence is also the first column in the matrix-exponential of the (lower triangular) Pascal-matrix, scaled by exp(-1): PE = exp(P) / exp(1) =
1
1 1
2 2 1
5 6 3 1
15 20 12 4 1
52 75 50 20 5 1
203 312 225 100 30 6 1
877 1421 1092 525 175 42 7 1
First 4 columns are A000110, A033306, A105479, A105480. The general case is mentioned in the two latter entries. PE is also the Hadamard-product Toeplitz(A000110) (X) P:
1
1 1
2 1 1
5 2 1 1
15 5 2 1 1 (X) P
52 15 5 2 1 1
203 52 15 5 2 1 1
877 203 52 15 5 2 1 1
(End)
The terms can also be computed with finite steps and precise integer arithmetic. Instead of exp(P)/exp(1) one can compute A = exp(P - I) where I is the identity-matrix of appropriate dimension since (P-I) is nilpotent to the order of its dimension. Then a(n)=A[n,1] where n is the row-index starting at 1. - Gottfried Helms, Apr 10 2007
When n is prime, a(n) == 2 (mod n), but the converse is not always true. Define a Bell pseudoprime to be a composite number n such that a(n) == 2 (mod n). W. F. Lunnon recently found the Bell pseudoprimes 21361 = 41*521 and C46 = 3*23*16218646893090134590535390526854205539989357 and conjectured that Bell pseudoprimes are extremely scarce. So the second Bell pseudoprime is unlikely to be known with certainty in the near future. I confirmed that 21361 is the first. - David W. Wilson, Aug 04 2007 and Sep 24 2007
This sequence and A000587 form a reciprocal pair under the list partition transform described in A133314. - Tom Copeland, Oct 21 2007
Starting (1, 2, 5, 15, 52, ...), equals row sums and right border of triangle A136789. Also row sums of triangle A136790. - Gary W. Adamson, Jan 21 2008
This is the exponential transform of A000012. - Thomas Wieder, Sep 09 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of idempotent order-decreasing full transformations (of an n-chain).
a(n) is also the number of nilpotent partial one-one order-decreasing transformations (of an n-chain).
a(n+1) is also the number of partial one-one order-decreasing transformations (of an n-chain). (End)
From Peter Bala, Oct 19 2008: (Start)
Bell(n) is the number of n-pattern sequences [Cooper & Kennedy]. An n-pattern sequence is a sequence of integers (a_1,...,a_n) such that a_i = i or a_i = a_j for some j < i. For example, Bell(3) = 5 since the 3-pattern sequences are (1,1,1), (1,1,3), (1,2,1), (1,2,2) and (1,2,3).
Bell(n) is the number of sequences of positive integers (N_1,...,N_n) of length n such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (see the comment by B. Blewett above). It is interesting to note that if we strengthen the latter condition to N_(i+1) <= 1 + N_i we get the Catalan numbers A000108 instead of the Bell numbers.
(End)
Equals the eigensequence of Pascal's triangle, A007318; and starting with offset 1, = row sums of triangles A074664 and A152431. - Gary W. Adamson, Dec 04 2008
The entries f(i, j) in the exponential of the infinite lower-triangular matrix of binomial coefficients b(i, j) are f(i, j) = b(i, j) e a(i - j). - David Pasino, Dec 04 2008
Equals lim_{k->oo} A071919^k. - Gary W. Adamson, Jan 02 2009
Equals A154107 convolved with A014182, where A014182 = expansion of exp(1-x-exp(-x)), the eigensequence of A007318^(-1). Starting with offset 1 = A154108 convolved with (1,2,3,...) = row sums of triangle A154109. - Gary W. Adamson, Jan 04 2009
Repeated iterates of (binomial transform of [1,0,0,0,...]) will converge upon (1, 2, 5, 15, 52, ...) when each result is prefaced with a "1"; such that the final result is the fixed limit: (binomial transform of [1,1,2,5,15,...]) = (1,2,5,15,52,...). - Gary W. Adamson, Jan 14 2009
From Karol A. Penson, May 03 2009: (Start)
Relation between the Bell numbers B(n) and the n-th derivative of 1/Gamma(1+x) evaluated at x=1:
a) produce a number of such derivatives through seq(subs(x=1, simplify((d^n/dx^n)GAMMA(1+x)^(-1))), n=1..5);
b) leave them expressed in terms of digamma (Psi(k)) and polygamma (Psi(k,n)) functions and unevaluated;
Examples of such expressions, for n=1..5, are:
n=1: -Psi(1),
n=2: -(-Psi(1)^2 + Psi(1,1)),
n=3: -Psi(1)^3 + 3*Psi(1)*Psi(1,1) - Psi(2,1),
n=4: -(-Psi(1)^4 + 6*Psi(1)^2*Psi(1,1) - 3*Psi(1,1)^2 - 4*Psi(1)*Psi(2,1) + Psi(3, 1)),
n=5: -Psi(1)^5 + 10*Psi(1)^3*Psi(1,1) - 15*Psi(1)*Psi(1,1)^2 - 10*Psi(1)^2*Psi(2,1) + 10*Psi(1,1)*Psi(2,1) + 5*Psi(1)*Psi(3,1) - Psi(4,1);
c) for a given n, read off the sum of absolute values of coefficients of every term involving digamma or polygamma functions.
This sum is equal to B(n). Examples: B(1)=1, B(2)=1+1=2, B(3)=1+3+1=5, B(4)=1+6+3+4+1=15, B(5)=1+10+15+10+10+5+1=52;
d) Observe that this decomposition of the Bell number B(n) apparently does not involve the Stirling numbers of the second kind explicitly.
(End)
The numbers given above by Penson lead to the multinomial coefficients A036040. - Johannes W. Meijer, Aug 14 2009
Column 1 of A162663. - Franklin T. Adams-Watters, Jul 09 2009
Asymptotic expansions (0!+1!+2!+...+(n-1)!)/(n-1)! = a(0) + a(1)/n + a(2)/n^2 + ... and (0!+1!+2!+...+n!)/n! = 1 + a(0)/n + a(1)/n^2 + a(2)/n^3 + .... - Michael Somos, Jun 28 2009
Starting with offset 1 = row sums of triangle A165194. - Gary W. Adamson, Sep 06 2009
a(n+1) = A165196(2^n); where A165196 begins: (1, 2, 4, 5, 7, 12, 14, 15, ...). such that A165196(2^3) = 15 = A000110(4). - Gary W. Adamson, Sep 06 2009
The divergent series g(x=1,m) = 1^m*1! - 2^m*2! + 3^m*3! - 4^m*4! + ..., m >= -1, which for m=-1 dates back to Euler, is related to the Bell numbers. We discovered that g(x=1,m) = (-1)^m * (A040027(m) - A000110(m+1) * A073003). We observe that A073003 is Gompertz's constant and that A040027 was published by Gould, see for more information A163940. - Johannes W. Meijer, Oct 16 2009
a(n) = E(X^n), i.e., the n-th moment about the origin of a random variable X that has a Poisson distribution with (rate) parameter, lambda = 1. - Geoffrey Critzer, Nov 30 2009
Let A000110 = S(x), then S(x) = A(x)/A(x^2) when A(x) = A173110; or (1, 1, 2, 5, 15, 52, ...) = (1, 1, 3, 6, 20, 60, ...) / (1, 0, 1, 0, 3, 0, 6, 0, 20, ...). - Gary W. Adamson, Feb 09 2010
The Bell numbers serve as the upper limit for the number of distinct homomorphic images from any given finite universal algebra. Every algebra homomorphism is determined by its kernel, which must be a congruence relation. As the number of possible congruence relations with respect to a finite universal algebra must be a subset of its possible equivalence classes (given by the Bell numbers), it follows naturally. - Max Sills, Jun 01 2010
For a proof of the o.g.f. given in the R. Stephan comment see, e.g., the W. Lang link under A071919. - Wolfdieter Lang, Jun 23 2010
Let B(x) = (1 + x + 2x^2 + 5x^3 + ...). Then B(x) is satisfied by A(x)/A(x^2) where A(x) = polcoeff A173110: (1 + x + 3x^2 + 6x^3 + 20x^4 + 60x^5 + ...) = B(x) * B(x^2) * B(x^4) * B(x^8) * .... - Gary W. Adamson, Jul 08 2010
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color without choosing any two colors the same positive number of times. (See related comments for A000108, A008277, A016098.) - Matthew Vandermast, Nov 22 2010
A binary counter with faulty bits starts at value 0 and attempts to increment by 1 at each step. Each bit that should toggle may or may not do so. a(n) is the number of ways that the counter can have the value 0 after n steps. E.g., for n=3, the 5 trajectories are 0,0,0,0; 0,1,0,0; 0,1,1,0; 0,0,1,0; 0,1,3,0. - David Scambler, Jan 24 2011
No Bell number is divisible by 8, and no Bell number is congruent to 6 modulo 8; see Theorem 6.4 and Table 1.7 in Lunnon, Pleasants and Stephens. - Jon Perry, Feb 07 2011, clarified by Eric Rowland, Mar 26 2014
a(n+1) is the number of (symmetric) positive semidefinite n X n 0-1 matrices. These correspond to equivalence relations on {1,...,n+1}, where matrix element M[i,j] = 1 if and only if i and j are equivalent to each other but not to n+1. - Robert Israel, Mar 16 2011
a(n) is the number of monotonic-labeled forests on n vertices with rooted trees of height less than 2. We note that a labeled rooted tree is monotonic-labeled if the label of any parent vertex is greater than the label of any offspring vertex. See link "Counting forests with Stirling and Bell numbers". - Dennis P. Walsh, Nov 11 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000772 and A094198. - Peter Bala, Nov 25 2011
B(n) counts the length n+1 rhyme schemes without repetitions. E.g., for n=2 there are 5 rhyme schemes of length 3 (aaa, aab, aba, abb, abc), and the 2 without repetitions are aba, abc. This is basically O. Munagi's result that the Bell numbers count partitions into subsets of nonconsecutive integers (see comment above dated Mar 20 2005). - Eric Bach, Jan 13 2012
Right and left borders and row sums of A212431 = A000110 or a shifted variant. - Gary W. Adamson, Jun 21 2012
Number of maps f: [n] -> [n] where f(x) <= x and f(f(x)) = f(x) (projections). - Joerg Arndt, Jan 04 2013
Permutations of [n] avoiding any given one of the 8 dashed patterns in the equivalence classes (i) 1-23, 3-21, 12-3, 32-1, and (ii) 1-32, 3-12, 21-3, 23-1. (See Claesson 2001 reference.) - David Callan, Oct 03 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Sum_{n>=0} a(n)/n! = e^(e-1) = 5.57494152476..., see A234473. - Richard R. Forberg, Dec 26 2013 (This is the e.g.f. for x=1. - Wolfdieter Lang, Feb 02 2015)
Sum_{j=0..n} binomial(n,j)*a(j) = (1/e)*Sum_{k>=0} (k+1)^n/k! = (1/e) Sum_{k=1..oo} k^(n+1)/k! = a(n+1), n >= 0, using the Dobinski formula. See the comment by Gary W. Adamson, Dec 04 2008 on the Pascal eigensequence. - Wolfdieter Lang, Feb 02 2015
In fact it is not really an eigensequence of the Pascal matrix; rather the Pascal matrix acts on the sequence as a shift. It is an eigensequence (the unique eigensequence with eigenvalue 1) of the matrix derived from the Pascal matrix by adding at the top the row [1, 0, 0, 0 ...]. The binomial sum formula may be derived from the definition in terms of partitions: label any element X of a set S of N elements, and let X(k) be the number of subsets of S containing X with k elements. Since each subset has a unique coset, the number of partitions p(N) of S is given by p(N) = Sum_{k=1..N} (X(k) p(N-k)); trivially X(k) = N-1 choose k-1. - Mason Bogue, Mar 20 2015
a(n) is the number of ways to nest n matryoshkas (Russian nesting dolls): we may identify {1, 2, ..., n} with dolls of ascending sizes and the sets of a set partition with stacks of dolls. - Carlo Sanna, Oct 17 2015
Number of permutations of [n] where the initial elements of consecutive runs of increasing elements are in decreasing order. a(4) = 15: `1234, `2`134, `23`14, `234`1, `24`13, `3`124, `3`2`14, `3`24`1, `34`12, `34`2`1, `4`123, `4`2`13, `4`23`1, `4`3`12, `4`3`2`1. - Alois P. Heinz, Apr 27 2016
Taking with alternating signs, the Bell numbers are the coefficients in the asymptotic expansion (Ramanujan): (-1)^n*(A000166(n) - n!/exp(1)) ~ 1/n - 2/n^2 + 5/n^3 - 15/n^4 + 52/n^5 - 203/n^6 + O(1/n^7). - Vladimir Reshetnikov, Nov 10 2016
Number of treeshelves avoiding pattern T231. See A278677 for definitions and examples. - Sergey Kirgizov, Dec 24 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. - N. J. A. Sloane, Feb 09 2017
a(n) = Sum(# of standard immaculate tableaux of shape m, m is a composition of n), where this sum is over all integer compositions m of n > 0. This formula is easily seen to hold by identifying standard immaculate tableaux of size n with set partitions of { 1, 2, ..., n }. For example, if we sum over integer compositions of 4 lexicographically, we see that 1+1+2+1+3+3+3+1 = 15 = A000110(4). - John M. Campbell, Jul 17 2017
a(n) is also the number of independent vertex sets (and vertex covers) in the (n-1)-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017
Even-numbered entries represent the numbers of configurations of identity and non-identity for alleles of a gene in n diploid individuals with distinguishable maternal and paternal alleles. - Noah A Rosenberg, Jan 28 2019
Number of partial equivalence relations (PERs) on a set with n elements (offset=1), i.e., number of symmetric, transitive (not necessarily reflexive) relations. The idea is to add a dummy element D to the set, and then take equivalence relations on the result; anything equivalent to D is then removed for the partial equivalence relation. - David Spivak, Feb 06 2019
Number of words of length n+1 with no repeated letters, when letters are unlabeled. - Thomas Anton, Mar 14 2019
Named by Becker and Riordan (1948) after the Scottish-American mathematician and writer Eric Temple Bell (1883 - 1960). - Amiram Eldar, Dec 04 2020
Also the number of partitions of {1,2,...,n+1} with at most one n+1 singleton. E.g., a(3)=5: {13|24, 12|34, 123|4, 14|23, 1234}. - Yuchun Ji, Dec 21 2020
a(n) is the number of sigma algebras on the set of n elements. Note that each sigma algebra is generated by a partition of the set. For example, the sigma algebra generated by the partition {{1}, {2}, {3,4}} is {{}, {1}, {2}, {1,2}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. - Jianing Song, Apr 01 2021
a(n) is the number of P_3-free graphs on n labeled nodes. - M. Eren Kesim, Jun 04 2021
a(n) is the number of functions X:([n] choose 2) -> {+,-} such that for any ordered 3-tuple abc we have X(ab)X(ac)X(bc) not in {+-+,++-,-++}. - Robert Lauff, Dec 09 2022
From Manfred Boergens, Mar 11 2025: (Start)
The partitions in the definition can be described as disjoint covers of the set. "Covers" in general give rise to the following amendments:
For disjoint covers which may include one empty set see A186021.
For arbitrary (including non-disjoint) covers see A003465.
For arbitrary (including non-disjoint) covers which may include one empty set see A000371. (End)

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 203*x^6 + 877*x^7 + 4140*x^8 + ...
From Neven Juric, Oct 19 2009: (Start)
The a(4)=15 rhyme schemes for n=4 are
  aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, abcd
The a(5)=52 rhyme schemes for n=5 are
  aaaaa, aaaab, aaaba, aaabb, aaabc, aabaa, aabab, aabac, aabba, aabbb, aabbc, aabca, aabcb, aabcc, aabcd, abaaa, abaab, abaac, ababa, ababb, ababc, abaca, abacb, abacc, abacd, abbaa, abbab, abbac, abbba, abbbb, abbbc, abbca, abbcb, abbcc, abbcd, abcaa, abcab, abcac, abcad, abcba, abcbb, abcbc, abcbd, abcca, abccb, abccc, abccd, abcda, abcdb, abcdc, abcdd, abcde
(End)
From _Joerg Arndt_, Apr 30 2011: (Start)
Restricted growth strings (RGS):
For n=0 there is one empty string;
for n=1 there is one string [0];
for n=2 there are 2 strings [00], [01];
for n=3 there are a(3)=5 strings [000], [001], [010], [011], and [012];
for n=4 there are a(4)=15 strings
1: [0000], 2: [0001], 3: [0010], 4: [0011], 5: [0012], 6: [0100], 7: [0101], 8: [0102], 9: [0110], 10: [0111], 11: [0112], 12: [0120], 13: [0121], 14: [0122], 15: [0123].
These are one-to-one with the rhyme schemes (identify a=0, b=1, c=2, etc.).
(End)
Consider the set S = {1, 2, 3, 4}. The a(4) = 1 + 3 + 6 + 4 + 1 = 15 partitions are: P1 = {{1}, {2}, {3}, {4}}; P21 .. P23 = {{a,4}, S\{a,4}} with a = 1, 2, 3; P24 .. P29 = {{a}, {b}, S\{a,b}} with 1 <= a < b <= 4;  P31 .. P34 = {S\{a}, {a}} with a = 1 .. 4; P4 = {S}. See the Bottomley link for a graphical illustration. - _M. F. Hasler_, Oct 26 2017
		

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  • Moser, Leo, and Max Wyman. An asymptotic formula for the Bell numbers. Trans. Royal Soc. Canada, 49 (1955), 49-53.
  • A. Murthy, Generalization of partition function, introducing Smarandache factor partition, Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000.
  • Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.4,1.8.
  • P. Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
  • A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 212.
  • G.-C. Rota, Finite Operator Calculus.
  • Frank Ruskey, Jennifer Woodcock and Yuji Yamauchi, Counting and computing the Rand and block distances of pairs of set partitions, Journal of Discrete Algorithms, Volume 16, October 2012, Pages 236-248.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge; see Section 1.4 and Example 5.2.4.
  • Abdullahi Umar, On the semigroups of order-decreasing finite full transformations, Proc. Roy. Soc. Edinburgh 120A (1992), 129-142.
  • Abdullahi Umar, On the semigroups of partial one-to-one order-decreasing finite transformations, Proc. Roy. Soc. Edinburgh 123A (1993), 355-363.

Crossrefs

Equals row sums of triangle A008277 (Stirling subset numbers).
Partial sums give A005001. a(n) = A123158(n, 0).
See A061462 for powers of 2 dividing a(n).
Rightmost diagonal of triangle A121207. A144293 gives largest prime factor.
Equals row sums of triangle A152432.
Row sums, right and left borders of A212431.
A diagonal of A011971. - N. J. A. Sloane, Jul 31 2012
Diagonal of A102661. - Manfred Boergens, Mar 11 2025
Cf. A054767 (period of this sequence mod n).
Row sums are A048993. - Wolfdieter Lang, Oct 16 2014
Sequences in the Erné (1974) paper: A000110, A000798, A001035, A001927, A001929, A006056, A006057, A006058, A006059.
Bell polynomials B(n,x): A001861 (x=2), A027710 (x=3), A078944 (x=4), A144180 (x=5), A144223 (x=6), A144263 (x=7), A221159 (x=8).
Cf. A243991 (sum of reciprocals), A085686 (inv. Euler Transf.).

Programs

  • Haskell
    type N = Integer
    n_partitioned_k :: N -> N -> N
    1 `n_partitioned_k` 1 = 1
    1 `n_partitioned_k` _ = 0
    n `n_partitioned_k` k = k * (pred n `n_partitioned_k` k) + (pred n `n_partitioned_k` pred k)
    n_partitioned :: N -> N
    n_partitioned 0 = 1
    n_partitioned n = sum $ map (\k -> n `n_partitioned_k` k) $ [1 .. n]
    -- Felix Denis, Oct 16 2012
    
  • Haskell
    a000110 = sum . a048993_row -- Reinhard Zumkeller, Jun 30 2013
    
  • Julia
    function a(n)
        t = [zeros(BigInt, n+1) for _ in 1:n+1]
        t[1][1] = 1
        for i in 2:n+1
            t[i][1] = t[i-1][i-1]
            for j in 2:i
                t[i][j] = t[i-1][j-1] + t[i][j-1]
            end
        end
        return [t[i][1] for i in 1:n+1]
    end
    print(a(28)) # Paul Muljadi, May 07 2024
    
  • Magma
    [Bell(n): n in [0..40]]; // Vincenzo Librandi, Feb 07 2011
    
  • Maple
    A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n-1,i)*A000110(n-1-i),i=0..n-1); fi; end: # version 1
    A := series(exp(exp(x)-1),x,60): A000110 := n->n!*coeff(A,x,n): # version 2
    A000110:= n-> add(Stirling2(n, k), k=0..n): seq(A000110(n), n=0..22); # version 3, from Zerinvary Lajos, Jun 28 2007
    A000110 := n -> combinat[bell](n): # version 4, from Peter Luschny, Mar 30 2011
    spec:= [S, {S=Set(U, card >= 1), U=Set(Z, card >= 1)}, labeled]: G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, unlabeled, x): seq(combstruct[count]([P, G, labeled], size=i), i=0..22);  # version 5, Zerinvary Lajos, Dec 16 2007
    BellList := proc(m) local A, P, n; A := [1, 1]; P := [1]; for n from 1 to m - 2 do
    P := ListTools:-PartialSums([A[-1], op(P)]); A := [op(A), P[-1]] od; A end: BellList(29); # Peter Luschny, Mar 24 2022
  • Mathematica
    f[n_] := Sum[ StirlingS2[n, k], {k, 0, n}]; Table[ f[n], {n, 0, 40}] (* Robert G. Wilson v *)
    Table[BellB[n], {n, 0, 40}] (* Harvey P. Dale, Mar 01 2011 *)
    B[0] = 1; B[n_] := 1/E Sum[k^(n - 1)/(k-1)!, {k, 1, Infinity}] (* Dimitri Papadopoulos, Mar 10 2015, edited by M. F. Hasler, Nov 30 2018 *)
    BellB[Range[0,40]] (* Eric W. Weisstein, Aug 10 2017 *)
    b[1] = 1; k = 1; Flatten[{1, Table[Do[j = k; k += b[m]; b[m] = j;, {m, 1, n-1}]; b[n] = k, {n, 1, 40}]}] (* Vaclav Kotesovec, Sep 07 2019 *)
    Table[j! Coefficient[Series[Exp[Exp[x] - 1], {x, 0, 20}], x, j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 01 2023 *)
    Table[(D[Exp[Exp[x]], {x, n}] /. x -> 0)/E, {n, 0, 20}] (* Joan Ludevid, Nov 05 2024 *)
  • Maxima
    makelist(belln(n),n,0,40); /* Emanuele Munarini, Jul 04 2011 */
    
  • PARI
    {a(n) = my(m); if( n<0, 0, m = contfracpnqn( matrix(2, n\2, i, k, if( i==1, -k*x^2, 1 - (k+1)*x))); polcoeff(1 / (1 - x + m[2,1] / m[1,1]) + x * O(x^n), n))}; /* Michael Somos */
    
  • PARI
    {a(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n)}; /* Michael Somos, Aug 22 2004 */
    
  • PARI
    a(n)=round(exp(-1)*suminf(k=0,1.0*k^n/k!)) \\ Gottfried Helms, Mar 30 2007 - WARNING! For illustration only: Gives silently a wrong result for n = 42 and an error for n > 42, with standard precision of 38 digits. - M. F. Hasler, Nov 30 2018
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n)) - 1), n))}; /* Michael Somos, Jun 28 2009 */
    
  • PARI
    Vec(serlaplace(exp(exp('x+O('x^66))-1))) \\ Joerg Arndt, May 26 2012
    
  • PARI
    A000110(n)=sum(k=0,n,stirling(n,k,2)) \\ M. F. Hasler, Nov 30 2018
    
  • Perl
    use bignum;sub a{my($n)=@;my@t=map{[(0)x($n+1)]}0..$n;$t[0][0]=1;for my$i(1..$n){$t[$i][0]=$t[$i-1][$i-1];for my$j(1..$i){$t[$i][$j]=$t[$i-1][$j-1]+$t[$i][$j-1]}}return map{$t[$][0]}0..$n-1}print join(", ",a(28)),"\n" # Paul Muljadi, May 08 2024
  • Python
    # The objective of this implementation is efficiency.
    # m -> [a(0), a(1), ..., a(m)] for m > 0.
    def A000110_list(m):
        A = [0 for i in range(m)]
        A[0] = 1
        R = [1, 1]
        for n in range(1, m):
            A[n] = A[0]
            for k in range(n, 0, -1):
                A[k-1] += A[k]
            R.append(A[0])
        return R
    A000110_list(40) # Peter Luschny, Jan 18 2011
    
  • Python
    # requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
    from itertools import accumulate
    A000110, blist, b = [1,1], [1], 1
    for _ in range(20):
        blist = list(accumulate([b]+blist))
        b = blist[-1]
        A000110.append(b) # Chai Wah Wu, Sep 02 2014, updated Chai Wah Wu, Sep 19 2014
    
  • Python
    from sympy import bell
    print([bell(n) for n in range(27)]) # Michael S. Branicky, Dec 15 2021
    
  • Python
    from functools import cache
    @cache
    def a(n, k=0): return int(n < 1) or k*a(n-1, k) + a(n-1, k+1)
    print([a(n) for n in range(27)])  # Peter Luschny, Jun 14 2022
    
  • Sage
    from sage.combinat.expnums import expnums2; expnums2(30, 1) # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    [bell_number(n) for n in (0..40)] # G. C. Greubel, Jun 13 2019
    

Formula

E.g.f.: exp(exp(x) - 1).
Recurrence: a(n+1) = Sum_{k=0..n} a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
a(n) = Sum_{j=0..n-1} (1/(n-1)!)*A000166(j)*binomial(n-1, j)*(n-j)^(n-1). - André F. Labossière, Dec 01 2004
G.f.: (Sum_{k>=0} 1/((1-k*x)*k!))/exp(1) = hypergeom([-1/x], [(x-1)/x], 1)/exp(1) = ((1-2*x)+LaguerreL(1/x, (x-1)/x, 1)+x*LaguerreL(1/x, (2*x-1)/x, 1))*Pi/(x^2*sin(Pi*(2*x-1)/x)), where LaguerreL(mu, nu, z) = (gamma(mu+nu+1)/(gamma(mu+1)*gamma(nu+1)))* hypergeom([-mu], [nu+1], z) is the Laguerre function, the analytic extension of the Laguerre polynomials, for mu not equal to a nonnegative integer. This generating function has an infinite number of poles accumulating in the neighborhood of x=0. - Karol A. Penson, Mar 25 2002
a(n) = exp(-1)*Sum_{k >= 0} k^n/k! [Dobinski]. - Benoit Cloitre, May 19 2002
a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(-1/2) exp(exp(r)-1) / r^n, where r is the positive root of r exp(r) = n. See, e.g., the Odlyzko reference.
a(n) is asymptotic to b^n*exp(b-n-1/2)*sqrt(b/(b+n)) where b satisfies b*log(b) = n - 1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493). - Benoit Cloitre, Oct 23 2002, corrected by Vaclav Kotesovec, Jan 06 2013
Lovasz (Combinatorial Problems and Exercises, North-Holland, 1993, Section 1.14, Problem 9) gives another asymptotic formula, quoted by Mezo and Baricz. - N. J. A. Sloane, Mar 26 2015
G.f.: Sum_{k>=0} x^k/(Product_{j=1..k} (1-j*x)) (see Klazar for a proof). - Ralf Stephan, Apr 18 2004
a(n+1) = exp(-1)*Sum_{k>=0} (k+1)^(n)/k!. - Gerald McGarvey, Jun 03 2004
For n>0, a(n) = Aitken(n-1, n-1) [i.e., a(n-1, n-1) of Aitken's array (A011971)]. - Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (-1)^(k-i)*binomial(k, i)*i^n + 0^n). - Paul Barry, Apr 18 2005
a(n) = A032347(n) + A040027(n+1). - Jon Perry, Apr 26 2005
a(n) = (2*n!/(Pi*e))*Im( Integral_{x=0..Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro]. - David Callan, Sep 03 2005
O.g.f.: 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(.../(1-n*x-n*x^2/(...)))))) (continued fraction due to Ph. Flajolet). - Paul D. Hanna, Jan 17 2006
From Karol A. Penson, Jan 14 2007: (Start)
Representation of Bell numbers B(n), n=1,2,..., as special values of hypergeometric function of type (n-1)F(n-1), in Maple notation: B(n)=exp(-1)*hypergeom([2,2,...,2],[1,1,...,1],1), n=1,2,..., i.e., having n-1 parameters all equal to 2 in the numerator, having n-1 parameters all equal to 1 in the denominator and the value of the argument equal to 1.
Examples:
B(1)=exp(-1)*hypergeom([],[],1)=1
B(2)=exp(-1)*hypergeom([2],[1],1)=2
B(3)=exp(-1)*hypergeom([2,2],[1,1],1)=5
B(4)=exp(-1)*hypergeom([2,2,2],[1,1,1],1)=15
B(5)=exp(-1)*hypergeom([2,2,2,2],[1,1,1,1],1)=52
(Warning: this formula is correct but its application by a computer may not yield exact results, especially with a large number of parameters.)
(End)
a(n+1) = 1 + Sum_{k=0..n-1} Sum_{i=0..k} binomial(k,i)*(2^(k-i))*a(i). - Yalcin Aktar, Feb 27 2007
a(n) = [1,0,0,...,0] T^(n-1) [1,1,1,...,1]', where T is the n X n matrix with main diagonal {1,2,3,...,n}, 1's on the diagonal immediately above and 0's elsewhere. [Meier]
a(n) = ((2*n!)/(Pi * e)) * ImaginaryPart(Integral[from 0 to Pi](e^e^e^(i*theta))*sin(n*theta) dtheta). - Jonathan Vos Post, Aug 27 2007
From Tom Copeland, Oct 10 2007: (Start)
a(n) = T(n,1) = Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * Lag(n,-1,j-n) = Sum_{j=0..n} [ E(n,j)/n! ] * [ n!*Lag(n,-1, j-n) ] where T(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m. Note that E(n,j)/n! = E(n,j) / (Sum_{k=0..n} E(n,k)).
The Eulerian numbers count the permutation ascents and the expression [n!*Lag(n,-1, j-n)] is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to n!*a(n) = Sum_{j=0..n} E(n,j) * [n!*Lag(n,-1, j-n)].
(End)
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then for Bell numbers B_n we have B_n=1/e*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! Sum_{k=1..n} a(n-k)/((n-k)! (k-1)!) where a(0)=1. - Thomas Wieder, Sep 09 2008
a(n+k) = Sum_{m=0..n} Stirling2(n,m) Sum_{r=0..k} binomial(k,r) m^r a(k-r). - David Pasino (davepasino(AT)yahoo.com), Jan 25 2009. (Umbrally, this may be written as a(n+k) = Sum_{m=0..n} Stirling2(n,m) (a+m)^k. - N. J. A. Sloane, Feb 07 2009)
Sum_{k=1..n-1} a(n)*binomial(n,k) = Sum_{j=1..n}(j+1)*Stirling2(n,j+1). - [Zhao] - R. J. Mathar, Jun 24 2024
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..n-i}...Sum_{k_n=0..1}
delta(k_1,k_2,...,k_i,...,k_n)
where delta(k_1,k_2,...,k_i,...,k_n) = 0 if any k_i > k_(i+1) and k_(i+1) <> 0
and delta(k_1,k_2,...,k_i,...,k_n) = 1 otherwise.
(End)
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]:=binomial(j-1,i-1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jul 08 2010
G.f. satisfies A(x) = (x/(1-x))*A(x/(1-x)) + 1. - Vladimir Kruchinin, Nov 28 2011
G.f.: 1 / (1 - x / (1 - 1*x / (1 - x / (1 - 2*x / (1 - x / (1 - 3*x / ... )))))). - Michael Somos, May 12 2012
a(n+1) = Sum_{m=0..n} Stirling2(n, m)*(m+1), n >= 0. Compare with the third formula for a(n) above. Here Stirling2 = A048993. - Wolfdieter Lang, Feb 03 2015
G.f.: (-1)^(1/x)*((-1/x)!/e + (!(-1-1/x))/x) where z! and !z are factorial and subfactorial generalized to complex arguments. - Vladimir Reshetnikov, Apr 24 2013
The following formulas were proposed during the period Dec 2011 - Oct 2013 by Sergei N. Gladkovskii: (Start)
E.g.f.: exp(exp(x)-1) = 1 + x/(G(0)-x); G(k) = (k+1)*Bell(k) + x*Bell(k+1) - x*(k+1)*Bell(k)*Bell(k+2)/G(k+1) (continued fraction).
G.f.: W(x) = (1-1/(G(0)+1))/exp(1); G(k) = x*k^2 + (3*x-1)*k - 2 + x - (k+1)*(x*k+x-1)^2/G(k+1); (continued fraction Euler's kind, 1-step).
G.f.: W(x) = (1 + G(0)/(x^2-3*x+2))/exp(1); G(k) = 1 - (x*k+x-1)/( ((k+1)!) - (((k+1)!)^2)*(1-x-k*x+(k+1)!)/( ((k+1)!)*(1-x-k*x+(k+1)!) - (x*k+2*x-1)*(1-2*x-k*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x) = 1/(1 - x/(1-x/(1 + x/G(0)))); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1); (continued fraction, 1-step).
G.f.: -1/U(0) where U(k) = x*k - 1 + x - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + x/U(0) where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + 1/(U(0) - x) where U(k) = 1 + x - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1 + x/(U(0)-x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1/G(0) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/G(k+1) ))); (continued fraction).
G.f.: G(0)/(1+x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (continued fraction).
G.f.: -(1+2*x) * Sum_{k >= 0} x^(2*k)*(4*x*k^2-2*k-2*x-1) / ((2*k+1) * (2*x*k-1)) * A(k) / B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1) * (2*x*p-x-1) * (2*x*p-2*x-1).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S-1) where S = Sum_{k>=0} ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k/Product_{i=0..k-1} (1-x-x*i)/(1-x).
G.f.: (G(0) - 2)/(2*x-1) where G(k) = 2 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: -G(0) where G(k) = 1 - (x*k - 2)/(x*k - 1 - x*(x*k - 1)/(x + (x*k - 2)/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 2 - (2*x*k - 1)/(x*k - 1 - x*(x*k - 1)/(x + (2*x*k - 1)/G(k+1) )); (continued fraction).
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-k*x)/(1-x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1-x)*G(0)) - 1/x where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )); (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 + x/( x*k - 1 )/Q(k+1); (continued fraction).
G.f.: 1+x/Q(0), where Q(k) = 1 - x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction).
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))-n-1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert W-function. - Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/W(n)-1-n) / (sqrt(1+W(n)) * W(n)^n). - Vaclav Kotesovec, Nov 13 2015
From Natalia L. Skirrow, Apr 13 2025: (Start)
By taking logarithmic derivatives of the equivalent to Kotesovec's asymptotic for Bell polynomials at x=1, we obtain properties of the nth row of A008277 as a statistical distribution (where W=W(n),X=W(n)+1)
a(n+1)/a(n) ~ n/W + W/(2*(W+1)^2) is 1 more than the expectation.
(2*a(n+1)+a(n+2))/a(n) - (a(n+1)/a(n))^2 - a(n+2)/a(n+1) ~ n/(W*X)+1/(2*X^2)-3/(2*X^3)+1/X^4 is 1 more than the variance.
(This is a complete asymptotic characterisation, since they converge to normal distributions; see Harper, 1967)
(End)
a(n) are the coefficients in the asymptotic expansion of -exp(-1)*(-1)^x*x*Gamma(-x,0,-1), where Gamma(a,z0,z1) is the generalized incomplete Gamma function. - Vladimir Reshetnikov, Nov 12 2015
a(n) = 1 + floor(exp(-1) * Sum_{k=1..2*n} k^n/k!). - Vladimir Reshetnikov, Nov 13 2015
a(p^m) == m+1 (mod p) when p is prime and m >= 1 (see Lemma 3.1 in the Hurst/Schultz reference). - Seiichi Manyama, Jun 01 2016
a(n) = Sum_{k=0..n} hypergeom([1, -k], [], 1)*Stirling2(n+1, k+1) = Sum_{k=0..n} A182386(k)*Stirling2(n+1, k+1). - Mélika Tebni, Jul 02 2022
For n >= 1, a(n) = Sum_{i=0..n-1} a(i)*A074664(n-i). - Davide Rotondo, Apr 21 2024
a(n) is the n-th derivative of e^e^x divided by e at point x=0. - Joan Ludevid, Nov 05 2024

Extensions

Edited by M. F. Hasler, Nov 30 2018

A000225 a(n) = 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727, 268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591
Offset: 0

Keywords

Comments

This is the Gaussian binomial coefficient [n,1] for q=2.
Number of rank-1 matroids over S_n.
Numbers k such that the k-th central binomial coefficient is odd: A001405(k) mod 2 = 1. - Labos Elemer, Mar 12 2003
This gives the (zero-based) positions of odd terms in the following convolution sequences: A000108, A007460, A007461, A007463, A007464, A061922.
Also solutions (with minimum number of moves) for the problem of Benares Temple, i.e., three diamond needles with n discs ordered by decreasing size on the first needle to place in the same order on the third one, without ever moving more than one disc at a time and without ever placing one disc at the top of a smaller one. - Xavier Acloque, Oct 18 2003
a(0) = 0, a(1) = 1; a(n) = smallest number such that a(n)-a(m) == 0 (mod (n-m+1)), for all m. - Amarnath Murthy, Oct 23 2003
Binomial transform of [1, 1/2, 1/3, ...] = [1/1, 3/2, 7/3, ...]; (2^n - 1)/n, n=1,2,3, ... - Gary W. Adamson, Apr 28 2005
Numbers whose binary representation is 111...1. E.g., the 7th term is (2^7) - 1 = 127 = 1111111 (in base 2). - Alexandre Wajnberg, Jun 08 2005
Number of nonempty subsets of a set with n elements. - Michael Somos, Sep 03 2006
For n >= 2, a(n) is the least Fibonacci n-step number that is not a power of 2. - Rick L. Shepherd, Nov 19 2007
Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which either x is a subset of y or y is a subset of x. - Ross La Haye, Jan 10 2008
A simpler way to state this is that it is the number of pairs (x,y) where at least one of x and y is the empty set. - Franklin T. Adams-Watters, Oct 28 2011
2^n-1 is the sum of the elements in a Pascal triangle of depth n. - Brian Lewis (bsl04(AT)uark.edu), Feb 26 2008
Sequence generalized: a(n) = (A^n -1)/(A-1), n >= 1, A integer >= 2. This sequence has A=2; A003462 has A=3; A002450 has A=4; A003463 has A=5; A003464 has A=6; A023000 has A=7; A023001 has A=8; A002452 has A=9; A002275 has A=10; A016123 has A=11; A016125 has A=12; A091030 has A=13; A135519 has A=14; A135518 has A=15; A131865 has A=16; A091045 has A=17; A064108 has A=20. - Ctibor O. Zizka, Mar 03 2008
a(n) is also a Mersenne prime A000668 when n is a prime number in A000043. - Omar E. Pol, Aug 31 2008
a(n) is also a Mersenne number A001348 when n is prime. - Omar E. Pol, Sep 05 2008
With offset 1, = row sums of triangle A144081; and INVERT transform of A009545 starting with offset 1; where A009545 = expansion of sin(x)*exp(x). - Gary W. Adamson, Sep 10 2008
Numbers n such that A000120(n)/A070939(n) = 1. - Ctibor O. Zizka, Oct 15 2008
For n > 0, sequence is equal to partial sums of A000079; a(n) = A000203(A000079(n-1)). - Lekraj Beedassy, May 02 2009
Starting with offset 1 = the Jacobsthal sequence, A001045, (1, 1, 3, 5, 11, 21, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009
Numbers n such that n=2*phi(n+1)-1. - Farideh Firoozbakht, Jul 23 2009
a(n) = (a(n-1)+1)-th odd numbers = A005408(a(n-1)) for n >= 1. - Jaroslav Krizek, Sep 11 2009
Partial sums of a(n) for n >= 0 are A000295(n+1). Partial sums of a(n) for n >= 1 are A000295(n+1) and A130103(n+1). a(n) = A006127(n) - (n+1). - Jaroslav Krizek, Oct 16 2009
If n is even a(n) mod 3 = 0. This follows from the congruences 2^(2k) - 1 ~ 2*2*...*2 - 1 ~ 4*4*...*4 - 1 ~ 1*1*...*1 - 1 ~ 0 (mod 3). (Note that 2*2*...*2 has an even number of terms.) - Washington Bomfim, Oct 31 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 26 2010
This is the sequence A(0,1;1,2;2) = A(0,1;3,-2;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
a(n) = S(n+1,2), a Stirling number of the second kind. See the example below. - Dennis P. Walsh, Mar 29 2011
Entries of row a(n) in Pascal's triangle are all odd, while entries of row a(n)-1 have alternating parities of the form odd, even, odd, even, ..., odd.
Define the bar operation as an operation on signed permutations that flips the sign of each entry. Then a(n+1) is the number of signed permutations of length 2n that are equal to the bar of their reverse-complements and avoid the set of patterns {(-2,-1), (-1,+2), (+2,+1)}. (See the Hardt and Troyka reference.) - Justin M. Troyka, Aug 13 2011
A159780(a(n)) = n and A159780(m) < n for m < a(n). - Reinhard Zumkeller, Oct 21 2011
This sequence is also the number of proper subsets of a set with n elements. - Mohammad K. Azarian, Oct 27 2011
a(n) is the number k such that the number of iterations of the map k -> (3k +1)/2 == 1 (mod 2) until reaching (3k +1)/2 == 0 (mod 2) equals n. (see the Collatz problem). - Michel Lagneau, Jan 18 2012
For integers a, b, denote by a<+>b the least c >= a such that Hd(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then a(n+1)=a(n)<+>1. Thus this sequence is the Hamming analog of nonnegative integers. - Vladimir Shevelev, Feb 13 2012
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... apparently A007733. - R. J. Mathar, Aug 10 2012
Start with n. Each n generates a sublist {n-1,n-2,...,1}. Each element of each sublist also generates a sublist. Take the sum of all. E.g., 3->{2,1} and 2->{1}, so a(3)=3+2+1+1=7. - Jon Perry, Sep 02 2012
This is the Lucas U(P=3,Q=2) sequence. - R. J. Mathar, Oct 24 2012
The Mersenne numbers >= 7 are all Brazilian numbers, as repunits in base two. See Proposition 1 & 5.2 in Links: "Les nombres brésiliens". - Bernard Schott, Dec 26 2012
Number of line segments after n-th stage in the H tree. - Omar E. Pol, Feb 16 2013
Row sums of triangle in A162741. - Reinhard Zumkeller, Jul 16 2013
a(n) is the highest power of 2 such that 2^a(n) divides (2^n)!. - Ivan N. Ianakiev, Aug 17 2013
In computer programming, these are the only unsigned numbers such that k&(k+1)=0, where & is the bitwise AND operator and numbers are expressed in binary. - Stanislav Sykora, Nov 29 2013
Minimal number of moves needed to interchange n frogs in the frogs problem (see for example the NRICH 1246 link or the Britton link below). - N. J. A. Sloane, Jan 04 2014
a(n) !== 4 (mod 5); a(n) !== 10 (mod 11); a(n) !== 2, 4, 5, 6 (mod 7). - Carmine Suriano, Apr 06 2014
After 0, antidiagonal sums of the array formed by partial sums of integers (1, 2, 3, 4, ...). - Luciano Ancora, Apr 24 2015
a(n+1) equals the number of ternary words of length n avoiding 01,02. - Milan Janjic, Dec 16 2015
With offset 0 and another initial 0, the n-th term of 0, 0, 1, 3, 7, 15, ... is the number of commas required in the fully-expanded von Neumann definition of the ordinal number n. For example, 4 := {0, 1, 2, 3} := {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}, which uses seven commas. Also, for n>0, a(n) is the total number of symbols required in the fully-expanded von Neumann definition of ordinal n - 1, where a single symbol (as usual) is always used to represent the empty set and spaces are ignored. E.g., a(5) = 31, the total such symbols for the ordinal 4. - Rick L. Shepherd, May 07 2016
With the quantum integers defined by [n+1]A001045%20are%20given%20by%20q%20=%20i%20*%20sqrt(2)%20for%20i%5E2%20=%20-1.%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Mersenne numbers are a(n+1) = q^n [n+1]_q with q = sqrt(2), whereas the signed Jacobsthal numbers A001045 are given by q = i * sqrt(2) for i^2 = -1. Cf. A239473. - _Tom Copeland, Sep 05 2016
For n>1: numbers n such that n - 1 divides sigma(n + 1). - Juri-Stepan Gerasimov, Oct 08 2016
This is also the second column of the Stirling2 triangle A008277 (see also A048993). - Wolfdieter Lang, Feb 21 2017
Except for the initial terms, the decimal representation of the x-axis of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 659", "Rule 721" and "Rule 734", based on the 5-celled von Neumann neighborhood initialized with a single on cell. - Robert Price, Mar 14 2017
a(n), n > 1, is the number of maximal subsemigroups of the monoid of order-preserving partial injective mappings on a set with n elements. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Also the number of independent vertex sets and vertex covers in the complete bipartite graph K_{n-1,n-1}. - Eric W. Weisstein, Sep 21 2017
Sum_{k=0..n} p^k is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*p + binomial(i+j-1, i), in this case p=2 (empirical observation). - Tony Foster III, May 11 2019
The rational numbers r(n) = a(n+1)/2^(n+1) = a(n+1)/A000079(n+1) appear also as root of the n-th iteration f^{[n]}(c; x) = 2^(n+1)*x - a(n+1)*c of f(c; x) = f^{[0]}(c; x) = 2*x - c as r(n)*c. This entry is motivated by a riddle of Johann Peter Hebel (1760 - 1826): Erstes Rechnungsexempel(Ein merkwürdiges Rechnungs-Exempel) from 1803, with c = 24 and n = 2, leading to the root r(2)*24 = 21 as solution. See the link and reference. For the second problem, also involving the present sequence, see a comment in A130330. - Wolfdieter Lang, Oct 28 2019
a(n) is the sum of the smallest elements of all subsets of {1,2,..,n} that contain n. For example, a(3)=7; the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 7. - Enrique Navarrete, Aug 21 2020
a(n-1) is the number of nonempty subsets of {1,2,..,n} which don't have an element that is the size of the set. For example, for n = 4, a(3) = 7 and the subsets are {2}, {3}, {4}, {1,3}, {1,4}, {3,4}, {1,2,4}. - Enrique Navarrete, Nov 21 2020
From Eric W. Weisstein, Sep 04 2021: (Start)
Also the number of dominating sets in the complete graph K_n.
Also the number of minimum dominating sets in the n-helm graph for n >= 3. (End)
Conjecture: except for a(2)=3, numbers m such that 2^(m+1) - 2^j - 2^k - 1 is composite for all 0 <= j < k <= m. - Chai Wah Wu, Sep 08 2021
a(n) is the number of three-in-a-rows passing through a corner cell in n-dimensional tic-tac-toe. - Ben Orlin, Mar 15 2022
From Vladimir Pletser, Jan 27 2023: (Start)
a(n) == 1 (mod 30) for n == 1 (mod 4);
a(n) == 7 (mod 120) for n == 3 (mod 4);
(a(n) - 1)/30 = (a(n+2) - 7)/120 for n odd;
(a(n) - 1)/30 = (a(n+2) - 7)/120 = A131865(m) for n == 1 (mod 4) and m >= 0 with A131865(0) = 0. (End)
a(n) is the number of n-digit numbers whose smallest decimal digit is 8. - Stefano Spezia, Nov 15 2023
Also, number of nodes in a perfect binary tree of height n-1, or: number of squares (or triangles) after the n-th step of the construction of a Pythagorean tree: Start with a segment. At each step, construct squares having the most recent segment(s) as base, and isosceles right triangles having the opposite side of the squares as hypotenuse ("on top" of each square). The legs of these triangles will serve as the segments which are the bases of the squares in the next step. - M. F. Hasler, Mar 11 2024
a(n) is the length of the longest path in the n-dimensional hypercube. - Christian Barrientos, Apr 13 2024
a(n) is the diameter of the n-Hanoi graph. Equivalently, a(n) is the largest minimum number of moves between any two states of the Towers of Hanoi problem (aka problem of Benares Temple described above). - Allan Bickle, Aug 09 2024

Examples

			For n=3, a(3)=S(4,2)=7, a Stirling number of the second kind, since there are 7 ways to partition {a,b,c,d} into 2 nonempty subsets, namely,
  {a}U{b,c,d}, {b}U{a,c,d}, {c}U{a,b,d}, {d}U{a,b,c}, {a,b}U{c,d}, {a,c}U{b,d}, and {a,d}U{b,c}. - _Dennis P. Walsh_, Mar 29 2011
From _Justin M. Troyka_, Aug 13 2011: (Start)
Since a(3) = 7, there are 7 signed permutations of 4 that are equal to the bar of their reverse-complements and avoid {(-2,-1), (-1,+2), (+2,+1)}. These are:
  (+1,+2,-3,-4),
  (+1,+3,-2,-4),
  (+1,-3,+2,-4),
  (+2,+4,-1,-3),
  (+3,+4,-1,-2),
  (-3,+1,-4,+2),
  (-3,-4,+1,+2). (End)
G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 63*x^6 + 127*x^7 + ...
For the Towers of Hanoi problem with 2 disks, the moves are as follows, so a(2) = 3.
12|_|_ -> 2|1|_ -> _|1|2 -> _|_|12  - _Allan Bickle_, Aug 07 2024
		

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 79.
  • Johann Peter Hebel, Gesammelte Werke in sechs Bänden, Herausgeber: Jan Knopf, Franz Littmann und Hansgeorg Schmidt-Bergmann unter Mitarbeit von Ester Stern, Wallstein Verlag, 2019. Band 3, S. 20-21, Loesung, S. 36-37. See also the link below.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 75-83.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, "Tower of Hanoi", Penguin Books, 1987, pp. 112-113.

Crossrefs

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
Cf. a(n)=A112492(n, 2). Rightmost column of A008969.
a(n) = A118654(n, 1) = A118654(n-1, 3), for n > 0.
Subsequence of A132781.
Smallest number whose base b sum of digits is n: this sequence (b=2), A062318 (b=3), A180516 (b=4), A181287 (b=5), A181288 (b=6), A181303 (b=7), A165804 (b=8), A140576 (b=9), A051885 (b=10).
Cf. A008277, A048993 (columns k=2), A000918, A130330.
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

Programs

  • Haskell
    a000225 = (subtract 1) . (2 ^)
    a000225_list = iterate ((+ 1) . (* 2)) 0
    -- Reinhard Zumkeller, Mar 20 2012
    
  • Maple
    A000225 := n->2^n-1; [ seq(2^n-1,n=0..50) ];
    A000225:=1/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, sequence starting at a(1)
  • Mathematica
    a[n_] := 2^n - 1; Table[a[n], {n, 0, 30}] (* Stefan Steinerberger, Mar 30 2006 *)
    Array[2^# - 1 &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
    NestList[2 # + 1 &, 0, 32] (* Robert G. Wilson v, Feb 28 2011 *)
    2^Range[0, 20] - 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {1, 3}, 20] (* Eric W. Weisstein, Sep 21 2017 *)
    CoefficientList[Series[1/(1 - 3 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    A000225(n) = 2^n-1  \\ Michael B. Porter, Oct 27 2009
    
  • PARI
    concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015
    
  • Python
    def A000225(n): return (1<Chai Wah Wu, Jul 06 2022
  • SageMath
    def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019
    

Formula

G.f.: x/((1-2*x)*(1-x)).
E.g.f.: exp(2*x) - exp(x).
E.g.f. if offset 1: ((exp(x)-1)^2)/2.
a(n) = Sum_{k=0..n-1} 2^k. - Paul Barry, May 26 2003
a(n) = a(n-1) + 2*a(n-2) + 2, a(0)=0, a(1)=1. - Paul Barry, Jun 06 2003
Let b(n) = (-1)^(n-1)*a(n). Then b(n) = Sum_{i=1..n} i!*i*Stirling2(n,i)*(-1)^(i-1). E.g.f. of b(n): (exp(x)-1)/exp(2x). - Mario Catalani (mario.catalani(AT)unito.it), Dec 19 2003
a(n+1) = 2*a(n) + 1, a(0) = 0.
a(n) = Sum_{k=1..n} binomial(n, k).
a(n) = n + Sum_{i=0..n-1} a(i); a(0) = 0. - Rick L. Shepherd, Aug 04 2004
a(n+1) = (n+1)*Sum_{k=0..n} binomial(n, k)/(k+1). - Paul Barry, Aug 06 2004
a(n+1) = Sum_{k=0..n} binomial(n+1, k+1). - Paul Barry, Aug 23 2004
Inverse binomial transform of A001047. Also U sequence of Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A099393(n-1) - A020522(n-1) for n > 0. - Reinhard Zumkeller, Feb 07 2006
a(n) = A119258(n,n-1) for n > 0. - Reinhard Zumkeller, May 11 2006
a(n) = 3*a(n-1) - 2*a(n-2); a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
Sum_{n>0} 1/a(n) = 1.606695152... = A065442, see A038631. - Philippe Deléham, Jun 27 2006
Stirling_2(n-k,2) starting from n=k+1. - Artur Jasinski, Nov 18 2006
a(n) = A125118(n,1) for n > 0. - Reinhard Zumkeller, Nov 21 2006
a(n) = StirlingS2(n+1,2). - Ross La Haye, Jan 10 2008
a(n) = A024036(n)/A000051(n). - Reinhard Zumkeller, Feb 14 2009
a(n) = A024088(n)/A001576(n). -Reinhard Zumkeller, Feb 15 2009
a(2*n) = a(n)*A000051(n); a(n) = A173787(n,0). - Reinhard Zumkeller, Feb 28 2010
For n > 0: A179857(a(n)) = A024036(n) and A179857(m) < A024036(n) for m < a(n). - Reinhard Zumkeller, Jul 31 2010
From Enrique Pérez Herrero, Aug 21 2010: (Start)
a(n) = J_n(2), where J_n is the n-th Jordan Totient function: (A007434, is J_2).
a(n) = Sum_{d|2} d^n*mu(2/d). (End)
A036987(a(n)) = 1. - Reinhard Zumkeller, Mar 06 2012
a(n+1) = A044432(n) + A182028(n). - Reinhard Zumkeller, Apr 07 2012
a(n) = A007283(n)/3 - 1. - Martin Ettl, Nov 11 2012
a(n+1) = A001317(n) + A219843(n); A219843(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2012
a(n) = det(|s(i+2,j+1)|, 1 <= i,j <= n-1), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013
G.f.: Q(0), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 - 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
E.g.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - (k+1)/Q(k+1))); (continued fraction).
G.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 23 2013
a(n) = A000203(2^(n-1)), n >= 1. - Ivan N. Ianakiev, Aug 17 2013
a(n) = Sum_{t_1+2*t_2+...+n*t_n=n} n*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n)/(t_1+t_2 +...+t_n). - Mircea Merca, Dec 06 2013
a(0) = 0; a(n) = a(n-1) + 2^(n-1) for n >= 1. - Fred Daniel Kline, Feb 09 2014
a(n) = A125128(n) - A000325(n) + 1. - Miquel Cerda, Aug 07 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Binomial transform of A057427.
Sum_{n>=0} a(n)/n! = A090142. (End)
a(n) = A000918(n) + 1. - Miquel Cerda, Aug 09 2016
a(n+1) = (A095151(n+1) - A125128(n))/2. - Miquel Cerda, Aug 12 2016
a(n) = (A079583(n) - A000325(n+1))/2. - Miquel Cerda, Aug 15 2016
Convolution of binomial coefficient C(n,a(k)) with itself is C(n,a(k+1)) for all k >= 3. - Anton Zakharov, Sep 05 2016
a(n) = (A083706(n-1) + A000325(n))/2. - Miquel Cerda, Sep 30 2016
a(n) = A005803(n) + A005408(n-1). - Miquel Cerda, Nov 25 2016
a(n) = A279396(n+2,2). - Wolfdieter Lang, Jan 10 2017
a(n) = n + Sum_{j=1..n-1} (n-j)*2^(j-1). See a Jun 14 2017 formula for A000918(n+1) with an interpretation. - Wolfdieter Lang, Jun 14 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k,i). - Wesley Ivan Hurt, Sep 21 2017
a(n+m) = a(n)*a(m) + a(n) + a(m). - Yuchun Ji, Jul 27 2018
a(n+m) = a(n+1)*a(m) - 2*a(n)*a(m-1). - Taras Goy, Dec 23 2018
a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*2 + binomial(i+j-1, i) (empirical observation). - Tony Foster III, May 11 2019
From Peter Bala, Jun 27 2025: (Start)
For n >= 1, a(3*n)/a(n) = A001576(n), a(4*n)/a(n) = A034496(n), a(5*n)/a(n) = A020514(n) a(6*n)/a(n) = A034665(n), a(7*n)/a(n) = A020516(n) and a(8*n)/a(n) = A034674(n).
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
Modulo differences in offsets, exp( Sum_{n >= 1} a(k*n)/a(n)*x^n/n ) is the o.g.f. of A006095 (k = 3), A006096 (k = 4), A006097 (k = 5), A006110 (k = 6), A022189 (k = 7), A022190 (k = 8), A022191 (k = 9) and A022192 (k = 10).
The following are all examples of telescoping series:
Sum_{n >= 1} 2^n/(a(n)*a(n+1)) = 1; Sum_{n >= 1} 2^n/(a(n)*a(n+1)*a(n+2)) = 1/9.
In general, for k >= 1, Sum_{n >= 1} 2^n/(a(n)*a(n+1)*...*a(n+k)) = 1/(a(1)*a(2)*...*a(k)*a(k)).
Sum_{n >= 1} 2^n/(a(n)*a(n+2)) = 4/9, since 2^n/(a(n)*a(n+2)) = b(n) - b(n+1), where b(n) = (2/3)*(3*2^(n-1) - 1)/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+2)) = -2/9, since (-2)^n/(a(n)*a(n+2)) = c(n) - c(n+1), where c(n) = (1/3)*(-2)^n/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} 2^n/(a(n)*a(n+4)) = 18/175, since 2^n/(a(n)*a(n+4)) = d(n) - d(n+1), where d(n) = (120*8^n - 140*4^n + 45*2^n - 4)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+4)) = -26/525, since (-2)^n/(a(n)*a(n+4)) = e(n) - e(n+1), where e(n) = (-1)^n*(40*8^n - 24*4^n + 5*2^n)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)). (End)

Extensions

Name partially edited by Eric W. Weisstein, Sep 04 2021

A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

Original entry on oeis.org

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112
Offset: 0

Comments

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013
Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
(1) (2,1) (3,2)
(1,1,1) (1,2,2)
(2,2,1)
(1,1,2,1)
(2,1,1,1)
(1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
From Michael Wallner, Jan 25 2022: (Start)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(2)=6:
3 4 2 4 3 4 3|4 4|3 2|4
1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
Example for a(2)=6:
3|4 2|4 4|3 3|4 4|3 4|2
1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
From Shel Kaphan, Jan 12 2023: (Start)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

Examples

			G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - _Michael B. Porter_, Jul 06 2016
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
  • Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
  • Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
  • Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
  • Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
  • Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

Programs

  • GAP
    List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018
  • Haskell
    a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011
    
  • Magma
    a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];
    
  • Maple
    A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
    with(combstruct); [seq(count([S,{S=Prod(Set(Z,card=i),Set(Z,card=i))}, labeled], size=(2*i)), i=0..20)];
    with(combstruct); [seq(count([S,{S=Sequence(Union(Arch,Arch)), Arch=Prod(Epsilon, Sequence(Arch),Z)},unlabeled],size=i), i=0..25)];
    with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
    A000984List := proc(m) local A, P, n; A := [1,2]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
    A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
    CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)
  • Maxima
    A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */
    
  • PARI
    A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014
    
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
    a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013
    
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
    a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013
    
  • Python
    from _future_ import division
    A000984_list, b = [1], 1
    for n in range(10**3):
        b = b*(4*n+2)//(n+1)
        A000984_list.append(b) # Chai Wah Wu, Mar 04 2016
    

Formula

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) = Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
2, 2, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 2014
0 = a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = (-4)^n*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Mar 31 2024: (Start)
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

A001006 Motzkin numbers: number of ways of drawing any number of nonintersecting chords joining n (labeled) points on a circle.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209, 593742784829, 1697385471211
Offset: 0

Keywords

Comments

Number of 4321-, (3412,2413)-, (3412,3142)- and 3412-avoiding involutions in S_n.
Number of sequences of length n-1 consisting of positive integers such that the first and last elements are 1 or 2 and the absolute difference between any 2 consecutive elements is 0 or 1. - Jon Perry, Sep 04 2003
From David Callan, Jul 15 2004: (Start)
Also number of Motzkin n-paths: paths from (0,0) to (n,0) in an n X n grid using only steps U = (1,1), F = (1,0) and D = (1,-1).
Number of Dyck n-paths with no UUU. (Given such a Dyck n-path, change each UUD to U, then change each remaining UD to F. This is a bijection to Motzkin n-paths. Example with n=5: U U D U D U U D D D -> U F U D D.)
Number of Dyck (n+1)-paths with no UDU. (Given such a Dyck (n+1)-path, mark each U that is followed by a D and each D that is not followed by a U. Then change each unmarked U whose matching D is marked to an F. Lastly, delete all the marked steps. This is a bijection to Motzkin n-paths. Example with n=6 and marked steps in small type: U U u d D U U u d d d D u d -> U U u d D F F u d d d D u d -> U U D F F D.) (End)
a(n) is the number of strings of length 2n+2 from the following recursively defined set: L contains the empty string and, for any strings a and b in L, we also find (ab) in L. The first few elements of L are e, (), (()), ((())), (()()), (((()))), ((()())), ((())()), (()(())) and so on. This proves that a(n) is less than or equal to C(n), the n-th Catalan number. See Orrick link (2024). - Saul Schleimer (saulsch(AT)math.rutgers.edu), Feb 23 2006 (Additional linked comment added by William P. Orrick, Jun 13 2024.)
a(n) = number of Dyck n-paths all of whose valleys have even x-coordinate (when path starts at origin). For example, T(4,2)=3 counts UDUDUUDD, UDUUDDUD, UUDDUDUD. Given such a path, split it into n subpaths of length 2 and transform UU->U, DD->D, UD->F (there will be no DUs for that would entail a valley with odd x-coordinate). This is a bijection to Motzkin n-paths. - David Callan, Jun 07 2006
Also the number of standard Young tableaux of height <= 3. - Mike Zabrocki, Mar 24 2007
a(n) is the number of RNA shapes of size 2n+2. RNA Shapes are essentially Dyck words without "directly nested" motifs of the form A[[B]]C, for A, B and C Dyck words. The first RNA Shapes are []; [][]; [][][], [[][]]; [][][][], [][[][]], [[][][]], [[][]][]; ... - Yann Ponty (ponty(AT)lri.fr), May 30 2007
The sequence is self-generated from top row A going to the left starting (1,1) and bottom row = B, the same sequence but starting (0,1) and going to the right. Take dot product of A and B and add the result to n-th term of A to get the (n+1)-th term of A. Example: a(5) = 21 as follows: Take dot product of A = (9, 4, 2, 1, 1) and (0, 1, 1, 2, 4) = (0, + 4 + 2 + 2 + 4) = 12; which is added to 9 = 21. - Gary W. Adamson, Oct 27 2008
Equals A005773 / A005773 shifted (i.e., (1,2,5,13,35,96,...) / (1,1,2,5,13,35,96,...)). - Gary W. Adamson, Dec 21 2008
Starting with offset 1 = iterates of M * [1,1,0,0,0,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
a(n) is the number of involutions of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(4)=9; indeed, p=3412=(13)(24) is the only involution of {1,2,3,4} with genus > 0. This follows easily from the fact that a permutation p of {1,2,...,n} has genus 0 if and only if the cycle decomposition of p gives a noncrossing partition of {1,2,...,n} and each cycle of p is increasing (see Lemma 2.1 of the Dulucq-Simion reference). [Also, redundantly, for p=3412=(13)(24) we have cp'=2341*3412=4123=(1432) and so g(p)=(1/2)(4+1-2-1)=1.] - Emeric Deutsch, May 29 2010
Let w(i,j,n) denote walks in N^2 which satisfy the multivariate recurrence w(i,j,n) = w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Then a(n) = Sum_{i = 0..n, j = 0..n} w(i,j,n) is the number of such walks of length n. - Peter Luschny, May 21 2011
a(n)/a(n-1) tends to 3.0 as N->infinity: (1+2*cos(2*Pi/N)) relating to longest odd N regular polygon diagonals, by way of example, N=7: Using the tridiagonal generator [cf. comment of Jan 07 2009], for polygon N=7, we extract an (N-1)/2 = 3 X 3 matrix, [0,1,0; 1,1,1; 0,1,1] with an e-val of 2.24697...; the longest Heptagon diagonal with edge = 1. As N tends to infinity, the diagonal lengths tend to 3.0, the convergent of the sequence. - Gary W. Adamson, Jun 08 2011
Number of (n+1)-length permutations avoiding the pattern 132 and the dotted pattern 23\dot{1}. - Jean-Luc Baril, Mar 07 2012
Number of n-length words w over alphabet {a,b,c} such that for every prefix z of w we have #(z,a) >= #(z,b) >= #(z,c), where #(z,x) counts the letters x in word z. The a(4) = 9 words are: aaaa, aaab, aaba, abaa, aabb, abab, aabc, abac, abca. - Alois P. Heinz, May 26 2012
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=1, r(k)<=k, and r(k)!=r(k-1); for example, the 9 RGS for n=4 are 1010, 1012, 1201, 1210, 1212, 1230, 1231, 1232, 1234. - Joerg Arndt, Apr 16 2013
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=0, r(k)<=k and r(k)-r(k-1) != 1; for example, the 9 RGS for n=4 are 0000, 0002, 0003, 0004, 0022, 0024, 0033, 0222, 0224. - Joerg Arndt, Apr 17 2013
Number of (4231,5276143)-avoiding involutions in S_n. - Alexander Burstein, Mar 05 2014
a(n) is the number of increasing unary-binary trees with n nodes that have an associated permutation that avoids 132. For more information about unary-binary trees with associated permutations, see A245888. - Manda Riehl, Aug 07 2014
a(n) is the number of involutions on [n] avoiding the single pattern p, where p is any one of the 8 (classical) patterns 1234, 1243, 1432, 2134, 2143, 3214, 3412, 4321. Also, number of (3412,2413)-, (3412,3142)-, (3412,2413,3142)-avoiding involutions on [n] because each of these 3 sets actually coincides with the 3412-avoiding involutions on [n]. This is a complete list of the 8 singles, 2 pairs, and 1 triple of 4-letter classical patterns whose involution avoiders are counted by the Motzkin numbers. (See Barnabei et al. 2011 reference.) - David Callan, Aug 27 2014
From Tony Foster III, Jul 28 2016: (Start)
A series created using 2*a(n) + a(n+1) has Hankel transform of F(2n), offset 3, F being the Fibonacci bisection, A001906 (empirical observation).
A series created using 2*a(n) + 3*a(n+1) + a(n+2) gives the Hankel transform of Sum_{k=0..n} k*Fibonacci(2*k), offset 3, A197649 (empirical observation). (End)
Conjecture: (2/n)*Sum_{k=1..n} (2k+1)*a(k)^2 is an integer for each positive integer n. - Zhi-Wei Sun, Nov 16 2017
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n." - Eric M. Schmidt, Dec 16 2017
Number of U_{k}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018
If tau_1 and tau_2 are two distinct permutation patterns chosen from the set {132,231,312}, then a(n) is the number of valid hook configurations of permutations of [n+1] that avoid the patterns tau_1 and tau_2. - Colin Defant, Apr 28 2019
Number of permutations of length n that are sorted to the identity by a consecutive-321-avoiding stack followed by a classical-21-avoiding stack. - Colin Defant, Aug 29 2020
From Helmut Prodinger, Dec 13 2020: (Start)
a(n) is the number of paths in the first quadrant starting at (0,0) and consisting of n steps from the infinite set {(1,1), (1,-1), (1,-2), (1,-3), ...}.
For example, denoting U=(1,1), D=(1,-1), D_ j=(1,-j) for j >= 2, a(4) counts UUUU, UUUD, UUUD_2, UUUD_3, UUDU, UUDD, UUD_2U, UDUU, UDUD.
This step set is inspired by {(1,1), (1,-1), (1,-3), (1,-5), ...}, suggested by Emeric Deutsch around 2000.
See Prodinger link that contains a bijection to Motzkin paths. (End)
Named by Donaghey (1977) after the Israeli-American mathematician Theodore Motzkin (1908-1970). In Sloane's "A Handbook of Integer Sequences" (1973) they were called "generalized ballot numbers". - Amiram Eldar, Apr 15 2021
Number of Motzkin n-paths a(n) is split into A107587(n), number of even Motzkin n-paths, and A343386(n), number of odd Motzkin n-paths. The value A107587(n) - A343386(n) can be called the "shadow" of a(n) (see A343773). - Gennady Eremin, May 17 2021
Conjecture: If p is a prime of the form 6m+1 (A002476), then a(p-2) is divisible by p. Currently, no counterexample exists for p < 10^7. Personal communication from Robert Gerbicz: mod such p this is equivalent to A066796 with comment: "Every A066796(n) from A066796((p-1)/2) to A066796(p-1) is divisible by prime p of form 6m+1". - Serge Batalov, Feb 08 2022
From Rob Burns, Nov 11 2024: (Start)
The conjecture is proved in the 2017 paper by Rob Burns in the Links below. The result is contained in Tables 4 and 5 of the paper, which show that a(p-2) == 0 (mod p) when p == 1 (mod 6) and a(p-2) == -1 (mod p) when p == -1 (mod 6).
In fact, the 2017 paper by Burns establishes more general congruences for a(p^k - 2) where k >= 1.
If p == 1 (mod 6) then a(p^k - 2) == 0 (mod p) for k >= 1.
If p == -1 (mod 6) then a(p^k - 2) == -1 (mod p) when k is odd and a(p^k - 2) == 0 (mod p) when k is even.
These are consequences of the transitions provided in Tables 4, 5 and 6 of the paper.
The 2024 paper by Nadav Kohen also proves the conjecture. Proposition 6 of the paper states that a prime p divides a(p-2) if and only if p = (1 mod 3). (End)
From Peter Bala, Feb 10 2022: (Start)
Conjectures:
(1) For prime p == 1 (mod 6) and n, r >= 1, a(n*p^r - 2) == -A005717(n-1) (mod p), where we take A005717(0) = 0 to match Batalov's conjecture above.
(2) For prime p == 5 (mod 6) and n >= 1, a(n*p - 2) == -A005773(n) (mod p).
(3) For prime p >= 3 and k >= 1, a(n + p^k) == a(n) (mod p) for 0 <= n <= (p^k - 3).
(4) For prime p >= 5 and k >= 2, a(n + p^k) == a(n) (mod p^2) for 0 <= n <= (p^(k-1) - 3). (End)
The Hankel transform of this sequence with a(0) omitted gives the period-6 sequence [1, 0, -1, -1, 0, 1, ...] which is A010892 with its first term omitted, while the Hankel transform of the current sequence is the all-ones sequence A000012, and also it is the unique sequence with this property which is similar to the unique Hankel transform property of the Catalan numbers. - Michael Somos, Apr 17 2022
The number of terms in which the exponent of any variable x_i is not greater than 2 in the expansion of Product_{j=1..n} Sum_{i=1..j} x_i. E.g.: a(4) = 9: 3*x1^2*x2^2, 4*x1^2*x2*x3, 2*x1^2*x2*x4, x1^2*x3^2, x1^2*x3*x4, 2*x1*x2^2*x3, x1*x2^2*x4, x1*x2*x3^2, x1*x2*x3*x4. - Elif Baser, Dec 20 2024

Examples

			G.f.: 1 + x + 2*x^2 + 4*x^3 + 9*x^4 + 21*x^5 + 51*x^6 + 127*x^7 + 323*x^8 + ...
.
The 21 Motzkin-paths of length 5: UUDDF, UUDFD, UUFDD, UDUDF, UDUFD, UDFUD, UDFFF, UFUDD, UFDUD, UFDFF, UFFDF, UFFFD, FUUDD, FUDUD, FUDFF, FUFDF, FUFFD, FFUDF, FFUFD, FFFUD, FFFFF.
		

References

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  • F. R. Bernhart, Catalan, Motzkin, and Riordan numbers, Discr. Math., 204 (1999) 73-112.
  • R. Bojicic and M. D. Petkovic, Orthogonal Polynomials Approach to the Hankel Transform of Sequences Based on Motzkin Numbers, Bulletin of the Malaysian Mathematical Sciences, 2015, doi:10.1007/s40840-015-0249-3.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pp. 24, 298, 618, 912.
  • A. J. Bu, Automated counting of restricted Motzkin paths, Enumerative Combinatorics and Applications, ECA 1:2 (2021) Article S2R12.
  • Naiomi Cameron, JE McLeod, Returns and Hills on Generalized Dyck Paths, Journal of Integer Sequences, Vol. 19, 2016, #16.6.1.
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  • Tomislav Doslic and Darko Veljan, Logarithmic behavior of some combinatorial sequences. Discrete Math. 308 (2008), no. 11, 2182-2212. MR2404544 (2009j:05019).
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  • M. Dziemianczuk, "Enumerations of plane trees with multiple edges and Raney lattice paths." Discrete Mathematics 337 (2014): 9-24.
  • Wenjie Fang, A partial order on Motzkin paths, Discrete Math., 343 (2020), #111802.
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  • Kris Hatch, Presentation of the Motzkin Monoid, Senior Thesis, Univ. Cal. Santa Barbara, 2012; http://ccs.math.ucsb.edu/senior-thesis/Kris-Hatch.pdf.
  • V. Jelinek, Toufik Mansour, and M. Shattuck, On multiple pattern avoiding set partitions, Advances in Applied Mathematics Volume 50, Issue 2, February 2013, pp. 292-326.
  • Hana Kim and R. P. Stanley, A refined enumeration of hex trees and related polynomials, http://www-math.mit.edu/~rstan/papers/hextrees.pdf, Preprint 2015.
  • S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011. See p. 399 Table A.7.
  • A. Kuznetsov et al., Trees associated with the Motzkin numbers, J. Combin. Theory, A 76 (1996), 145-147.
  • T. Lengyel, On divisibility properties of some differences of Motzkin numbers, Annales Mathematicae et Informaticae, 41 (2013) pp. 121-136.
  • W. A. Lorenz, Y. Ponty, and P. Clote, Asymptotics of RNA Shapes, Journal of Computational Biology. 2008, 15(1): 31-63. doi:10.1089/cmb.2006.0153.
  • Piera Manara and Claudio Perelli Cippo, The fine structure of 4321 avoiding involutions and 321 avoiding involutions, PU. M. A. Vol. 22 (2011), 227-238; http://www.mat.unisi.it/newsito/puma/public_html/22_2/manara_perelli-cippo.pdf.
  • Toufik Mansour, Restricted 1-3-2 permutations and generalized patterns, Annals of Combin., 6 (2002), 65-76.
  • Toufik Mansour, Matthias Schork, and Mark Shattuck, Catalan numbers and pattern restricted set partitions. Discrete Math. 312(2012), no. 20, 2979-2991. MR2956089.
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  • Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
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Crossrefs

Bisections: A026945, A099250.
Sequences related to chords in a circle: A001006, A054726, A006533, A006561, A006600, A007569, A007678. See also entries for chord diagrams in Index file.
a(n) = A005043(n)+A005043(n+1).
A086246 is another version, although this is the main entry. Column k=3 of A182172.
Motzkin numbers A001006 read mod 2,3,4,5,6,7,8,11: A039963, A039964, A299919, A258712, A299920, A258711, A299918, A258710.
Cf. A004148, A004149, A023421, A023422, A023423, A290277 (inv. Euler Transf.).

Programs

  • Haskell
    a001006 n = a001006_list !! n
    a001006_list = zipWith (+) a005043_list $ tail a005043_list
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    # Three different Maple scripts for this sequence:
    A001006 := proc(n)
        add(binomial(n,2*k)*A000108(k),k=0..floor(n/2)) ;
    end proc:
    A001006 := proc(n) option remember; local k; if n <= 1 then 1 else procname(n-1) + add(procname(k)*procname(n-k-2),k=0..n-2); end if; end proc:
    # n -> [a(0),a(1),..,a(n)]
    A001006_list := proc(n) local w, m, j, i; w := proc(i,j,n) option remember;
    if min(i,j,n) < 0 or max(i,j) > n then 0
    elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi else
    w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) fi end:
    [seq( add( add( w(i, j, m), i = 0..m), j = 0..m), m = 0..n)] end:
    A001006_list(29); # Peter Luschny, May 21 2011
  • Mathematica
    a[0] = 1; a[n_Integer] := a[n] = a[n - 1] + Sum[a[k] * a[n - 2 - k], {k, 0, n - 2}]; Array[a, 30]
    (* Second program: *)
    CoefficientList[Series[(1 - x - (1 - 2x - 3x^2)^(1/2))/(2x^2), {x, 0, 29}], x] (* Jean-François Alcover, Nov 29 2011 *)
    Table[Hypergeometric2F1[(1-n)/2, -n/2, 2, 4], {n,0,29}] (* Peter Luschny, May 15 2016 *)
    Table[GegenbauerC[n,-n-1,-1/2]/(n+1),{n,0,100}] (* Emanuele Munarini, Oct 20 2016 *)
    MotzkinNumber = DifferenceRoot[Function[{y, n}, {(-3n-3)*y[n] + (-2n-5)*y[n+1] + (n+4)*y[n+2] == 0, y[0] == 1, y[1] == 1}]];
    Table[MotzkinNumber[n], {n, 0, 29}] (* Jean-François Alcover, Oct 27 2021 *)
  • Maxima
    a[0]:1$
    a[1]:1$
    a[n]:=((2*n+1)*a[n-1]+(3*n-3)*a[n-2])/(n+2)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • Maxima
    M(n) := coeff(expand((1+x+x^2)^(n+1)),x^n)/(n+1);
    makelist(M(n),n,0,60); /* Emanuele Munarini, Apr 04 2012 */
    
  • Maxima
    makelist(ultraspherical(n,-n-1,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */
    
  • PARI
    {a(n) = polcoeff( ( 1 - x - sqrt((1 - x)^2 - 4 * x^2 + x^3 * O(x^n))) / (2 * x^2), n)}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x / (1 + x + x^2) + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp(x + x * O(x^n)) * besseli(1, 2 * x + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • Python
    from gmpy2 import divexact
    A001006 = [1, 1]
    for n in range(2, 10**3):
        A001006.append(divexact(A001006[-1]*(2*n+1)+(3*n-3)*A001006[-2],n+2))
    # Chai Wah Wu, Sep 01 2014
    
  • Python
    def mot():
        a, b, n = 0, 1, 1
        while True:
            yield b//n
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
    A001006 = mot()
    print([next(A001006) for n in range(30)]) # Peter Luschny, May 16 2016
    
  • Python
    # A simple generator of Motzkin-paths (see the first comment of David Callan).
    C = str.count
    def aGen(n: int):
        a = [""]
        for w in a:
            if len(w) == n:
                if C(w, "U") == C(w, "D"): yield w
            else:
                for j in "UDF":
                    u = w + j
                    if C(u, "U") >= C(u, "D"): a += [u]
        return a
    for n in range(6):
        MP = [w for w in aGen(n)];
        print(len(MP), ":", MP)  # Peter Luschny, Dec 03 2024

Formula

G.f.: A(x) = ( 1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2).
G.f. A(x) satisfies A(x) = 1 + x*A(x) + x^2*A(x)^2.
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2). - Joerg Arndt, Oct 23 2012
a(n) = (-1/2) Sum_{i+j = n+2, i >= 0, j >= 0} (-3)^i*C(1/2, i)*C(1/2, j).
a(n) = (3/2)^(n+2) * Sum_{k >= 1} 3^(-k) * Catalan(k-1) * binomial(k, n+2-k). [Doslic et al.]
a(n) ~ 3^(n+1)*sqrt(3)*(1 + 1/(16*n))/((2*n+3)*sqrt((n+2)*Pi)). [Barcucci, Pinzani and Sprugnoli]
Limit_{n->infinity} a(n)/a(n-1) = 3. [Aigner]
a(n+2) - a(n+1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). [Bernhart]
a(n) = (1/(n+1)) * Sum_{i} (n+1)!/(i!*(i+1)!*(n-2*i)!). [Bernhart]
From Len Smiley: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k+1), inv. Binomial Transform of A000108.
a(n) = (1/(n+1))*Sum_{k=0..ceiling((n+1)/2)} binomial(n+1, k)*binomial(n+1-k, k-1);
D-finite with recurrence: (n+2)*a(n) = (2*n+1)*a(n-1) + (3*n-3)*a(n-2). (End)
a(n) = Sum_{k=0..n} C(n, 2k)*A000108(k). - Paul Barry, Jul 18 2003
E.g.f.: exp(x)*BesselI(1, 2*x)/x. - Vladeta Jovovic, Aug 20 2003
a(n) = A005043(n) + A005043(n+1).
The Hankel transform of this sequence gives A000012 = [1, 1, 1, 1, 1, 1, ...]. E.g., Det([1, 1, 2, 4; 1, 2, 4, 9; 2, 4, 9, 21; 4, 9, 21, 51]) = 1. - Philippe Deléham, Feb 23 2004
a(m+n) = Sum_{k>=0} A064189(m, k)*A064189(n, k). - Philippe Deléham, Mar 05 2004
a(n) = (1/(n+1))*Sum_{j=0..floor(n/3)} (-1)^j*binomial(n+1, j)*binomial(2*n-3*j, n). - Emeric Deutsch, Mar 13 2004
a(n) = A086615(n) - A086615(n-1) (n >= 1). - Emeric Deutsch, Jul 12 2004
G.f.: A(x)=(1-y+y^2)/(1-y)^2 where (1+x)*(y^2-y)+x=0; A(x)=4*(1+x)/(1+x+sqrt(1-2*x-3*x^2))^2; a(n)=(3/4)*(1/2)^n*Sum_(k=0..2*n, 3^(n-k)*C(k)*C(k+1, n+1-k) ) + 0^n/4 [after Doslic et al.]. - Paul Barry, Feb 22 2005
G.f.: c(x^2/(1-x)^2)/(1-x), c(x) the g.f. of A000108. - Paul Barry, May 31 2006
Asymptotic formula: a(n) ~ sqrt(3/4/Pi)*3^(n+1)/n^(3/2). - Benoit Cloitre, Jan 25 2007
a(n) = A007971(n+2)/2. - Zerinvary Lajos, Feb 28 2007
a(n) = (1/(2*Pi))*Integral_{x=-1..3} x^n*sqrt((3-x)*(1+x)) is the moment representation. - Paul Barry, Sep 10 2007
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, Phi([1]) is the Catalan numbers A000108. The present sequence is Phi([0,1,1]), see the 6th formula. - Gary W. Adamson, Oct 27 2008
G.f.: 1/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/.... (continued fraction). - Paul Barry, Dec 06 2008
G.f.: 1/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 08 2009
a(n) = (-3)^(1/2)/(6*(n+2)) * (-1)^n*(3*hypergeom([1/2, n+1],[1],4/3) - hypergeom([1/2, n+2],[1],4/3)). - Mark van Hoeij, Nov 12 2009
G.f.: 1/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-... (continued fraction). - Paul Barry, Mar 02 2010
G.f.: 1/(1-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, Jan 26 2011 [Adds apparently a third '1' in front. - R. J. Mathar, Jan 29 2011]
Let A(x) be the g.f., then B(x)=1+x*A(x) = 1 + 1*x + 1*x^2 + 2*x^3 + 4*x^4 + 9*x^5 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x) (continued fraction); more generally B(x)=C(x/(1+x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = (2/Pi)*Integral_{x=-1..1} (1+2*x)^n*sqrt(1-x^2). - Peter Luschny, Sep 11 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = 1/2/(x^2)-1/2/x-1/2/(x^2)*G(0); G(k) = 1+(4*k-1)*x*(2+3*x)/(4*k+2-x*(2+3*x)*(4*k+1)*(4*k+2) /(x*(2+3*x)*(4*k+1)+(4*k+4)/G(k+1))), if -1 < x < 1/3; (continued fraction). - Sergei N. Gladkovskii, Dec 01 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = (-1 + 1/G(0))/(2*x); G(k) = 1-2*x/(1+x/(1+x/(1-2*x/(1-x/(2-x/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Dec 11 2011
0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * ( -3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) unless n=-2. - Michael Somos, Mar 23 2012
a(n) = (-1)^n*hypergeometric([-n,3/2],[3],4). - Peter Luschny, Aug 15 2012
Representation in terms of special values of Jacobi polynomials P(n,alpha,beta,x), in Maple notation: a(n)= 2*(-1)^n*n!*JacobiP(n,2,-3/2-n,-7)/(n+2)!, n>=0. - Karol A. Penson, Jun 24 2013
G.f.: Q(0)/x - 1/x, where Q(k) = 1 + (4*k+1)*x/((1+x)*(k+1) - x*(1+x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1+x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
Catalan(n+1) = Sum_{k=0..n} binomial(n,k)*a(k). E.g.: 42 = 1*1 + 4*1 + 6*2 + 4*4 + 1*9. - Doron Zeilberger, Mar 12 2015
G.f. A(x) with offset 1 satisfies: A(x)^2 = A( x^2/(1-2*x) ). - Paul D. Hanna, Nov 08 2015
a(n) = GegenbauerPoly(n,-n-1,-1/2)/(n+1). - Emanuele Munarini, Oct 20 2016
a(n) = a(n-1) + A002026(n-1). Number of Motzkin paths that start with an F step plus number of Motzkin paths that start with an U step. - R. J. Mathar, Jul 25 2017
G.f. A(x) satisfies A(x)*A(-x) = F(x^2), where F(x) is the g.f. of A168592. - Alexander Burstein, Oct 04 2017
G.f.: A(x) = exp(int((E(x)-1)/x dx)), where E(x) is the g.f. of A002426. Equivalently, E(x) = 1 + x*A'(x)/A(x). - Alexander Burstein, Oct 05 2017
G.f. A(x) satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^k. - Ilya Gutkovskiy, Apr 11 2019
From Gennady Eremin, May 08 2021: (Start)
G.f.: 2/(1 - x + sqrt(1-2*x-3*x^2)).
a(n) = A107587(n) + A343386(n) = 2*A107587(n) - A343773(n) = 2*A343386(n) + A343773(n). (End)
Revert transform of A049347 (after Michael Somos). - Gennady Eremin, Jun 11 2021
Sum_{n>=0} 1/a(n) = 2.941237337631025604300320152921013604885956025483079699366681494505960039781389... - Vaclav Kotesovec, Jun 17 2021
Let a(-1) = (1 - sqrt(-3))/2 and a(n) = a(-3-n)*(-3)^(n+3/2) for all n in Z. Then a(n) satisfies my previous formula relation from Mar 23 2012 now for all n in Z. - Michael Somos, Apr 17 2022
Let b(n) = 1 for n <= 1, otherwise b(n) = Sum_{k=2..n} b(k-1) * b(n-k), then a(n) = b(n+1) (conjecture). - Joerg Arndt, Jan 16 2023
From Peter Bala, Feb 03 2024: (Start)
G.f.: A(x) = 1/(1 + x)*c(x/(1 + x))^2 = 1 + x/(1 + x)*c(x/(1 + x))^3, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
A(x) = 1/(1 - 3*x)*c(-x/(1 -3*x))^2.
a(n+1) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n, k)*A000245(k+1).
a(n) = 3^n * Sum_{k = 0..n} (-3)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 3^n * hypergeom([3/2, -n], [3], 4/3). (End)
G.f. A(x) satisfies A(x) = exp( x*A(x) + Integral x*A(x)/(1 - x^2*A(x)) dx ). - Paul D. Hanna, Mar 04 2024
a(n) = hypergeom([-n/2,1/2-n/2],[2],4). - Karol A. Penson, May 18 2025

A001700 a(n) = binomial(2*n+1, n+1): number of ways to put n+1 indistinguishable balls into n+1 distinguishable boxes = number of (n+1)-st degree monomials in n+1 variables = number of monotone maps from 1..n+1 to 1..n+1.

Original entry on oeis.org

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Keywords

Comments

To show for example that C(2n+1, n+1) is the number of monotone maps from 1..n + 1 to 1..n + 1, notice that we can describe such a map by a nondecreasing sequence of length n + 1 with entries from 1 to n + 1. The number k of increases in this sequence is anywhere from 0 to n. We can specify these increases by throwing k balls into n+1 boxes, so the total is Sum_{k = 0..n} C((n+1) + k - 1, k) = C(2*n+1, n+1).
Also number of ordered partitions (or compositions) of n + 1 into n + 1 parts. E.g., a(2) = 10: 003, 030, 300, 012, 021, 102, 120, 210, 201, 111. - Mambetov Bektur (bektur1987(AT)mail.ru), Apr 17 2003
Also number of walks of length n on square lattice, starting at origin, staying in first and second quadrants. - David W. Wilson, May 05 2001. (E.g., for n = 2 there are 10 walks, all starting at 0, 0: 0, 1 -> 0, 0; 0, 1 -> 1, 1; 0, 1 -> 0, 2; 1, 0 -> 0, 0; 1, 0 -> 1, 1; 1, 0 -> 2, 0; 1, 0 -> 1, -1; -1, 0 -> 0, 0; -1, 0 -> -1, 1; -1, 0-> -2, 0.)
Also total number of leaves in all ordered trees with n + 1 edges.
Also number of digitally balanced numbers [A031443] from 2^(2*n+1) to 2^(2*n+2). - Naohiro Nomoto, Apr 07 2001
Also number of ordered trees with 2*n + 2 edges having root of even degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of paths of length 2*d(G) connecting two neighboring nodes in optimal chordal graph of degree 4, G(2*d(G)^2 + 2*d(G) + 1, 2d(G) + 1), where d(G) = diameter of graph G. - S. Bujnowski (slawb(AT)atr.bydgoszcz.pl), Feb 11 2002
Define an array by m(1, j) = 1, m(i, 1) = i, m(i, j) = m(i, j-1) + m(i-1, j); then a(n) = m(n, n), diagonal of A165257 - Benoit Cloitre, May 07 2002
Also the numerator of the constant term in the expansion of cos^(2*n)(x) or sin^(2*n)(x) when the denominator is 2^(2*n-1). - Robert G. Wilson v
Consider the expansion of cos^n(x) as a linear combination of cosines of multiple angles. If n is odd, then the expansion is a combination of a*cos((2*k-1)*x)/2^(n-1) for all 2*k - 1 <= n. If n is even, then the expansion is a combination of a*cos(2k*x)/2^(n-1) terms plus a constant. "The constant term, [a(n)/2^(2n-1)], is due to the fact that [cos^2n(x)] is never negative, i.e., electrical engineers would say the average or 'dc value' of [cos^(2*n)(x)] is [a(n)/2^(2*n-1)]. The dc value of [cos^(2*n-1)(x)] on the other hand, is zero because it is symmetrical about the horizontal axis, i.e., it is negative and positive equally." Nahin[62] - Robert G. Wilson v, Aug 01 2002
Also number of times a fixed Dyck word of length 2*k occurs in all Dyck words of length 2*n + 2*k. Example: if the fixed Dyck word is xyxy (k = 2), then it occurs a(1) = 3 times in the 5 Dyck words of length 6 (n = 1): (xy[xy)xy], xyxxyy, xxyyxy, x(xyxy)y, xxxyyy (placed between parentheses). - Emeric Deutsch, Jan 02 2003
a(n+1) is the determinant of the n X n matrix m(i, j) = binomial(2*n-i, j). - Benoit Cloitre, Aug 26 2003
a(n-1) = (2*n)!/(2*n!*n!), formula in [Davenport] used by Gauss for the special case prime p = 4*n + 1: x = a(n-1) mod p and y = x*(2n)! mod p are solutions of p = x^2 + y^2. - Frank Ellermann. Example: For prime 29 = 4*7 + 1 use a(7-1) = 1716 = (2*7)!/(2*7!*7!), 5 = 1716 mod 29 and 2 = 5*(2*7)! mod 29, then 29 = 5*5 + 2*2.
The number of compositions of 2*n, say c_1 + c_2 + ... + c_k = 2n, satisfy that Sum_{i = 1..j} c_i < 2*j for all j = 1..k, or equivalently, the number of subsets, say S, of [2*n-1] = {1, 2, ..., 2*n-1} with at least n elements such that if 2k is in S, then there must be at least k elements in S smaller than 2k. E.g., a(2) = 3 because we can write 4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1. - Ricky X. F. Chen (ricky_chen(AT)mail.nankai.edu.cn), Jul 30 2006
The number of walks of length 2*n + 1 on an infinite linear lattice that begin at the origin and end at node (1). Also the number of paths on a square lattice from the origin to (n+1, n) that use steps (1,0) and (0,1). Also number of binary numbers of length 2*n + 1 with n + 1 ones and n zeros. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
If Y is a 3-subset of an 2*n-set X then, for n >= 3, a(n-1) is the number of n-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Also the number of rankings (preferential arrangements) of n unlabeled elements onto n levels when empty levels are allowed. - Thomas Wieder, May 24 2008
Also the Catalan transform of A000225 shifted one index, i.e., dropping A000225(0). - R. J. Mathar, Nov 11 2008
With offset 1. The number of solutions in nonnegative integers to X1 + X2 + ... + Xn = n. The number of terms in the expansion of (X1 + X2 + ... + Xn)^n. The coefficient of x^n in the expansion of (1 + x + x^2 + ...)^n. The number of distinct image sets of all functions taking [n] into [n]. - Geoffrey Critzer, Feb 22 2009
The Hankel transform of the aerated sequence 1, 0, 3, 0, 10, 0, ... is 1, 3, 3, 5, 5, 7, 7, ... (A109613(n+1)). - Paul Barry, Apr 21 2009
Also the number of distinct network topologies for a network of n items with 1 to n - 1 unidirectional connections to other objects in the network. - Anthony Bachler, May 05 2010
Equals INVERT transform of the Catalan numbers starting with offset 1. E.g.: a(3) = 35 = (1, 2, 5) dot (10, 3, 1) + 14 = 21 + 14 = 35. - Gary W. Adamson, May 15 2009
The integral of 1/(1+x^2)^(n+1) is given by a(n)/2^(2*n - 1) * (x/(1 + x^2)^n*P(x) + arctan(x)), where P(x) is a monic polynomial of degree 2*n - 2 with rational coefficients. - Christiaan van de Woestijne, Jan 25 2011
a(n) is the number of Schroder paths of semilength n in which the (2,0)-steps at level 0 come in 2 colors and there are no (2,0)-steps at a higher level. Example: a(2) = 10 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 2^2 = 4 paths of shape HH, 2 paths of shape HUD, 2 paths of shape UDH, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
a(n) is the number of Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=35 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 3^3 = 27 paths of shape HHH, 3 paths of shape HUD, 3 paths of shape UDH, and 2 paths of shape UHD. - Emeric Deutsch, May 02 2011
Also number of digitally balanced numbers having length 2*(n + 1) in binary representation: a(n) = #{m: A070939(A031443(m)) = 2*(n + 1)}. - Reinhard Zumkeller, Jun 08 2011
a(n) equals 2^(2*n + 3) times the coefficient of Pi in 2F1([1/2, n+2]; [3/2]; -1). - John M. Campbell, Jul 17 2011
For positive n, a(n) equals 4^(n+2) times the coefficient of Pi^2 in Integral_{x = 0..Pi/2} x sin^(2*n + 2)x. - John M. Campbell, Jul 19 2011 [Apparently, the contributor means Integral_{x = 0..Pi/2} x * (sin(x))^(2*n + 2).]
a(n-1) = C(2*n, n)/2 is the number of ways to assign 2*n people into 2 (unlabeled) groups of size n. - Dennis P. Walsh, Nov 09 2011
Equals row sums of triangle A205945. - Gary W. Adamson, Feb 01 2012
a(n-1) gives the number of n-regular sequences defined by Erdős and Gallai in 1960 in connection with the degree sequences of simple graphs. - Matuszka Tamás, Mar 06 2013
a(n) is the sum of falling diagonals of squares in the comment in A085812 (equivalent to the Cloitre formula of Aug 2002). - John Molokach, Sep 26 2013
For n > 0: largest terms of Zigzag matrices as defined in A088961. - Reinhard Zumkeller, Oct 25 2013
Also the number of different possible win/loss round sequences (from the perspective of the eventual winner) in a "best of 2*n + 1" two-player game. For example, a(2) = 10 means there are 10 different win/loss sequences in a "best of 5" game (like a tennis match in which the first player to win 3 sets, out of a maximum of 5, wins the match); the 10 sequences are WWW, WWLW, WWLLW, WLWW, WLWLW, WLLWW, LWWW, LWWLW, LWLWW, LLWWW. See also A072600. - Philippe Beaudoin, May 14 2014; corrected by Jon E. Schoenfield, Nov 23 2014
When adding 1 to the beginning of the sequence: Convolving a(n)/2^n with itself equals 2^(n+1). For example, when n = 4: convolving {1, 1/1, 3/2, 10/4, 35/8, 126/16} with itself is 32 = 2^5. - Bob Selcoe, Jul 16 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to a family of arrays associated to the Catalan A000108 (t = 1), and Riordan, or Motzkin sums A005043 (t = 0), with the o.g.f. [1 - sqrt(1 - 4x/(1 + (1 - t)x))]/2 and inverse x*(1 - x)/[1 + (t - 1)*x*(1 - x)]. See A091867 for more info on this family. Here is t = -3 (mod signs in the results).
Let C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1 + t*x) with inverse P(x, -t).
O.g.f: G(x) = [-1 + sqrt(1 + 4*x/(1 - 4*x))]/2 = -C[P(-x, 4)].
Inverse o.g.f: Ginv(x) = x*(1 + x)/(1 + 4*x*(1 + x)) = -P(Cinv(-x), -4) (shifted signed A001792). A088218(x) = 1 + G(x).
Equals A001813/2 omitting the leading 1 there. (End)
Placing n distinguishable balls into n indistinguishable boxes gives A000110(n) (the number of set partitions). - N. J. A. Sloane, Jun 19 2015
The sequence is the INVERTi transform of A049027: (1, 4, 17, 74, 326, ...). - Gary W. Adamson, Jun 23 2015
a(n) is the number of compositions of 2*n + 2 such that the sum of the elements at odd positions is equal to the sum of the elements at even positions. a(2) = 10 because there are 10 such compositions of 6: (3, 3), (1, 3, 2), (2, 3, 1), (1, 1, 2, 2), (1, 2, 2, 1), (2, 2, 1, 1), (2, 1, 1, 2), (1, 2, 1, 1, 1), (1, 1, 1, 2, 1), (1, 1, 1, 1, 1, 1). - Ran Pan, Oct 08 2015
a(n-1) is also the Schur function of the partition (n) of n evaluated at x_1 = x_2 = ... = x_n = 1, i.e., the number of semistandard Young tableaux of shape (n) (weakly increasing rows with n boxes with numbers from {1, 2, ..., n}). - Wolfdieter Lang, Oct 11 2015
Also the number of ordered (rooted planar) forests with a total of n+1 edges and no trivial trees. - Nachum Dershowitz, Mar 30 2016
a(n) is the number of sets (i1,...in) of length n so that n >= i1 >= i2 >= ...>= in >= 1. For instance, n=3 as there are only 10 such sets (3,3,3) (3,3,2) (3,3,1) (3,2,2) (3,2,1) (3,1,1) (2,2,2) (2,2,1) (2,1,1) (1,1,1,) 3,2,1 is each used 10 times respectively. - Anton Zakharov, Jul 04 2016
The repeated middle term in the odd rows of Pascal's triangle, or half the central binomial coefficient in the even rows of Pascal's triangle, n >= 2. - Enrique Navarrete, Feb 12 2018
a(n) is the number of walks of length 2n+1 from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n+1 from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Total number of nodes summed over all Dyck paths of semilength n. - Alois P. Heinz, Mar 08 2020
a(n-1) is the determinant of the n X n matrix m(i, j) = binomial(n+i-1, j). - Fabio Visonà, May 21 2022
Let X_i be iid standard Gaussian random variable N(0,1), and S_n be the partial sum S_n = X_1+...+X_n. Then P(S_1>0,S_2>0,...,S_n>0) = a(n+1)/2^(2n-1) = a(n+1) / A004171(n+1). For example, P(S_1>0) = 1/2, P(S_1>0,S_2>0) = 3/8, P(S_1>0,S_2>0,S_3>0) = 5/16, etc. This probability is also equal to the volume of the region x_1 > 0, x_2 > -x_1, x_3 > -(x_1+x_2), ..., x_n > -(x_1+x_2+...+x_(n-1)) in the hypercube [-1/2, 1/2]^n. This also holds for the Cauchy distribution and other stable distributions with mean 0, skew 0 and scale 1. - Xiaohan Zhang, Nov 01 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 132, 213, and 321. - Lara Pudwell, Apr 10 2023
Number of vectors in (Z_>=0)^(n+1) such that the sum of the components is n+1. binomial(2*n-1, n) provides this property for n. - Michael Richard, Jun 12 2023
Also number of discrete negations on the finite chain L_n={0,1,...,n-1,n}, i.e., monotone decreasing unary operators such that N(0)=n and N(n)=0. - Marc Munar, Oct 10 2023
a(n) is the number of Dyck paths of semilength n+1 having one of its peaks marked. - Juan B. Gil, Jan 03 2024
a(n) is the dimension of the (n+1)-st symmetric power of an (n+1)-dimensional vector space. - Mehmet A. Ates, Feb 15 2024
a(n) is the independence number of the twisted odd graph O^(sigma)(n+2). - _Miquel A. Fiol, Aug 26 2024
a(n) is the number of non-descending sequences with length n and the last number is less or equal to n. a(n) is also the number of integer partitions (of any positive integer) with length n and largest part is less or equal to n. - Zlatko Damijanic, Dec 06 2024
a(n) is the number of triangulations of a once-punctured (n+1)-gon [from Fontaine & Plamondon's Theorem 3.6]. - Esther Banaian, May 06 2025

Examples

			There are a(2)=10 ways to put 3 indistinguishable balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
a(2) = 10: Semistandard Young tableaux for partition (3) of 3 (the indeterminates x_i, i = 1, 2, 3 are omitted and only their indices are given): 111, 112, 113, 122, 123, 133, 222, 223, 233, 333. - _Wolfdieter Lang_, Oct 11 2015
		

References

  • H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th ed., 1999, ch. V.3 (p. 122).
  • A. Frosini, R. Pinzani, and S. Rinaldi, About half the middle binomial coefficient, Pure Math. Appl., 11 (2000), 497-508.
  • Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • Paul J. Nahin, "An Imaginary Tale, The Story of [Sqrt(-1)]," Princeton University Press, Princeton, NJ 1998, p. 62.
  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equals A000984(n+1)/2.
a(n) = (2*n+1)*Catalan(n) [A000108] = A035324(n+1, 1) (first column of triangle).
Row sums of triangles A028364, A050166, A039598.
Bisections: a(2*k) = A002458(k), a(2*k+1) = A001448(k+1)/2, k >= 0.
Other versions of the same sequence: A088218, A110556, A138364.
Diagonals 1 and 2 of triangle A100257.
Second row of array A102539.
Column of array A073165.
Row sums of A103371. - Susanne Wienand, Oct 22 2011
Cf. A002054: C(2*n+1, n-1). - Bruno Berselli, Jan 20 2014

Programs

  • GAP
    List([0..30],n->Binomial(2*n+1,n+1)); # Muniru A Asiru, Feb 26 2019
  • Haskell
    a001700 n = a007318 (2*n+1) (n+1)  -- Reinhard Zumkeller, Oct 25 2013
    
  • Magma
    [Binomial(2*n, n)/2: n in [1..40]]; // Vincenzo Librandi, Nov 10 2014
    
  • Maple
    A001700 := n -> binomial(2*n+1,n+1); seq(A001700(n), n=0..20);
    A001700List := proc(m) local A, P, n; A := [1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
    A := [op(A), P[-1]] od; A end: A001700List(27); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[ Binomial[2n + 1, n + 1], {n, 0, 23}]
    CoefficientList[ Series[2/((Sqrt[1 - 4 x] + 1)*Sqrt[1 - 4 x]), {x, 0, 22}], x] (* Robert G. Wilson v, Aug 08 2011 *)
  • Maxima
    B(n,a,x):=coeff(taylor(exp(x*t)*(t/(exp(t)-1))^a,t,0,20),t,n)*n!;
    makelist((-1)^(n)*B(n,n+1,-n-1)/n!,n,0,10); /* Vladimir Kruchinin, Apr 06 2016 */
    
  • PARI
    a(n)=binomial(2*n+1,n+1)
    
  • PARI
    z='z+O('z^50); Vec((1/sqrt(1-4*z)-1)/(2*z)) \\ Altug Alkan, Oct 11 2015
    
  • Python
    from _future_ import division
    A001700_list, b = [], 1
    for n in range(10**3):
        A001700_list.append(b)
        b = b*(4*n+6)//(n+2) # Chai Wah Wu, Jan 26 2016
    
  • Sage
    [rising_factorial(n+1,n+1)/factorial(n+1) for n in (0..22)] # Peter Luschny, Nov 07 2011
    

Formula

a(n-1) = binomial(2*n, n)/2 = A000984(n)/2 = (2*n)!/(2*n!*n!).
D-finite with recurrence: a(0) = 1, a(n) = 2*(2*n+1)*a(n-1)/(n+1) for n > 0.
G.f.: (1/sqrt(1 - 4*x) - 1)/(2*x).
L.g.f.: log((1 - sqrt(1 - 4*x))/(2*x)) = Sum_{n >= 0} a(n)*x^(n+1)/(n+1). - Vladimir Kruchinin, Aug 10 2010
G.f.: 2F1([1, 3/2]; [2]; 4*x). - Paul Barry, Jan 23 2009
G.f.: 1/(1 - 2*x - x/(1 - x/(1 - x/(1 - x/(1 - ... (continued fraction). - Paul Barry, May 06 2009
G.f.: c(x)^2/(1 - x*c(x)^2), c(x) the g.f. of A000108. - Paul Barry, Sep 07 2009
O.g.f.: c(x)/sqrt(1 - 4*x) = (2 - c(x))/(1 - 4*x), with c(x) the o.g.f. of A000108. Added second formula. - Wolfdieter Lang, Sep 02 2012
Convolution of A000108 (Catalan) and A000984 (central binomial): Sum_{k=0..n} C(k)*binomial(2*(n-k), n-k), C(k) Catalan. - Wolfdieter Lang, Dec 11 1999
a(n) = Sum_{k=0..n} C(n+k, k). - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n+1, k+1). - Benoit Cloitre, Oct 19 2002
a(n) = Sum_{k = 0..n+1} binomial(2*n+2, k)*cos((n - k + 1)*Pi). - Paul Barry, Nov 02 2004
a(n) = 4^n*binomial(n+1/2, n)/(n+1). - Paul Barry, May 10 2005
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n + 1)/(2*n + 1)! = BesselI(1, 2*x). - Michael Somos, Jun 22 2005
E.g.f. in Maple notation: exp(2*x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). Integral representation as n-th moment of a positive function on [0, 4]: a(n) = Integral_{x = 0..4} x^n * (x/(4 - x))^(1/2)/(2*Pi) dx, n >= 0. This representation is unique. - Karol A. Penson, Oct 11 2001
Narayana transform of [1, 2, 3, ...]. Let M = the Narayana triangle of A001263 as an infinite lower triangular matrix and V = the Vector [1, 2, 3, ...]. Then A001700 = M * V. - Gary W. Adamson, Apr 25 2006
a(n) = A122366(n,n). - Reinhard Zumkeller, Aug 30 2006
a(n) = C(2*n, n) + C(2*n, n-1) = A000984(n) + A001791(n). - Zerinvary Lajos, Jan 23 2007
a(n-1) = (n+1)*(n+2)*...*(2*n-1)/(n-1)! (product of n-1 consecutive integers, divided by (n-1)!). - Jonathan Vos Post, Apr 09 2007; [Corrected and shortened by Giovanni Ciriani, Mar 26 2019]
a(n-1) = (2*n - 1)!/(n!*(n - 1)!). - William A. Tedeschi, Feb 27 2008
a(n) = (2*n + 1)*A000108(n). - Paul Barry, Aug 21 2007
Binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double binomial transform of A001405. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A132813. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A134285. - Gary W. Adamson, Nov 19 2007
a(n) = 2*A000984(n) - A000108(n), that is, a(n) = 2*C(2*n, n) - n-th Catalan number. - Joseph Abate, Jun 11 2010
Conjectured: 4^n GaussHypergeometric(1/2,-n; 2; 1) -- Solution for the path which stays in the first and second quadrant. - Benjamin Phillabaum, Feb 20 2011
a(n)= Sum_{k=0..n} A038231(n,k) * (-1)^k * A000108(k). - Philippe Deléham, Nov 27 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1] = -1, A[i,j] = Catalan(j-i), (i <= j), and A[i,j] = 0, otherwise. Then, for n >= 1, a(n) = (-1)^n * charpoly(A,-2). - Milan Janjic, Jul 08 2010
a(n) is the upper left term of M^(n+1), where M is the infinite matrix in which a column of (1,2,3,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros, as follows:
1, 1, 0, 0, 0, ...
2, 1, 1, 0, 0, ...
3, 1, 1, 1, 0, ...
4, 1, 1, 1, 1, ...
...
Alternatively, a(n) is the upper left term of M^n where M is the infinite matrix:
3, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
...
- Gary W. Adamson, Jul 14 2011
a(n) = (n + 1)*hypergeom([-n, -n], [2], 1). - Peter Luschny, Oct 24 2011
a(n) = Pochhammer(n+1, n+1)/(n+1)!. - Peter Luschny, Nov 07 2011
E.g.f.: 1 + 6*x/(U(0) - 6*x); U(k) = k^2 + (4*x + 3)*k + 6*x + 2 - 2*x*(k + 1)*(k + 2)*(2*k + 5)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
a(n) = 2*A000984(n) - A000108(n). [Abate & Whitt]
a(n) = 2^(2*n+1)*binomial(n+1/2, -1/2). - Peter Luschny, May 06 2014
For n > 1: a(n-1) = A166454(2*n, n), central terms in A166454. - Reinhard Zumkeller, Mar 04 2015
a(n) = 2*4^n*Gamma(3/2 + n)/(sqrt(Pi)*Gamma(2+n)). - Peter Luschny, Dec 14 2015
a(n) ~ 2*4^n*(1 - (5/8)/n + (73/128)/n^2 - (575/1024)/n^3 + (18459/32768)/n^4)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
a(n) = (-1)^(n)*B(n, n+1, -n-1)/n!, where B(n,a,x) is a generalized Bernoulli polynomial. - Vladimir Kruchinin, Apr 06 2016
a(n) = Gamma(2 + 2*n)/(n!*Gamma(2 + n)). Andres Cicuttin, Apr 06 2016
a(n) = (n + (n + 1))!/(Gamma(n)*Gamma(1 + n)*A002378(n)), for n > 0. Andres Cicuttin, Apr 07 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
Sum_{n >= 0} 1/a(n) = 2*(9 + 2*sqrt(3)*Pi)/27 = A248179.
Sum_{n >= 0} (-1)^n/a(n) = 2*(5 + 4*sqrt(5)*arcsinh(1/2))/25 = 2*(5*A145433 - 1).
Sum_{n >= 0} (-1)^n*a(n)/n! = BesselI(2,2)*exp(-2) = A229020*A092553. (End)
Conjecture: a(n) = Sum_{k=2^n..2^(n+1)-1} A178244(k). - Mikhail Kurkov, Feb 20 2021
a(n-1) = 1 + (1/n)*Sum_{t=1..n/2} (2*cos((2*t-1)*Pi/(2*n)))^(2*n). - Greg Dresden, Oct 11 2022
a(n) = Product_{1 <= i <= j <= n} (i + j + 1)/(i + j - 1). Cf. A006013. - Peter Bala, Feb 21 2023
Sum_{n >= 0} a(n)*x^(n+1)/(n+1) = x + 3*x^2/2 + 10*x^3/3 + 35*x^4/4 + ... = the series reversion of exp(-x)*(1 - exp(-x)). - Peter Bala, Sep 06 2023

Extensions

Name corrected by Paul S. Coombes, Jan 11 2012
Name corrected by Robert Tanniru, Feb 01 2014

A056040 Swinging factorial, a(n) = 2^(n-(n mod 2))*Product_{k=1..n} k^((-1)^(k+1)).

Original entry on oeis.org

1, 1, 2, 6, 6, 30, 20, 140, 70, 630, 252, 2772, 924, 12012, 3432, 51480, 12870, 218790, 48620, 923780, 184756, 3879876, 705432, 16224936, 2704156, 67603900, 10400600, 280816200, 40116600, 1163381400, 155117520, 4808643120, 601080390, 19835652870, 2333606220
Offset: 0

Author

Labos Elemer, Jul 25 2000

Keywords

Comments

a(n) is the number of 'swinging orbitals' which are enumerated by the trinomial n over [floor(n/2), n mod 2, floor(n/2)].
Similar to but different from A001405(n) = binomial(n, floor(n/2)), a(n) = lcm(A001405(n-1), A001405(n)) (for n>0).
A055773(n) divides a(n), A001316(floor(n/2)) divides a(n).
Exactly p consecutive multiples of p follow the least positive multiple of p if p is an odd prime. Compare with the similar property of A100071. - Peter Luschny, Aug 27 2012
a(n) is the number of vertices of the polytope resulting from the intersection of an n-hypercube with the hyperplane perpendicular to and bisecting one of its long diagonals. - Didier Guillet, Jun 11 2018 [Edited by Peter Munn, Dec 06 2022]

Examples

			a(10) = 10!/5!^2 = trinomial(10,[5,0,5]);
a(11) = 11!/5!^2 = trinomial(11,[5,1,5]).
		

Crossrefs

Programs

  • Magma
    [(Factorial(n)/(Factorial(Floor(n/2)))^2): n in [0..40]]; // Vincenzo Librandi, Sep 11 2011
    
  • Maple
    SeriesCoeff := proc(s,n) series(s(w,n),w,n+2);
    convert(%,polynom); coeff(%,w,n) end;
    a1 := proc(n) local k;
    2^(n-(n mod 2))*mul(k^((-1)^(k+1)),k=1..n) end:
    a2 := proc(n) option remember;
    `if`(n=0,1,n^irem(n,2)*(4/n)^irem(n+1,2)*a2(n-1)) end;
    a3 := n -> n!/iquo(n,2)!^2;
    g4 := z -> BesselI(0,2*z)*(1+z);
    a4 := n -> n!*SeriesCoeff(g4,n);
    g5 := z -> (1+z/(1-4*z^2))/sqrt(1-4*z^2);
    a5 := n -> SeriesCoeff(g5,n);
    g6 := (z,n) -> (1+z^2)^n+n*z*(1+z^2)^(n-1);
    a6 := n -> SeriesCoeff(g6,n);
    a7 := n -> combinat[multinomial](n,floor(n/2),n mod 2,floor(n/2));
    h := n -> binomial(n,floor(n/2)); # A001405
    a8 := n -> ilcm(h(n-1),h(n));
    F := [a1, a2, a3, a4, a5, a6, a7, a8];
    for a in F do seq(a(i), i=0..32) od;
  • Mathematica
    f[n_] := 2^(n - Mod[n, 2])*Product[k^((-1)^(k + 1)), {k, n}]; Array[f, 33, 0] (* Robert G. Wilson v, Aug 02 2010 *)
    f[n_] := If[OddQ@n, n*Binomial[n - 1, (n - 1)/2], Binomial[n, n/2]]; Array[f, 33, 0] (* Robert G. Wilson v, Aug 10 2010 *)
    sf[n_] := With[{f = Floor[n/2]}, Pochhammer[f+1, n-f]/f!]; (* or, twice faster: *) sf[n_] := n!/Quotient[n, 2]!^2; Table[sf[n], {n, 0, 32}] (* Jean-François Alcover, Jul 26 2013, updated Feb 11 2015 *)
  • PARI
    a(n)=n!/(n\2)!^2 \\ Charles R Greathouse IV, May 02 2011
    
  • Sage
    def A056040():
        r, n = 1, 0
        while True:
            yield r
            n += 1
            r *= 4/n if is_even(n) else n
    a = A056040(); [next(a) for i in range(36)]  # Peter Luschny, Oct 24 2013

Formula

a(n) = n!/floor(n/2)!^2. [Essentially the original name.]
a(0) = 1, a(n) = n^(n mod 2)*(4/n)^(n+1 mod 2)*a(n-1) for n>=1.
E.g.f.: (1+x)*BesselI(0, 2*x). - Vladeta Jovovic, Jan 19 2004
O.g.f.: a(n) = SeriesCoeff_{n}((1+z/(1-4*z^2))/sqrt(1-4*z^2)).
P.g.f.: a(n) = PolyCoeff_{n}((1+z^2)^n+n*z*(1+z^2)^(n-1)).
a(2n+1) = A046212(2n+1) = A100071(2n+1). - M. F. Hasler, Jan 25 2012
a(2*n) = binomial(2*n,n); a(2*n+1) = (2*n+1)*binomial(2*n,n). Central terms of triangle A211226. - Peter Bala, Apr 10 2012
D-finite with recurrence: n*a(n) + (n-2)*a(n-1) + 4*(-2*n+3)*a(n-2) + 4*(-n+1)*a(n-3) + 16*(n-3)*a(n-4) = 0. - Alexander R. Povolotsky, Aug 17 2012
Sum_{n>=0} 1/a(n) = 4/3 + 8*Pi/(9*sqrt(3)). - Alexander R. Povolotsky, Aug 18 2012
E.g.f.: U(0) where U(k)= 1 + x/(1 - x/(x + (k+1)*(k+1)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
Central column of the coefficients of the swinging polynomials A162246. - Peter Luschny, Oct 22 2013
a(n) = Sum_{k=0..n} A189231(n, 2*k). (Cf. A212303 for the odd case.) - Peter Luschny, Oct 30 2013
a(n) = hypergeometric([-n,-n-1,1/2],[-n-2,1],2)*2^(n-1)*(n+2). - Peter Luschny, Sep 22 2014
a(n) = 4^floor(n/2)*hypergeometric([-floor(n/2), (-1)^n/2], [1], 1). - Peter Luschny, May 19 2015
Sum_{n>=0} (-1)^n/a(n) = 4/3 - 4*Pi/(9*sqrt(3)). - Amiram Eldar, Mar 10 2022

Extensions

Extended and edited by Peter Luschny, Jun 28 2009
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