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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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

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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

A005773 Number of directed animals of size n (or directed n-ominoes in standard position).

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 96, 267, 750, 2123, 6046, 17303, 49721, 143365, 414584, 1201917, 3492117, 10165779, 29643870, 86574831, 253188111, 741365049, 2173243128, 6377181825, 18730782252, 55062586341, 161995031226, 476941691177, 1405155255055, 4142457992363
Offset: 0

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Keywords

Comments

This sequence, with first term a(0) deleted, appears to be determined by the conditions that the diagonal and first superdiagonal of U are {1,1,1,1,...} and {2,3,4,5,...,n+1,...} respectively, where A=LU is the LU factorization of the Hankel matrix A given by [{a(1),a(2),...}, {a(2),a(3),...}, ..., {a(n),a(n+1),...}, ...]. - John W. Layman, Jul 21 2000
Also the number of base 3 n-digit numbers (not starting with 0) with digit sum n. For the analogous sequence in base 10 see A071976, see example. - John W. Layman, Jun 22 2002
Also number of paths in an n X n grid from (0,0) to the line x=n-1, using only steps U=(1,1), H=(1,0) and D=(1,-1) (i.e., left factors of length n-1 of Motzkin paths, palindromic Motzkin paths of length 2n-2 or 2n-1). Example: a(3)=5, namely, HH, UD, HU, UH and UU. Also number of ordered trees with n edges and having nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 01 2002
Number of symmetric Dyck paths of semilength 2n-1 with no peaks at even level. Example: a(3)=5 because we have UDUDUDUDUD, UDUUUDDDUD, UUUUUDDDDD, UUUDUDUDDD and UUUDDUUDDD, where U=(1,1) and D=(1,-1). Also number of symmetric Dyck paths of semilength 2n with no peaks at even level. Example: a(3)=5 because we have UDUDUDUDUDUD, UDUUUDUDDDUD, UUUDUDUDUDDD, UUUUUDUDDDDD and UUUDDDUUUDDD. - Emeric Deutsch, Nov 21 2003
a(n) is the sum of the (n-1)-st central trinomial coefficient and its predecessor. Example: a(4) = 6 + 7 and (1 + x + x^2)^3 = ... + 6*x^2 + 7*x^3 + ... . - David Callan, Feb 07 2004
a(n) is the number of UDU-free paths of n upsteps (U) and n downsteps (D) that start U (n>=1). Example: a(2)=2 counts UUDD, UDDU. - David Callan, Aug 18 2004
a(n) is also the number of Grand-Dyck paths of semilength n starting with an up-step and avoiding the pattern DUD. - David Bevan, Nov 19 2019
Hankel transform of a(n+1) = [1,2,5,13,35,96,...] gives A000012 = [1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
Equals row sums of triangle A136787 starting (1, 2, 5, 13, 35, ...). - Gary W. Adamson, Jan 21 2008
a(n) is the number of permutations on [n] that avoid the patterns 1-23-4 and 1-3-2, where the omission of a dash in a pattern means the permutation entries must be adjacent. Example: a(4) = 13 counts all 14 (Catalan number) (1-3-2)-avoiding permutations on [4] except 1234. - David Callan, Jul 22 2008
a(n) is also the number of involutions of length 2n-2 which are invariant under the reverse-complement map and have no decreasing subsequences of length 4. - Eric S. Egge, Oct 21 2008
Hankel transform is A010892. - Paul Barry, Jan 19 2009
a(n) is the number of Dyck words of semilength n with no DUUU. For example, a(4) = 14-1 = 13 because there is only one Dyck 4-word containing DUUU, namely UDUUUDDD. - Eric Rowland, Apr 21 2009
Inverse binomial transform of A024718. - Philippe Deléham, Dec 13 2009
Let w(i, j, n) denote walks in N^2 which satisfy the multivariate recurrence
w(i, j, n) = w(i - 1, j, n - 1) + w(i, j - 1, 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. Let alpha(n) the number of such walks of length n, alpha(n) = Sum_{i = 0..n, j=0..n} w(i, j, n). Then a(n+1) = alpha(n). - Peter Luschny, May 21 2011
Number of length-n strings [d(0),d(1),d(2),...,d(n-1)] where 0 <= d(k) <= k and abs(d(k) - d(k-1)) <= 1 (smooth factorial numbers, see example). - Joerg Arndt, Nov 10 2012
a(n) is the number of n-multisets of {1,...,n} containing no pair of consecutive integers (e.g., 111, 113, 133, 222, 333 for n=3). - David Bevan, Jun 10 2013
a(n) is also the number of n-multisets of [n] in which no integer except n occurs exactly once (e.g., 111, 113, 222, 223, 333 for n=3). - David Bevan, Nov 19 2019
Number of minimax elements in the affine Weyl group of the Lie algebra so(2n+1) or the Lie algebra sp(2n). See Panyushev 2005. Cf. A245455. - Peter Bala, Jul 22 2014
The shifted, signed array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating (here t=-2) o.g.f. G(x,t) = (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse o.g.f. Ginv(x,t) = x(1-x)/(1+(t-1)x(1-x)) (A057682). See A091867 for more info on this family. - Tom Copeland, Nov 09 2014
Alternatively, this sequence corresponds to the number of positive walks with n steps {-1,0,1} starting at the origin, ending at any altitude, and staying strictly above the x-axis. - David Nguyen, Dec 01 2016
Let N be a squarefree number with n prime factors: p_1 < p_2 < ... < p_n. Let D be its set of divisors, E the subset of D X D made of the (d_1, d_2) for which, provided that we know which p_i are in d_1, which p_i are in d_2, d_1 <= d_2 is provable without needing to know the numerical values of the p_i. It appears that a(n+1) is the number of (d_1, d_2) in E such that d_1 and d_2 are coprime. - Luc Rousseau, Aug 21 2017
Number of ordered rooted trees with n non-root nodes and all non-root nodes having outdegrees 1 or 2. - Andrew Howroyd, Dec 04 2017
a(n) is the number of compositions (ordered partitions) of n where there are A001006(k-1) sorts of part k (see formula by Andrew Howroyd, Dec 04 2017). - Joerg Arndt, Jan 26 2024

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + 96*x^6 + 267*x^7 + ...
a(3) = 5, a(4) = 13; since the top row of M^3 = (5, 5, 2, 1, ...)
From _Eric Rowland_, Sep 25 2021: (Start)
There are a(4) = 13 directed animals of size 4:
  O
  O    O    O    OO              O         O
  O    O    OO   O    OO   O    OO   OOO   O    O    OO    O
  O    OO   O    O    OO   OOO  O    O    OO   OOO  OO   OOO  OOOO
(End)
From _Joerg Arndt_, Nov 10 2012: (Start)
There are a(4)=13 smooth factorial numbers of length 4 (dots for zeros):
[ 1]   [ . . . . ]
[ 2]   [ . . . 1 ]
[ 3]   [ . . 1 . ]
[ 4]   [ . . 1 1 ]
[ 5]   [ . . 1 2 ]
[ 6]   [ . 1 . . ]
[ 7]   [ . 1 . 1 ]
[ 8]   [ . 1 1 . ]
[ 9]   [ . 1 1 1 ]
[10]   [ . 1 1 2 ]
[11]   [ . 1 2 1 ]
[12]   [ . 1 2 2 ]
[13]   [ . 1 2 3 ]
(End)
From _Joerg Arndt_, Nov 22 2012: (Start)
There are a(4)=13 base 3 4-digit numbers (not starting with 0) with digit sum 4:
[ 1]   [ 2 2 . . ]
[ 2]   [ 2 1 1 . ]
[ 3]   [ 1 2 1 . ]
[ 4]   [ 2 . 2 . ]
[ 5]   [ 1 1 2 . ]
[ 6]   [ 2 1 . 1 ]
[ 7]   [ 1 2 . 1 ]
[ 8]   [ 2 . 1 1 ]
[ 9]   [ 1 1 1 1 ]
[10]   [ 1 . 2 1 ]
[11]   [ 2 . . 2 ]
[12]   [ 1 1 . 2 ]
[13]   [ 1 . 1 2 ]
(End)
		

References

  • J. E. Goodman and J. O'Rourke, editors, Handbook of Discrete and Computational Geometry, CRC Press, 1997, p. 237.
  • T. Mansour, Combinatorics of Set Partitions, Discrete Mathematics and Its Applications, CRC Press, 2013, p. 377.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.46a.
  • R. P. Stanley, Catalan Numbers, Cambridge, 2015, p. 132.

Crossrefs

See also A005775. Inverse of A001006. Also sum of numbers in row n+1 of array T in A026300. Leading column of array in A038622.
The right edge of the triangle A062105.
Column k=3 of A295679.
Interpolates between Motzkin numbers (A001006) and Catalan numbers (A000108). Cf. A054391, A054392, A054393, A055898.
Except for the first term a(0), sequence is the binomial transform of A001405.
a(n) = A002426(n-1) + A005717(n-1) if n > 0. - Emeric Deutsch, Aug 14 2002

Programs

  • Haskell
    a005773 n = a005773_list !! n
    a005773_list = 1 : f a001006_list [] where
       f (x:xs) ys = y : f xs (y : ys) where
         y = x + sum (zipWith (*) a001006_list ys)
    -- Reinhard Zumkeller, Mar 30 2012
    
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 2*x/(3*x-1+Sqrt(1-2*x-3*x^2)) )); // G. C. Greubel, Apr 05 2019
  • Maple
    seq( sum(binomial(i-1, k)*binomial(i-k, k), k=0..floor(i/2)), i=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    A005773:=proc(n::integer)
    local i, j, A, istart, iend, KartProd, Liste, Term, delta;
        A:=0;
        for i from 0 to n do
            Liste[i]:=NULL;
            istart[i]:=0;
            iend[i]:=n-i+1:
            for j from istart[i] to iend[i] do
                Liste[i]:=Liste[i], j;
            end do;
            Liste[i]:=[Liste[i]]:
        end do;
        KartProd:=cartprod([seq(Liste[i], i=1..n)]);
        while not KartProd[finished] do
            Term:=KartProd[nextvalue]();
            delta:=1;
            for i from 1 to n-1 do
                if (op(i, Term) - op(i+1, Term))^2 >= 2 then
                    delta:=0;
                    break;
                end if;
            end do;
            A:=A+delta;
        end do;
    end proc; # Thomas Wieder, Feb 22 2009:
    # n -> [a(0),a(1),..,a(n)]
    A005773_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-1,j,n-1)+W(i,j-1,n-1)+W(i+1,j-1,n-1) fi end:
    [1,seq(add(add(W(i,j,m),i=0..m),j=0..m),m=0..n-1)] end:
    A005773_list(27); # Peter Luschny, May 21 2011
    A005773 := proc(n)
        option remember;
        if n <= 1 then
            1 ;
        else
            2*n*procname(n-1)+3*(n-2)*procname(n-2) ;
            %/n ;
        end if;
    end proc:
    seq(A005773(n),n=0..10) ; # R. J. Mathar, Jul 25 2017
  • Mathematica
    CoefficientList[Series[(2x)/(3x-1+Sqrt[1-2x-3x^2]), {x,0,40}], x] (* Harvey P. Dale, Apr 03 2011 *)
    a[0]=1; a[n_] := Sum[k/n*Sum[Binomial[n, j]*Binomial[j, 2*j-n-k], {j, 0, n}], {k, 1, n}]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 31 2015, after Vladimir Kruchinin *)
    A005773[n_] := 2 (-1)^(n+1) JacobiP[n - 1, 3, -n -1/2, -7] / (n^2 + n); A005773[0] := 1; Table[A005773[n], {n, 0, 27}] (* Peter Luschny, May 25 2021 *)
  • PARI
    a(n)=if(n<2,n>=0,(2*n*a(n-1)+3*(n-2)*a(n-2))/n)
    
  • PARI
    for(n=0, 27, print1(if(n==0, 1, sum(k=0, n-1, (-1)^(n - 1 + k)*binomial(n - 1, k)*binomial(2*k + 1, k + 1))),", ")) \\ Indranil Ghosh, Mar 14 2017
    
  • PARI
    Vec(1/(1-serreverse(x*(1-x)/(1-x^3) + O(x*x^25)))) \\ Andrew Howroyd, Dec 04 2017
    
  • Sage
    def da():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        yield 1
        while True:
            yield b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)//n
    A005773 = da()
    print([next(A005773) for  in range(28)]) # _Peter Luschny, May 16 2016
    
  • Sage
    (2*x/(3*x-1+sqrt(1-2*x-3*x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 05 2019
    

Formula

G.f.: 2*x/(3*x-1+sqrt(1-2*x-3*x^2)). - Len Smiley
Also a(0)=1, a(n) = Sum_{k=0..n-1} M(k)*a(n-k-1), where M(n) are the Motzkin numbers (A001006).
D-finite with recurrence n*a(n) = 2*n*a(n-1) + 3*(n-2)*a(n-2), a(0)=a(1)=1. - Michael Somos, Feb 02 2002
G.f.: 1/2+(1/2)*((1+x)/(1-3*x))^(1/2). Related to Motzkin numbers A001006 by a(n+1) = 3*a(n) - A001006(n-1) [see Yaqubi Lemma 2.6].
a(n) = Sum_{q=0..n} binomial(q, floor(q/2))*binomial(n-1, q) for n > 0. - Emeric Deutsch, Aug 15 2002
From Paul Barry, Jun 22 2004: (Start)
a(n+1) = Sum_{k=0..n} (-1)^(n+k)*C(n, k)*C(2*k+1, k+1).
a(n) = 0^n + Sum_{k=0..n-1} (-1)^(n+k-1)*C(n-1, k)*C(2*k+1, k+1). (End)
a(n+1) = Sum_{k=0..n} (-1)^k*3^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
Starting (1, 2, 5, 13, ...) gives binomial transform of A001405 and inverse binomial transform of A001700. - Gary W. Adamson, Aug 31 2007
Starting (1, 2, 5, 13, 35, 96, ...) gives row sums of triangle A132814. - Gary W. Adamson, Aug 31 2007
G.f.: 1/(1-x/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 19 2009
G.f.: 1+x/(1-2*x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-.... (continued fraction). - Paul Barry, Jan 19 2009
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))^2 >= 2 for i=1..n-1 and delta(l_1,l_2,..., l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
INVERT transform of offset Motzkin numbers (A001006): (a(n)){n>=1}=(1,1,2,4,9,21,...). - _David Callan, Aug 27 2009
A005773(n) = ((n+3)*A001006(n+1) + (n-3)*A001006(n)) * (n+2)/(18*n) for n > 0. - Mark van Hoeij, Jul 02 2010
a(n) = Sum_{k=1..n} (k/n * Sum_{j=0..n} binomial(n,j)*binomial(j,2*j-n-k)). - Vladimir Kruchinin, Sep 06 2010
a(0) = 1; a(n+1) = Sum_{t=0..n} n!/((n-t)!*ceiling(t/2)!*floor(t/2)!). - Andrew S. Hays, Feb 02 2011
a(n) = leftmost column term of M^n*V, where M = an infinite quadradiagonal matrix with all 1's in the main, super and subdiagonals, [1,0,0,0,...] in the diagonal starting at position (2,0); and rest zeros. V = vector [1,0,0,0,...]. - Gary W. Adamson, Jun 16 2011
From Gary W. Adamson, Jul 29 2011: (Start)
a(n) = upper left term of M^n, a(n+1) = sum of top row terms of M^n; M = an infinite square production matrix in which the main diagonal is (1,1,0,0,0,...) as follows:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 0, 1, 0, 0, ...
1, 1, 1, 0, 1, 0, ...
1, 1, 1, 1, 0, 1, ...
1, 1, 1, 1, 1, 0, ... (End)
Limit_{n->oo} a(n+1)/a(n) = 3.0 = lim_{n->oo} (1 + 2*cos(Pi/n)). - Gary W. Adamson, Feb 10 2012
a(n) = A025565(n+1) / 2 for n > 0. - Reinhard Zumkeller, Mar 30 2012
With first term deleted: E.g.f.: a(n) = n! * [x^n] exp(x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
G.f.: G(0)/2 + 1/2, where G(k) = 1 + 2*x*(4*k+1)/( (2*k+1)*(1+x) - x*(1+x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1+x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) ~ 3^(n-1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Jul 30 2013
For n > 0, a(n) = (-1)^(n+1) * hypergeom([3/2, 1-n], [2], 4). - Vladimir Reshetnikov, Apr 25 2016
a(n) = GegenbauerC(n-2,-n+1,-1/2) + GegenbauerC(n-1,-n+1,-1/2) for n >= 1. - Peter Luschny, May 12 2016
0 = a(n)*(+9*a(n+1) + 18*a(n+2) - 9*a(n+3)) + a(n+1)*(-6*a(n+1) + 7*a(n+2) - 2*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for n >= 0. - Michael Somos, Dec 01 2016
G.f.: 1/(1-x*G(x)) where G(x) is g.f. of A001006. - Andrew Howroyd, Dec 04 2017
a(n) = (-1)^(n + 1)*2*JacobiP(n - 1, 3, -n - 1/2, -7)/(n^2 + n). - Peter Luschny, May 25 2021
a(n+1) = A005043(n) + 2*A005717(n) for n >= 1. - Peter Bala, Feb 11 2022
a(n) = Sum_{k=0..n-1} A064189(n-1,k) for n >= 1. - Alois P. Heinz, Aug 29 2022

A061384 Numbers n such that sum of digits = number of digits.

Original entry on oeis.org

1, 11, 20, 102, 111, 120, 201, 210, 300, 1003, 1012, 1021, 1030, 1102, 1111, 1120, 1201, 1210, 1300, 2002, 2011, 2020, 2101, 2110, 2200, 3001, 3010, 3100, 4000, 10004, 10013, 10022, 10031, 10040, 10103, 10112, 10121, 10130, 10202, 10211
Offset: 1

Views

Author

Amarnath Murthy, May 03 2001

Keywords

Comments

Number of d-digit entries is A071976(d). - Robert Israel, Apr 06 2016
Equivalently, numbers n > 0 for which the arithmetic mean of the digits equals 1. - M. F. Hasler, Dec 07 2018

Examples

			120 is a term as the arithmetic mean of the digits is (1+2+0)/3 = 1.
		

Crossrefs

Totally balanced subset: A071154. Cf. also A061383-A061388, A061423-A061425.
Cf. A071976.
Cf. A007953 (sum of digits), A055642 (number of digits).

Programs

  • Magma
    [ n: n in [1..10215] | &+Intseq(n) eq #Intseq(n) ]; // Bruno Berselli, Jun 30 2011
    
  • Maple
    Q:= proc(n,s) option remember;
    # n-digit integers with digit sum s
    if s = 0 then []
    elif s = 1 then [10^(n-1)]
    elif n = 1 then
       if s <= 9 then [s]
       else []
       fi
    else
      map(op,[seq(map(t -> 10*t+i, procname(n-1,s-i)), i=0..min(9,s-1))])
    fi
    end proc:
    map(op, [seq(sort(Q(n,n)),n=1..5)]); # Robert Israel, Apr 06 2016
  • Mathematica
    Select[Range[15000], Total[IntegerDigits[#]] == IntegerLength[#]&] (* Harvey P. Dale, Jan 08 2011 *)
  • PARI
    isok(n) = (sumdigits(n)/#Str(n) == 1); \\ Michel Marcus, Mar 28 2016
    
  • PARI
    is_A061384(n)={sumdigits(n)==logint(n+!n,10)+1} \\ M. F. Hasler, Dec 07 2018
    
  • PARI
    A061384_row(n)={my(L=List(), u=vector(n, i, i==1), d); forvec(v=vector(n+1, i, [if(i>n,n, 1), if(i>1, n, 1)]), vecmax(d=v[^1]-v[^-1]+u)<10 && listput(L,fromdigits(d)),1);Vec(L)} \\ Return the list of all n-digit terms. - M. F. Hasler, Dec 07 2018
    
  • Python
    from itertools import count, islice
    def Q(n, s): # length-n strings of 0..9 with sum s, after Robert Israel
        if s == 0: yield "0"*n
        elif n == 1: yield (str(s) if s <= 9 else "")
        else:
            m = min(9, s) + 1
            yield from (str(i)+t for i in range(m) for t in Q(n-1, s-i))
    def agen():
        yield from (int(t) for n in count(1) for t in Q(n, n) if t[0] != "0")
    print(list(islice(agen(), 43))) # Michael S. Branicky, May 26 2022
    
  • Python
    from itertools import count, islice
    from collections import Counter
    from sympy.utilities.iterables import partitions, multiset_permutations
    def A061384_gen(): # generator of terms
        for l in count(1):
            for i in range(1,min(l,9)+1):
                yield from sorted(int(str(i)+''.join(map(str,j))) for s,p in partitions(l-i,k=9,size=True) for j in multiset_permutations([0]*(l-1-s)+list(Counter(p).elements())))
    A061384_list = list(islice(A061384_gen(),30)) # Chai Wah Wu, Nov 28 2023

Formula

{n > 0 | A007953(n) = A055642(n)}. - M. F. Hasler, Dec 07 2018

Extensions

More terms from Erich Friedman, May 08 2001

A289410 Irregular triangular array T(m,k) with m (row) >= 1 and k (column) >= 1 read by rows: number of m-digit numbers whose digit sum is k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 3, 6, 10, 15, 21, 28, 36, 45, 54, 61, 66, 69, 70, 69, 66, 61, 54, 45, 36, 28, 21, 15, 10, 6, 3, 1, 1, 4, 10, 20, 35, 56, 84, 120, 165, 219, 279, 342, 405, 465, 519, 564, 597, 615, 615, 597, 564, 519, 465, 405, 342, 279, 219, 165, 120, 84
Offset: 1

Views

Author

Miquel Cerda, Jul 05 2017

Keywords

Comments

The m-th row is palindromic; T(m,k) = T(m,9*m+1-k).

Examples

			The irregular triangle T(m,k) begins:
m\k  1  2  3  4  5   6   7   8   9   10   11  12   13   14  15  16  17  18  19
1    1  1  1  1  1   1   1   1   1;
2    1  2  3  4  5   6   7   8   9    9    8   7    6    5   4   3   2   1;
3    1  3  6  10 15  21  28  36  45   54   61  66   69   70  69  66  61  54 45,...;
4    1  4  10 20 35  56  84  120 165  219  279 342  405  465,...;
5    1  5  15 35 70  126 210 330 495  714  992 1330 1725,...;
6    1  6  21 56 126 252 462 792 1287 2001 2992,...;
etc.
Row m(2), column k(4) there are 4 numbers of 2-digits whose digits sum = 4: 13, 22, 31, 40.
		

Crossrefs

The row sums = 9*10^(m-1) = A052268(n). The row lengths = 9*m = A008591(n). The middle diagonal = A071976. (row m=3) = A071817, (row m=4) = A090579, (row m=5) = A090580, (row m=6) = A090581, (row m=7) = A278969, (row m=8) = A278971, (row m=9) = A289354, (column k=3) = A000217, (column k=4) = A000292, (column k=5) = A000332, (column k=6) = A000389, (column k=7) = A000579, (column k=8) = A000580, (column k=9) = A000581, (column k=10) = A035927.

Programs

  • Maple
    row:= proc(m) local g; g:= normal((1 - x^10)^(m-1)*(x - x^10)/(1 - x)^m);
    seq(coeff(g,x,j),j=1..9*m) end proc:
    seq(row(k),k=1..5); # Robert Israel, Jul 19 2017

Formula

G.f. of row m: (1 - x^10)^(m-1)*(x - x^10)/(1 - x)^m.
G.f. as array: (1+x+x^2)*(1+x^3+x^6)*x*y/(1-y*(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)). - Robert Israel, Jul 19 2017

Extensions

Edited by Robert Israel, Jul 19 2017

A382531 Number of n-digit base-10 numbers whose digit sum is equal to ceiling(9*n/2).

Original entry on oeis.org

1, 9, 70, 615, 5520, 50412, 468448, 4379055, 41395240, 392406145, 3748943890, 35866068766, 345143007910, 3323483518810, 32150758083580, 311088525668335, 3021445494584902, 29344719005694973, 285904843977651598, 2785022004925340460, 27203012941819689340
Offset: 1

Views

Author

Miquel Cerda, Mar 30 2025

Keywords

Comments

Digit sum ceiling(9*n/2) = A130877(n+1) has highest frequency among all n-digit base-10 numbers.
The count excludes numbers with leading zeros.

Examples

			a(2) = 9, the 2-digit numbers with digit sum 9 are: 18, 27, 36, 45, 54, 63, 72, 81, 90.
		

Crossrefs

Cf. A210736 (analogous for base-2 digits).
Cf. A025015 (maximal coefficient of (1+...+x^9)^n).

Programs

  • Maple
    b:= proc(n, s, t) option remember; `if`(9*n b(n, ceil(9*n/2), 1):
    seq(a(n), n=1..23);  # Alois P. Heinz, Apr 12 2025

Formula

a(n) = [x^ceiling(9*n/2)] (f^n - f^(n-1)) with f = (x^10-1)/(x-1). - Alois P. Heinz, Apr 12 2025

A071646 Number of base 4 n-digit numbers with digit sum n.

Original entry on oeis.org

1, 2, 6, 19, 61, 201, 672, 2269, 7723, 26452, 91058, 314766, 1091884, 3798900, 13251136, 46325285, 162268775, 569385098, 2001012474, 7042014879, 24813529581, 87533417037, 309107111536, 1092585807044, 3865270781236
Offset: 1

Views

Author

John W. Layman, Jun 22 2002

Keywords

Crossrefs

Programs

  • Maxima
    a(n):=sum((sum(binomial(j,n-3*k+2*j)*binomial(k,j),j,0,k))*binomial(n-1,n-k),k,1,n); /* Vladimir Kruchinin, Nov 07 2013 */
  • PARI
    a(n)=local(y=(x^4-1)/(x-1)); if(n<0,0,polcoeff(y^n-y^(n-1),n))
    

Formula

a(n)=[x^n] (1+x+x^2+x^3)^n-(1+x+x^2+x^3)^(n-1). - Michael Somos, Jul 19 2002
a(n)*790*(2*n^2-n) = a(n-1)*(-16328*n^4+137200*n^3-400977*n^2+489925*n-207450)+a(n-2)*(44902*n^4-399751*n^3+1267117*n^2-1672482*n+769980)+a(n-3)*4*(n-3)*(8164*n^3-52272*n^2+115397*n-81223)+a(n-4)*16*(n-4)*(n-3)*(4082*n^2-11849*n+8529), n>2. - Michael Somos, Jul 19 2002
a(n) = sum(k=1..n, (sum(j=0..k, binomial(j,n-3*k+2*j) *binomial(k,j))) *binomial(n-1,n-k)). [Vladimir Kruchinin, Nov 07 2013]
Showing 1-6 of 6 results.