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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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

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Author

N. J. A. Sloane

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.

Links

Crossrefs

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
Cf. A000079, A001045, A009545, A016189, A052955, A083329, A085104, A144081.
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. A000203, A239473, A279396.
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

A005408 The odd numbers: a(n) = 2*n + 1.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172).
Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.
The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).
Also continued fraction for coth(1) (A073747 is decimal expansion). - Rick L. Shepherd, Aug 07 2002
a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003
Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003
Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010). - Lekraj Beedassy, Aug 27 2004
Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005
For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.
First differences of squares (A000290). - Lekraj Beedassy, Jul 15 2006
The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006
2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007
For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007
A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454. - Reinhard Zumkeller, Oct 27 2007
Numbers n such that sigma(2n) = 3*sigma(n). - Farideh Firoozbakht, Feb 26 2008
a(n) = A139391(A016825(n)) = A006370(A016825(n)). - Reinhard Zumkeller, Apr 17 2008
Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Equals INVERT transform of A078050 (signed - cf. comments); and row sums of triangle A144106. - Gary W. Adamson, Sep 11 2008
Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008
A000035(a(n))=1, A059841(a(n))=0. - Reinhard Zumkeller, Sep 29 2008
Multiplicative closure of A065091. - Reinhard Zumkeller, Oct 14 2008
a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009
Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0). - Philippe Deléham, Sep 17 2009
Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009
Or n without 2 as prime factor. - Juri-Stepan Gerasimov, Nov 19 2009
Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009
A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)). - Reinhard Zumkeller, Apr 13 2010
For n >= 1, a(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010
Union of A179084 and A179085. - Reinhard Zumkeller, Jun 28 2010
For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010
Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010
A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 2011
A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012
a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012
A007814(a(n)) = 0; A037227(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2012
1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012
Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012
A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013
This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013
These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014
a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014
The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015
a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016
Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017
Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017
Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023

Examples

			G.f. = q + 3*q^3 + 5*q^5 + 7*q^7 + 9*q^9 + 11*q^11 + 13*q^13 + 15*q^15 + ...

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
  • T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology, p. 264.
  • D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
  • E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203-205.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000027, A005843, A065091.
See A120062 for sequences related to integer-sided triangles with integer inradius n.
Cf. A128200, A000290, A078050, A144106, A109613, A167875.
Cf. A001651 (n=1 or 2 mod 3), A047209 (n=1 or 4 mod 5).
Cf. A003558, A216371, A179480 (relating to the Coach theorem).
Cf. A000754 (boustrophedon transform).

Programs

Formula

a(n) = 2*n + 1. a(-1 - n) = -a(n). a(n+1) = a(n) + 2.
G.f.: (1 + x) / (1 - x)^2.
E.g.f.: (1 + 2*x) * exp(x).
G.f. with interpolated zeros: (x^3+x)/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(-x))/2. - Geoffrey Critzer, Aug 25 2012
a(n) = L(n,-2)*(-1)^n, where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
Euler transform of length 2 sequence [3, -1]. - Michael Somos, Mar 30 2007
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = v * (1 + 2*u) * (1 - 2*u + 16*v) - (u - 4*v)^2 * (1 + 2*u + 2*u^2). - Michael Somos, Mar 30 2007
a(n) = b(2*n + 1) where b(n) = n if n is odd is multiplicative. [This seems to say that A000027 is multiplicative? - R. J. Mathar, Sep 23 2011]
From Hieronymus Fischer, May 25 2007: (Start)
a(n) = (n+1)^2 - n^2.
G.f. g(x) = Sum_{k>=0} x^floor(sqrt(k)) = Sum_{k>=0} x^A000196(k). (End)
a(0) = 1, a(1) = 3, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(n) = A000330(A016777(n))/A000217(A016777(n)). - Pierre CAMI, Sep 27 2008
a(n) = A034856(n+1) - A000217(n) = A005843(n) + A000124(n) - A000217(n) = A005843(n) + 1. - Jaroslav Krizek, Sep 05 2009
a(n) = (n - 1) + n (sum of two sequential integers). - Dominick Cancilla, Aug 09 2010
a(n) = 4*A000217(n)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. - Bruno Berselli, Nov 17 2010
n*a(2n+1)^2+1 = (n+1)*a(2n)^2; e.g., 3*15^2+1 = 4*13^2. - Charlie Marion, Dec 31 2010
arctanh(x) = Sum_{n>=0} x^(2n+1)/a(n). - R. J. Mathar, Sep 23 2011
a(n) = det(f(i-j+1))A113311(n);%20for%20n%20%3C%200%20we%20have%20f(n)=0.%20-%20_Mircea%20Merca">{1<=i,j<=n}, where f(n) = A113311(n); for n < 0 we have f(n)=0. - _Mircea Merca, Jun 23 2012
G.f.: Q(0), where Q(k) = 1 + 2*(k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
a(n) = floor(sqrt(2*A000384(n+1))). - Ivan N. Ianakiev, Jun 17 2013
a(n) = 3*A000330(n)/A000217(n), n > 0. - Ivan N. Ianakiev, Jul 12 2013
a(n) = Product_{k=1..2*n} 2*sin(Pi*k/(2*n+1)) = Product_{k=1..n} (2*sin(Pi*k/(2*n+1)))^2, n >= 0 (undefined product = 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013
Noting that as n -> infinity, sqrt(n^2 + n) -> n + 1/2, let f(n) = n + 1/2 - sqrt(n^2 + n). Then for n > 0, a(n) = round(1/f(n))/4. - Richard R. Forberg, Feb 16 2014
a(n) = Sum_{k=0..n+1} binomial(2*n+1,2*k)*4^(k)*bernoulli(2*k). - Vladimir Kruchinin, Feb 24 2015
a(n) = Sum_{k=0..n} binomial(6*n+3, 6*k)*Bernoulli(6*k). - Michel Marcus, Jan 11 2016
a(n) = A000225(n+1) - A005803(n+1). - Miquel Cerda, Nov 25 2016
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
Sum_{n>=0} 1/a(n)^2 = Pi^2/8 = A111003. - Bernard Schott, Dec 10 2020
Sum_{n >= 1} (-1)^n/(a(n)*a(n+1)) = Pi/4 - 1/2 = 1/(3 + (1*3)/(4 + (3*5)/(4 + ... + (4*n^2 - 1)/(4 + ... )))). Cf. A016754. - Peter Bala, Mar 28 2024
a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024
a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024
a(n) = A000217(a(n))/n for n > 0. - Stefano Spezia, Feb 15 2025

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010
Peripheral comments deleted by N. J. A. Sloane, May 09 2022

A000918 a(n) = 2^n - 2.

Original entry on oeis.org

-1, 0, 2, 6, 14, 30, 62, 126, 254, 510, 1022, 2046, 4094, 8190, 16382, 32766, 65534, 131070, 262142, 524286, 1048574, 2097150, 4194302, 8388606, 16777214, 33554430, 67108862, 134217726, 268435454, 536870910, 1073741822, 2147483646, 4294967294, 8589934590, 17179869182, 34359738366, 68719476734, 137438953470
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

For n > 1, a(n) is the expected number of tosses of a fair coin to get n-1 consecutive heads. - Pratik Poddar, Feb 04 2011
For n > 2, Sum_{k=1..a(n)} (-1)^binomial(n, k) = A064405(a(n)) + 1 = 0. - Benoit Cloitre, Oct 18 2002
For n > 0, the number of nonempty proper subsets of an n-element set. - Ross La Haye, Feb 07 2004
Numbers j such that abs( Sum_{k=0..j} (-1)^binomial(j, k)*binomial(j + k, j - k) ) = 1. - Benoit Cloitre, Jul 03 2004
For n > 2 this formula also counts edge-rooted forests in a cycle of length n. - Woong Kook (andrewk(AT)math.uri.edu), Sep 08 2004
For n >= 1, conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4, 5, 6 and 7 as a digit. - Alexandre Wajnberg, Apr 25 2005
Beginning with a(2) = 2, these are the partial sums of the subsequence of A000079 = 2^n beginning with A000079(1) = 2. Hence for n >= 2 a(n) is the smallest possible sum of exactly one prime, one semiprime, one triprime, ... and one product of exactly n-1 primes. A060389 (partial sums of the primorials, A002110, beginning with A002110(1)=2) is the analog when all the almost primes must also be squarefree. - Rick L. Shepherd, May 20 2005
From the second term on (n >= 1), the binary representation of these numbers is a 0 preceded by (n - 1) 1's. This pattern (0)111...1110 is the "opposite" of the binary 2^n+1: 1000...0001 (cf. A000051). - Alexandre Wajnberg, May 31 2005
The numbers 2^n - 2 (n >= 2) give the positions of 0's in A110146. Also numbers k such that k^(k + 1) = 0 mod (k + 2). - Zak Seidov, Feb 20 2006
Partial sums of A155559. - Zerinvary Lajos, Oct 03 2007
Number of surjections from an n-element set onto a two-element set, with n >= 2. - Mohamed Bouhamida, Dec 15 2007
It appears that these are the numbers n such that 3*A135013(n) = n*(n + 1), thus answering Problem 2 on the Mathematical Olympiad Foundation of Japan, Final Round Problems, Feb 11 1993 (see link Japanese Mathematical Olympiad).
Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x is a proper subset of y or y is a proper subset of x and x and y are disjoint. Then a(n+1) = |R|. - Ross La Haye, Mar 19 2009
The permutohedron Pi_n has 2^n - 2 facets [Pashkovich]. - Jonathan Vos Post, Dec 17 2009
First differences of A005803. - Reinhard Zumkeller, Oct 12 2011
For n >= 1, a(n + 1) is the smallest even number with bit sum n. Cf. A069532. - Jason Kimberley, Nov 01 2011
a(n) is the number of branches of a complete binary tree of n levels. - Denis Lorrain, Dec 16 2011
For n>=1, a(n) is the number of length-n words on alphabet {1,2,3} such that the gap(w)=1. For a word w the gap g(w) is the number of parts missing between the minimal and maximal elements of w. Generally for words on alphabet {1,2,...,m} with g(w)=g>0 the e.g.f. is Sum_{k=g+2..m} (m - k + 1)*binomial((k - 2),g)*(exp(x) - 1)^(k - g). a(3)=6 because we have: 113, 131, 133, 311, 313, 331. Cf. A240506. See the Heubach/Mansour reference. - Geoffrey Critzer, Apr 13 2014
For n > 0, a(n) is the minimal number of internal nodes of a red-black tree of height 2*n-2. See the Oct 02 2015 comment under A027383. - Herbert Eberle, Oct 02 2015
Conjecture: For n>0, a(n) is the least m such that A007814(A000108(m)) = n-1. - L. Edson Jeffery, Nov 27 2015
Actually this follows from the procedure for determining the multiplicity of prime p in C(n) given in A000108 by Franklin T. Adams-Watters: For p=2, the multiplicity is the number of 1 digits minus 1 in the binary representation of n+1. Obviously, the smallest k achieving "number of 1 digits" = k is 2^k-1. Therefore C(2^k-2) is divisible by 2^(k-1) for k > 0 and there is no smaller m for which 2^(k-1) divides C(m) proving the conjecture. - Peter Schorn, Feb 16 2020
For n >= 0, a(n) is the largest number you can write in bijective base-2 (a.k.a. the dyadic system, A007931) with n digits. - Harald Korneliussen, May 18 2019
The terms of this sequence are also the sum of the terms in each row of Pascal's triangle other than the ones. - Harvey P. Dale, Apr 19 2020
For n > 1, binomial(a(n),k) is odd if and only if k is even. - Charlie Marion, Dec 22 2020
For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 2. - David desJardins, Oct 27 2022
For n >= 1, a(n+1) is the number of roots of unity in the unique degree-n unramified extension of the 2-adic field Q_2. Note that for each p, the unique degree-n unramified extension of Q_p is the splitting field of x^(p^n) - x, hence containing p^n - 1 roots of unity for p > 2 and 2*(2^n - 1) for p = 2. - Jianing Song, Nov 08 2022

Examples

			a(4) = 14 because the 14 = 6 + 4 + 4 rationals (in lowest terms) for n = 3 are (see the Jun 14 2017 formula above): 1/2, 1, 3/2, 2, 5/2, 3; 1/4, 3/4, 5/4, 7/4; 1/8, 3/8, 5/8, 7/8. - _Wolfdieter Lang_, Jun 14 2017

References

  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134. - Mohammad K. Azarian, Oct 27 2011
  • S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2009 page 86, Exercise 3.16.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • 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).
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.

Links

Crossrefs

Row sums of triangle A026998.
Cf. A000919, A001117, A001118, A095121, A110146.
Cf. A033484, A000225, A000325, A095151, A125128.
Cf. A058809 (3^n-3, n>0).

Programs

  • Haskell
    a000918 = (subtract 2) . (2 ^)
a000918_list = iterate ((subtract 2) . (* 2) . (+ 2)) (- 1)
-- Reinhard Zumkeller, Apr 23 2013
  • Magma
    [2^n - 2: n in [0..40]]; // Vincenzo Librandi, Jun 23 2011
  • Maple
    seq(2^n-2,n=0..20) ;
  • Mathematica
    Table[2^n - 2, {n, 0, 29}] (* Alonso del Arte, Dec 01 2012 *)
  • PARI
    a(n)=2^n-2 \\ Charles R Greathouse IV, Jun 16 2011
  • Python
    def A000918(n): return (1<Chai Wah Wu, Jun 10 2025

Formula

a(n) = 2*A000225(n-1).
G.f.: 1/(1 - 2*x) - 2/(1 - x), e.g.f.: (e^x - 1)^2 - 1. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
For n >= 1, a(n) = A008970(n + 1, 2). - Philippe Deléham, Feb 21 2004
G.f.: (3*x - 1)/((2*x - 1)*(x - 1)). - Simon Plouffe in his 1992 dissertation for the sequence without the leading -1
a(n) = 2*a(n - 1) + 2. - Alexandre Wajnberg, Apr 25 2005
a(n) = A000079(n) - 2. - Omar E. Pol, Dec 16 2008
a(n) = A058896(n)/A052548(n). - Reinhard Zumkeller, Feb 14 2009
a(n) = A164874(n - 1, n - 1) for n > 1. - Reinhard Zumkeller, Aug 29 2009
a(n) = A173787(n,1); a(n) = A028399(2*n)/A052548(n), n > 0. - Reinhard Zumkeller, Feb 28 2010
a(n + 1) = A027383(2*n - 1). - Jason Kimberley, Nov 02 2011
G.f.: U(0) - 1, where U(k) = 1 + x/(2^k + 2^k/(x - 1 - x^2*2^(k + 1)/(x*2^(k + 1) - (k + 1)/U(k + 1) ))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012
a(n+1) is the sum of row n in triangle A051601. - Reinhard Zumkeller, Aug 05 2013
a(n+1) = A127330(n,0). - Reinhard Zumkeller, Nov 16 2013
a(n) = Sum_{k=1..n-1} binomial(n, k) for n > 0. - Dan McCandless, Nov 14 2015
From Miquel Cerda, Aug 16 2016: (Start)
a(n) = A000225(n) - 1.
a(n) = A125128(n-1) - A000325(n).
a(n) = A095151(n) - A125128(n) - 1. (End)
a(n+1) = 2*(n + Sum_{j=1..n-1} (n-j)*2^(j-1)), n >= 1. This is the number of the rationals k/2, k = 1..2*n for n >= 1 and (2*k+1)/2^j for j = 2..n, n >= 2, and 2*k+1 < n-(j-1). See the example for n = 3 below. Motivated by the proposal A287012 by Mark Rickert. - Wolfdieter Lang, Jun 14 2017

Extensions

Maple programs fixed by Vaclav Kotesovec, Dec 13 2014

A000325 a(n) = 2^n - n.

Original entry on oeis.org

1, 1, 2, 5, 12, 27, 58, 121, 248, 503, 1014, 2037, 4084, 8179, 16370, 32753, 65520, 131055, 262126, 524269, 1048556, 2097131, 4194282, 8388585, 16777192, 33554407, 67108838, 134217701, 268435428, 536870883, 1073741794, 2147483617
Offset: 0

Views

Author

Rosario Salamone (Rosario.Salamone(AT)risc.uni-linz.ac.at)

Keywords

Comments

Number of permutations of degree n with at most one fall; called "Grassmannian permutations" by Lascoux and Schützenberger. - Axel Kohnert (Axel.Kohnert(AT)uni-bayreuth.de)
Number of different permutations of a deck of n cards that can be produced by a single shuffle. [DeSario]
Number of Dyck paths of semilength n having at most one long ascent (i.e., ascent of length at least two). Example: a(4)=12 because among the 14 Dyck paths of semilength 4, the only paths that have more than one long ascent are UUDDUUDD and UUDUUDDD (each with two long ascents). Here U = (1, 1) and D = (1, -1). Also number of ordered trees with n edges having at most one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004
Number of {12,1*2*,21*}-avoiding signed permutations in the hyperoctahedral group.
Number of 1342-avoiding circular permutations on [n+1].
2^n - n is the number of ways to partition {1, 2, ..., n} into arithmetic progressions, where in each partition all the progressions have the same common difference and have lengths at least 1. - Marty Getz (ffmpg1(AT)uaf.edu) and Dixon Jones (fndjj(AT)uaf.edu), May 21 2005
if b(0) = x and b(n) = b(n-1) + b(n-1)^2*x^(n-2) for n > 0, then b(n) is a polynomial of degree a(n). - Michael Somos, Nov 04 2006
The chromatic invariant of the Mobius ladder graph M_n for n >= 2. - Jonathan Vos Post, Aug 29 2008
Dimension sequence of the dual alternative operad (i.e., associative and satisfying the identity xyz + yxz + zxy + xzy + yzx + zyx = 0) over the field of characteristic 3. - Pasha Zusmanovich, Jun 09 2009
An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 26 and 176, lead to this sequence (without the first leading 1). For the central square these vectors lead to the companion sequence A168604. - Johannes W. Meijer, Aug 15 2010
a(n+1) is also the number of order-preserving and order-decreasing partial isometries (of an n-chain). - Abdullahi Umar, Jan 13 2011
A040001(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012
A130103(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012
The number of labeled graphs with n vertices whose vertex set can be partitioned into a clique and a set of isolated points. - Alex J. Best, Nov 20 2012
For n > 0, a(n) is a B_2 sequence. - Thomas Ordowski, Sep 23 2014
See coefficients of the linear terms of the polynomials of the table on p. 10 of the Getzler link. - Tom Copeland, Mar 24 2016
Consider n points lying on a circle, then for n>=2 a(n-1) is the maximum number of ways to connect two points with non-intersecting chords. - Anton Zakharov, Dec 31 2016
Also the number of cliques in the (n-1)-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) < e(k). [Martinez and Savage, 2.7]
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i), e(j), e(k) pairwise distinct. [Martinez and Savage, 2.7]
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(j) >= e(k) and e(i) != e(k) pairwise distinct. [Martinez and Savage, 2.7]
(End)
Number of F-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are F-equivalent iff the positions of pattern F are identical in these paths. - Sergey Kirgizov, Apr 08 2018
From Gus Wiseman, Feb 10 2019: (Start)
Also the number of connected partitions of an n-cycle. For example, the a(1) = 1 through a(4) = 12 connected partitions are:
{{1}} {{12}} {{123}} {{1234}}
{{1}{2}} {{1}{23}} {{1}{234}}
{{12}{3}} {{12}{34}}
{{13}{2}} {{123}{4}}
{{1}{2}{3}} {{124}{3}}
{{134}{2}}
{{14}{23}}
{{1}{2}{34}}
{{1}{23}{4}}
{{12}{3}{4}}
{{14}{2}{3}}
{{1}{2}{3}{4}}
(End)
Number of subsets of n-set without the single-element subsets. - Yuchun Ji, Jul 16 2019
For every prime p, there are infinitely many terms of this sequence that are divisible by p (see IMO Compendium link and Doob reference). Corresponding indices n are: for p = 2, even numbers A299174; for p = 3, A047257; for p = 5, A349767. - Bernard Schott, Dec 10 2021
Primes are in A081296 and corresponding indices in A048744. - Bernard Schott, Dec 12 2021

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 12*x^4 + 27*x^5 + 58*x^6 + 121*x^7 + ...

References

  • Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1983, page 158, 1993.

Links

Crossrefs

Column 1 of triangle A008518.
Row sum of triangles A184049 and A184050.
Cf. A000108, A002064, A133116, A160692, A005803, A006127, A186947.
Cf. A001610, A066982, A323950, A323951.
Cf. A047257, A299174, A349767.
Cf. A048744, A081296.

Programs

  • Haskell
    a000325 n = 2 ^ n - n
a000325_list = zipWith (-) a000079_list [0..]
-- Reinhard Zumkeller, Jul 17 2012
  • Magma
    [2^n - n: n in [0..35]]; // Vincenzo Librandi, May 13 2011
  • Maple
    A000325 := proc(n) option remember; if n <=1 then n+1 else 2*A000325(n-1)+n-1; fi; end;
g:=1/(1-2*z): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-n, n=0..31); # Zerinvary Lajos, Jan 09 2009
  • Mathematica
    Table[2^n - n, {n, 0, 39}] (* Alonso del Arte, Sep 15 2014 *)
LinearRecurrence[{4, -5, 2}, {1, 2, 5}, {0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)
  • PARI
    {a(n) = 2^n - n}; /* Michael Somos, Nov 04 2006 */
  • Python
    def A000325(n): return (1<Chai Wah Wu, Jan 11 2023

Formula

a(n+1) = 2*a(n) + n - 1, a(0) = 1. - Reinhard Zumkeller, Apr 12 2003
Binomial transform of 1, 0, 1, 1, 1, .... The sequence starting 1, 2, 5, ... has a(n) = 1 + n + 2*Sum_{k=2..n} binomial(n, k) = 2^(n+1) - n - 1. This is the binomial transform of 1, 1, 2, 2, 2, 2, .... a(n) = 1 + Sum_{k=2..n} C(n, k). - Paul Barry, Jun 06 2003
G.f.: (1-3x+3x^2)/((1-2x)*(1-x)^2). - Emeric Deutsch, Feb 22 2004
A107907(a(n+2)) = A000051(n+2) for n > 0. - Reinhard Zumkeller, May 28 2005
a(n+1) = sum of n-th row of the triangle in A109128. - Reinhard Zumkeller, Jun 20 2005
Row sums of triangle A133116. - Gary W. Adamson, Sep 14 2007
G.f.: 1 / (1 - x / (1 - x / ( 1 - x / (1 + x / (1 - 2*x))))). - Michael Somos, May 12 2012
First difference is A000225. PSUM transform is A084634. - Michael Somos, May 12 2012
a(n) = [x^n](B(x)^n-B(x)^(n-1)), n>0, a(0)=1, where B(x) = (1+2*x+sqrt(1+4*x^2))/2. - Vladimir Kruchinin, Mar 07 2014
E.g.f.: (exp(x) - x)*exp(x). - Ilya Gutkovskiy, Aug 07 2016
a(n) = A125128(n) - A000225(n) + 1. - Miquel Cerda, Aug 12 2016
a(n) = 2*A125128(n) - A095151(n) + 1. - Miquel Cerda, Aug 12 2016
a(n) = A079583(n-1) - A000225(n-1). - Miquel Cerda, Aug 15 2016
a(n)^2 - 4*a(n-1)^2 = (n-2)*(a(n)+2*a(n-1)). - Yuchun Ji, Jul 13 2018
a(n) = 2^(-n) * A186947(n) = 2^n * A002064(-n) for all n in Z. - Michael Somos, Jul 18 2018
a(2^n) = (2^a(n) - 1)*2^n. - Lorenzo Sauras Altuzarra, Feb 01 2022

A008517 Second-order Eulerian triangle T(n,k), 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 8, 6, 1, 22, 58, 24, 1, 52, 328, 444, 120, 1, 114, 1452, 4400, 3708, 720, 1, 240, 5610, 32120, 58140, 33984, 5040, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Second-order Eulerian numbers <> = T(n,k+1) count the permutations of the multiset {1,1,2,2,...,n,n} with k ascents with the restriction that for all m, all integers between the two copies of m are less than m. In particular, the two 1s are always next to each other.
When seen as coefficients of polynomials with descending exponents, evaluations are in A000311 (x=2) and A001662 (x=-1).
The row reversed triangle is A112007. There one can find comments on the o.g.f.s for the diagonals of the unsigned Stirling1 triangle |A008275|.
Stirling2(n,n-k) = Sum_{m=0..k-1} T(k,m+1)*binomial(n+k-1+m, 2*k), k>=1. See the Graham et al. reference p. 271 eq. (6.43).
This triangle is the coefficient triangle of the numerator polynomials appearing in the o.g.f. for the k-th diagonal (k >= 1) of the Stirling2 triangle A048993.
The o.g.f. for column k satisfies the recurrence G(k,x) = x*(2*x*(d/dx)G(k-1,x) + (2-k)*G(k-1,x))/(1-k*x), k >= 2, with G(1,x) = 1/(1-x). - Wolfdieter Lang, Oct 14 2005
This triangle is in some sense generated by the differential equation y' = 1 - 2/(1+x+y). (This is the differential equation satisfied by the function defined implicitly as x+y=exp(x-y).) If we take y = a(0) + a(1)x + a(2)x^2 + a(3)x^3 + ... and assume a(0)=c then all the a's may be calculated formally in terms of c and we have a(1) = (c-1)/(c+1) and, for n > 1, a(n) = 2^n/n! (1+c)^(1-2n)( T(n,1)c - T(n,2)c^2 + T(n,3)c^3 - ... + (-1)^(n-1) T(n,n)c^n ). - Moshe Shmuel Newman, Aug 08 2007
From the recurrence relation, the generating function F(x,y) := 1 + Sum_{n>=1, 1<=k<=n} [T(n,k)x^n/n!*y^k] satisfies the partial differential equation F = (1/y-2x)F_x + (y-1)F_y, with (non-elementary) solution F(x,y) = (1-y)/(1-Phi(w)) where w = y*exp(x(y-1)^2-y) and Phi(x) is defined by Phi(x) = x*exp(Phi(x)). By Lagrange inversion (see Wilf's book "generatingfunctionology", page 168, Example 1), Phi(x) = Sum_{n>=1} n^(n-1)*x^n/n!. Thus Phi(x) can alternatively be described as the e.g.f. for rooted labeled trees on n vertices A000169. - David Callan, Jul 25 2008
A method for solving PDEs such as the one above for F(x,y) is described in the Klazar reference (see pages 207-208). In his case, the auxiliary ODE dy/dx = b(x,y)/a(x,y) is exact; in this case it is not exact but has an integrating factor depending on y alone, namely y-1. The e.g.f. for the row sums A001147 is 1/sqrt(1-2*x) and the demonstration that F(x,1) = 1/sqrt(1-2*x) is interesting: two applications of l'Hopital's rule to lim_{y->1}F(x,y) yield F(x,1) = 1/(1-2x)*1/F(x,1). So l'Hopital's rule doesn't directly yield F(x,1) but rather an equation to be solved for F(x,1)!. - David Callan, Jul 25 2008
From Tom Copeland, Oct 12 2008; May 19 2010: (Start)
Let P(0,t)= 0, P(1,t)= 1, P(2,t)= t, P(3,t)= t + 2 t^2, P(4,t)= t + 8 t^2 + 6 t^3, ... be the row polynomials of the present array, then
exp(x*P(.,t)) = ( u + Tree(t*exp(u)) ) / (1-t) = WD(x*(1-t), t/(1-t)) / (1-t)
where u = x*(1-t)^2 - t, Tree(x) is the e.g.f. of A000169 and WD(x,t) is the e.g.f. for A134991, relating the Ward and 2-Eulerian polynomials by a simple transformation.
Note also apparently P(4,t) / (1-t)^3 = Ward Poly(4, t/(1-t)) = essentially an e.g.f. for A093500.
The compositional inverse of f(x,t) = exp(P(.,t)*x) about x=0 is
g(x,t) = ( x - (t/(1-t)^2)*(exp(x*(1-t))-x*(1-t)-1) )
= x - t*x^2/2! - t*(1-t)*x^3/3! - t*(1-t)^2*x^4/4! - t*(1-t)^3*x^5/5! - ... .
Can apply A134685 to these coefficients to generate f(x,t). (End)
Triangle A163936 is similar to the one given above except for an extra right hand column [1, 0, 0, 0, ... ] and that its row order is reversed. - Johannes W. Meijer, Oct 16 2009
From Tom Copeland, Sep 04 2011: (Start)
Let h(x,t) = 1/(1-(t/(1-t))*(exp(x*(1-t))-1)), an e.g.f. in x for row polynomials in t of A008292, then the n-th row polynomial in t of the table A008517 is given by ((h(x,t)*D_x)^(n+1))x with the derivative evaluated at x=0.
Also, df(x,t)/dx = h(f(x,t),t) where f(x,t) is an e.g.f. in x of the row polynomials in t of A008517, i.e., exp(x*P(.,t)) in Copeland's 2008 comment. (End)
The rows are the h-vectors of A134991. - Tom Copeland, Oct 03 2011
Hilbert series of the pre-WDVV ring, thus h-vectors of the Whitehouse simplicial complex (cf. Readdy, Table 1). - Tom Copeland, Sep 20 2014
Arises in Buckholtz's analysis of the error term in the series for exp(nz). - N. J. A. Sloane, Jul 05 2016

Examples

			Triangle begins:
  1;
  1,   2;
  1,   8,   6;
  1,  22,  58,  24;
  1,  52, 328, 444, 120; ...
Row 3: There are three plane increasing 0-1-2-3 trees on 3 vertices. The number of colors are shown to the right of a vertex.
.
    1o (2*t+1)         1o t*(t+2)      1o t*(t+2)
     |                 / \             / \
     |                /   \           /   \
    2o (2*t+1)      2o    3o        3o    2o
     |
     |
    3o
.
The total number of trees is (2*t+1)^2 + t*(t+2) + t*(t+2) = 1 + 8*t + 6*t^2.

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270. [with offsets [0,0]: see A201637]

Links

Crossrefs

Columns include A005803, A004301, A006260.
Right-hand columns include A000142, A002538, A002539.
Row sums are A001147.
For a (0,0) based version as used in 'Concrete Mathematics' and by Maple see A201637. For a (0,0) based version which has this triangle as a subtriangle see A340556.

Programs

  • Maple
    with(combinat): A008517 := proc(n, m) local k: add((-1)^(n+k)* binomial(2*n+1, k)* stirling1(2*n-m-k+1, n-m-k+1), k=0..n-m) end: seq(seq(A008517(n, m), m=1..n), n=1..8);
# Johannes W. Meijer, Oct 16 2009, revised Nov 22 2012
A008517 := proc(n,k) option remember; `if`(n=1,`if`(k=0,1,0), A008517(n-1,k)* (k+1) + A008517(n-1,k-1)*(2*n-k-1)) end: seq(print(seq(A008517(n,k), k=0..n-1)), n=1..9);
# Peter Luschny, Apr 20 2011
  • Mathematica
    a[n_, m_] = Sum[(-1)^(n + k)*Binomial[2 n + 1, k]*StirlingS1[2n-m-k+1, n-m-k+1], {k, 0, n-m}]; Flatten[Table[a[n, m], {n, 1, 9}, {m, 1, n}]][[1 ;; 44]] (* Jean-François Alcover, May 18 2011, after Johannes W. Meijer *)
  • PARI
    {T(n, k) = my(z); if( n<1, 0, z = 1 + O(x); for( k=1, n, z = 1 + intformal( z^2 * (z+y-1))); n! * polcoeff( polcoeff(z, n),k))}; /* Michael Somos, Oct 13 2002 */
  • PARI
    {T(n,k)=polcoeff((1-x)^(2*n+1)*sum(j=0,2*n+1,j^(n+j)*x^j/j!*exp(-j*x +x*O(x^k))),k)} \\ Paul D. Hanna, Oct 31 2012
for(n=1,10,for(k=1,n,print1(T(n,k),", "));print(""))
  • PARI
    T(n, m) = sum(k=0, n-m, (-1)^(n+k)*binomial(2*n+1, k)*stirling(2*n-m-k+1, n-m-k+1, 1)); \\ Michel Marcus, Dec 07 2021
  • Sage
    @CachedFunction
def A008517(n, k):
    if n==1: return 1 if k==0 else 0
    return A008517(n-1,k)*(k+1)+A008517(n-1,k-1)*(2*n-k-1)
for n in (1..9): [A008517(n, k) for k in(0..n-1)] # Peter Luschny, Oct 31 2012

Formula

T(n,k) = 0 if n < k, T(1,1) = 1, T(n,-1) = 0, T(n,k) = k*T(n-1,k) + (2*n-k)*T(n-1,k-1).
a(n,m) = Sum_{k=0..n-m} (-1)^(n+k)*binomial(2*n+1, k)*Stirling1(2*n-m-k+1, n-m-k+1). - Johannes W. Meijer, Oct 16 2009
From Peter Bala, Sep 29 2011: (Start)
For k = 0,1,2,... put G(k,x,t) := x-(1+2^k*t)*x^2/2+(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 1 and for the Eulerian numbers A008292 when k = 0.
Let v = -t*exp((1-t)^2*x-t) and let B(x,t) = -(1+1/t*LambertW(v))/(1+LambertW(v)). From the e.g.f. given by Copeland above we find B(x,t) = compositional inverse with respect to x of G(1,x,t) = Sum_{n>=1} R(n,t)*x^n/n! = x+(1+2*t)*x^2/2!+(1+8*t+6*t^2)*x^3/3!+.... The function B(x,t) satisfies the differential equation dB/dx = (1+B)*(1+t*B)^2 = 1 + (2*t+1)*B + t*(t+2)*B^2 + t^2*B^3.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees where each vertex has outdegree <= 3, the vertices of outdegree 1 come in 2*t+1 colors, the vertices of outdegree 2 come in t*(t+2) colors and the vertices of outdegree 3 come in t^2 colors. An example is given below. Cf. A008292. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x,t) = (1+x)*(1+t*x)^2 and let D be the operator f(x,t)*d/dx. Then R(n+1,t) = D^n(f(x,t)) evaluated at x = 0. (End)
From Tom Copeland, Oct 03 2011: (Start)
a(n,k) = Sum_{i=0..k} (-1)^(k-i) binomial(n-i,k-i) A134991(n,i), offsets 0.
P(n+1,t) = (1-t)^(2n+1) Sum_{k>=1} k^(n+k) [t*exp(-t)]^k / k! for n>0; consequently, Sum_{k>=1} (-1)^k k^(n+k) x^k/k!= [1+LW(x)]^(-(2n+1))P[n+1,-LW(x)] where LW(x) is the Lambert W-Function and P(n,t), for n > 0, are the row polynomials as given in Copeland's 2008 comment. (End)
The e.g.f. A(x,t) = -v * { Sum_{j>=1} D(j-1,u) (-z)^j / j! } where u=x*(1-t)^2-t, v=(1+u)/(1-t), z=(t+u)/[(1+u)^2] and D(j-1,u) are the polynomials of A042977. dA(x,t)/dx=(1-t)/[1+u-(1-t)A(x,t)]=(1-t)/{1+LW[-t exp(u)]}, (Copeland's e.g.f. in 2008 comment). - Tom Copeland, Oct 06 2011
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=-t, and (a_n)=-P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 = 0. - Tom Copeland, Oct 08 2011
The compositional inverse (with respect to x) of y = y(t;x) = (x-t*(exp(x)-1)) is 1/(1-t)*y + t/(1-t)^3*y^2/2! + (t+2*t^2)/(1-t)^5*y^3/3! + (t+8*t^2+6*t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed in the Comments section, the rational functions in t are the generating functions for the diagonals of the triangle of Stirling numbers of the second kind (A048993). See the Bala link for a proof. Cf. A112007 and A134991. - Peter Bala, Dec 04 2011
O.g.f. of row n: (1-x)^(2*n+1) * Sum_{k>=0} k^(n+k) * exp(-k*x) * x^k/k!. - Paul D. Hanna, Oct 31 2012
T(n, k) = n!*[x^n][t^k](egf) where egf = (1-t)/(1 + LambertW(-exp(t^2*x - 2*t*x - t + x)*t)) and after expansion W(-exp(-t)t) is substituted by (-t). - Shamil Shakirov, Feb 17 2025

A340556 E2(n, k), the second-order Eulerian numbers with E2(0, k) = δ_{0, k}. Triangle read by rows, E2(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 8, 6, 0, 1, 22, 58, 24, 0, 1, 52, 328, 444, 120, 0, 1, 114, 1452, 4400, 3708, 720, 0, 1, 240, 5610, 32120, 58140, 33984, 5040, 0, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
Offset: 0

Views

Author

Peter Luschny, Feb 05 2021

Keywords

Comments

The second-order Eulerian number E2(n, k) is the number of Stirling permutations of order n with exactly k descents; here the last index is defined to be a descent. More formally, let Q_n denote the set of permutations of the multiset {1,1,2,2, ..., n,n} in which, for all j, all entries between two occurrences of j are larger than j, then E2(n, k) = card({s in Q_n with des(s) = k}), where des(s) = card({j: s(j) > s(j+1)}) is the number of descents of s.
Also the number of Riordan trapezoidal words of length n with k distinct letters (see Riordan 1976, p. 9).
Also the number of rooted plane trees on n + 1 vertices with k leaves (see Janson 2008, p. 543).
Let b(n) = (1/2)*Sum_{k=0..n-1} (-1)^k*E2(n-1, k+1) / C(2*n-1, k+1). Apparently b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1}F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials (A278075). (See Rzadkowski and Urlinska, example 4.)

Examples

			Triangle starts:
  [0] 1;
  [1] 0, 1;
  [2] 0, 1, 2;
  [3] 0, 1, 8,    6;
  [4] 0, 1, 22,   58,    24;
  [5] 0, 1, 52,   328,   444,     120;
  [6] 0, 1, 114,  1452,  4400,    3708,    720;
  [7] 0, 1, 240,  5610,  32120,   58140,   33984,    5040;
  [8] 0, 1, 494,  19950, 195800,  644020,  785304,   341136,   40320;
  [9] 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880.
To illustrate the generating function for row 3: The expansion of (1 - x)^7*(x*exp(-x) + 16*x^2*exp(-x)^2 + (243*x^3*exp(-x)^3)/2) gives the polynomial x + 8*x^2 + 6*x^3. The coefficients of this polynomial give row 3.
.
Stirling permutations of order 3 with exactly k descents: (When counting the descents one may assume an invisible '0' appended to the permutations.)
  T[3, k=0]:
  T[3, k=1]: 112233;
  T[3, k=2]: 331122; 223311; 221133; 133122; 122331; 122133; 113322; 112332;
  T[3, k=3]: 332211; 331221; 233211; 221331; 133221; 123321.

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.

Links

Crossrefs

Indexing the second-order Eulerian numbers comes in three flavors: A008517 (following Riordan and Comtet), A201637 (following Graham, Knuth, and Patashnik) and this indexing, extending the definition of Gessel and Stanley. (A008517 is the main entry of the numbers.) The corresponding triangles of the first-order Eulerian numbers are A008292, A173018, and A123125.
Row reversed: A163936 (with offset = 0).
Values: E2poly(n, 1) = A001147(n), E2poly(n, -1) ~ -A001662(n+1), E2poly(n, 2) = A112487(n), 2^n*E2poly(n, 1/2) = A000311(n+1), 2^n*E2poly(n, -1/2) = A341106(n).
Diagonals: A000142, A002538, A002539, A112008, A112485.
Columns: A000007, A000012, A005803, A004301, A006260.
Cf. A008299, A137375, A008306, A106828, A269939, A278075, A331998.

Programs

  • Maple
    # Using the recurrence:
E2 := proc(n, k) option remember;
if k = 0 and n = 0 then return 1 fi; if n < 0 then return 0 fi;
E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) end: seq(seq(E2(n, k), k = 0..n), n = 0..9);
# Using the row generating function:
E2egf := n -> (1-x)^(2*n+1)*add(k^(n+k)/k!*(x*exp(-x))^k, k=0..n);
T := (n, k) -> coeftayl(E2egf(n), x=0, k): seq(print(seq(T(n, j),j=0..n)), n=0..7);
# Using the built-in function:
E2 := (n, k) -> `if`(k=0, k^n, combinat:-eulerian2(n, k-1)):
# Using the compositional inverse (series reversion):
E2triangle := proc(N) local r, s, C; Order := N + 2;
s := solve(y = series(x - t*(exp(x) - 1), x), x):
r := n -> -n!*(t - 1)^(2*n - 1)*coeff(s, y, n); C := [seq(expand(r(n)), n = 1..N)];
seq(print(seq(coeff(C[n+1], t, k), k = 0..n)), n = 0..N-1) end: E2triangle(10);
  • Mathematica
    T[n_, k_] := T[n, k] = If[k == 0, Boole[n == 0], If[n < 0, 0, k T[n - 1, k] + (2 n - k) T[n - 1, k - 1]]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
(* Via row polynomials: *)
E2poly[n_] := If[n == 0, 1,
  Expand[Simplify[x (x - 1)^(2 n) D[((1 - x)^(1 - 2 n) E2poly[n - 1]), x]]]];
Table[CoefficientList[E2poly[n], x], {n, 0, 9}] // Flatten
(* Series reversion *)
Revert[gf_, len_] := Module[{S = InverseSeries[Series[gf, {x, 0, len + 1}], x]},
Table[CoefficientList[(n + 1)! (1 - t)^(2 n + 1) Coefficient[S, x, n + 1], t],
{n, 0, len}] // Flatten]; Revert[x + t - t Exp[x], 6]
  • PARI
    E2poly(n) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1)));
{ for(n = 0, 9, print(Vecrev(E2poly(n)))) }
  • PARI
    T(n, k) = sum(j=0, n-k, (-1)^(n-j)*binomial(2*n+1, j)*stirling(2*n-k-j+1, n-k-j+1, 1)); \\ Michel Marcus, Feb 11 2021
  • SageMath
    # See also link to notebook.
@cached_function
def E2(n, k):
    if n < 0: return 0
    if k == 0: return k^n
    return k * E2(n - 1, k) + (2*n - k) * E2(n - 1, k - 1)  # Peter Luschny, Mar 08 2025

Formula

E2(n, k) = E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) for n > 0 and 0 <= k <= n, and E2(0, 0) = 1; in all other cases E(n, k) = 0.
E2(n, k) = Sum_{j=0..n-k}(-1)^(n-j)*binomial(2*n+1, j)*Stirling1(2*n-k-j+1, n-k-j+1).
E2(n, k) = Sum_{j=0..k}(-1)^(k-j)*binomial(2*n + 1, k - j)*Stirling2(n + j, j).
Stirling1(x, x - n) = (-1)^n*Sum_{k=0..n} E2(n, k)*binomial(x + k - 1, 2*n).
Stirling2(x, x - n) = Sum_{k=0..n} E2(n, k)*binomial(x + n - k, 2*n).
E2poly(n, x) = Sum_{k=0..n} E2(n, k)*x^k, as row polynomials.
E2poly(n, x) = x*(x-1)^(2*n)*d_{x}((1-x)^(1-2*n)*E2poly(n-1)) for n>=1 and E2poly(0)=1.
E2poly(n, x) = (1 - x)^(2*n + 1)*Sum_{k=0..n}(k^(n + k)/k!)*(x*exp(-x))^k.
W(n, k) = [x^k] (1+x)^n*E2poly(n, x/(1 + x)) are the Ward numbers A269939.
E2(n, k) = [x^k] (1-x)^n*Wpoly(n, x/(1 - x)); Wpoly(n, x) = Sum_{k=0..n}W(n, k)*x^k.
W(n, k) = Sum_{j=0..k} E2(n, j)*binomial(n - j, n - k).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*W(n, j)*binomial(n - j, k - j).
The compositional inverse with respect to x of x - t*(exp(x) - 1) (see B. Drake):
T(n, k) = [t^k](n+1)!*(1-t)^(2*n+1)*[x^(n+1)] InverseSeries(x - t*(exp(x)-1), x).
AS1(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, j+1), where AS1(n, k) are the associated Stirling numbers of the first kind (A008306, A106828).
E2(n, k) = Sum_{j=0..n-k+1} (-1)^(n-k-j+1)*AS1(n+j, j)*binomial(n-j, n-k-j+1), for n >= 1.
AS2(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, n-k-j) for n >=1, where AS2(n, k) are the associated Stirling numbers of the second kind (A008299, A137375).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*AS2(n + j, j)*binomial(n - j, k - j).

A163936 Triangle related to the o.g.f.s. of the right-hand columns of A130534 (E(x,m=1,n)).

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 6, 8, 1, 0, 24, 58, 22, 1, 0, 120, 444, 328, 52, 1, 0, 720, 3708, 4400, 1452, 114, 1, 0, 5040, 33984, 58140, 32120, 5610, 240, 1, 0, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 0, 362880, 3733920, 11026296, 12440064, 5765500, 1062500
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The asymptotic expansions of the higher-order exponential integral E(x,m=1,n) lead to triangle A130524, see A163931 for information on E(x,m,n). The o.g.f.s. of the right-hand columns of triangle A130534 have a nice structure: gf(p) = W1(z,p)/(1-z)^(2*p-1) with p = 1 for the first right-hand column, p = 2 for the second right-hand column, etc. The coefficients of the W1(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. Our triangle is the same as A112007 with an extra right-hand column, see also the second Eulerian triangle A008517. The row sums of our triangle lead to A001147.
We observe that the row sums of the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4) for z=1 lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four left-hand columns of the triangle of the Bessel coefficients A001497 or, if one wishes, the right-hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next left- (right-) hand columns of A001497 (A001498). An interesting phenomenon.
If one assumes the triangle not (1,1) based but (0,0) based, one has T(n, k) = E2(n, n-k), where E2(n, k) are the second-order Eulerian numbers A340556. - Peter Luschny, Feb 12 2021

Examples

			Triangle starts:
[ 1]      1;
[ 2]      1,       0;
[ 3]      2,       1,      0;
[ 4]      6,       8,      1,      0;
[ 5]     24,      58,     22,      1,      0;
[ 6]    120,     444,    328,     52,      1,     0;
[ 7]    720,    3708,   4400,   1452,    114,     1,   0;
[ 8]   5040,   33984,  58140,  32120,   5610,   240,   1,  0;
[ 9]  40320,  341136, 785304, 644020, 195800, 19950, 494,  1, 0;
The first few W1(z,p) polynomials are
W1(z,p=1) = 1/(1-z);
W1(z,p=2) = (1 + 0*z)/(1-z)^3;
W1(z,p=3) = (2 + 1*z + 0*z^2)/(1-z)^5;
W1(z,p=4) = (6 + 8*z + 1*z^2 + 0*z^3)/(1-z)^7.

Links

Crossrefs

Row sums equal A001147.
A000142, A002538, A002539, A112008, A112485 are the first few left hand columns.
A000007, A000012, A005803(n+2), A004301, A006260 are the first few right hand columns.
Cf. A163931 (E(x,m,n)), A048994 (Stirling1) and A008517 (Euler).
Cf. A112007, A163937 (E(x,m=2,n)), A163938 (E(x,m=3,n)) and A163939 (E(x,m=4,n)).
Cf. A001497 (Bessel), A001498 (Bessel), A001147 (m=1), A001147 (m=2), A001879 (m=3) and A000457 (m=4), A001880 (m=5), A001881 (m=6) and A038121 (m=7).
Cf. A340556.

Programs

  • Maple
    with(combinat): a := proc(n, m): add((-1)^(n+k+1)*binomial(2*n-1, k)*stirling1(m+n-k-1, m-k), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..9);  # Johannes W. Meijer, revised Nov 27 2012
  • Mathematica
    Table[Sum[(-1)^(n + k + 1)*Binomial[2*n - 1, k]*StirlingS1[m + n - k - 1, m - k], {k, 0, m - 1}], {n, 1, 10}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
  • PARI
    for(n=1,10, for(m=1,n, print1(sum(k=0,m-1,(-1)^(n+k+1)* binomial(2*n-1,k)*stirling(m+n-k-1,m-k, 1)), ", "))) \\ G. C. Greubel, Aug 13 2017
  • PARI
    \\ assuming offset = 0:
E2poly(n,x) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1,x)));
{ for(n = 0, 9, print(Vec(E2poly(n,x)))) } \\ Peter Luschny, Feb 12 2021

Formula

a(n, m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(2*n-1,k)*Stirling1(m+n-k-1,m-k), for 1 <= m <= n.
Assuming offset = 0 the T(n, k) are the coefficients of recursively defined polynomials. T(n, k) = [x^k] x^n*E2poly(n, 1/x), where E2poly(n, x) = x*(x - 1)^(2*n)*d_{x}((1 - x)^(1 - 2*n)*E2poly(n - 1, x))) for n >= 1 and E2poly(0, x) = 1. - Peter Luschny, Feb 12 2021

A187059 The exponent of highest power of 2 dividing the product of the elements of the n-th row of Pascal's triangle (A001142).

Original entry on oeis.org

0, 0, 1, 0, 5, 2, 4, 0, 17, 10, 12, 4, 18, 8, 11, 0, 49, 34, 36, 20, 42, 24, 27, 8, 58, 36, 39, 16, 47, 22, 26, 0, 129, 98, 100, 68, 106, 72, 75, 40, 122, 84, 87, 48, 95, 54, 58, 16, 162, 116, 119, 72, 127, 78, 82, 32, 147, 94, 98, 44, 108, 52, 57, 0, 321, 258, 260, 196, 266, 200, 203, 136, 282, 212, 215, 144, 223, 150, 154, 80, 322, 244, 247, 168, 255, 174, 178, 96, 275, 190, 194, 108, 204, 116, 121, 32, 418, 324, 327, 232, 335
Offset: 0

Views

Author

Bruce Reznick, Mar 05 2011

Keywords

Comments

The exponent of the highest power of 2 which divides Product_{k=0..n} binomial(n, k). This can be computed using de Polignac's formula.
This is the function ord_2(Ḡ_n) extensively studied in Lagarias-Mehta (2014), and plotted in Fig. 1.1. - Antti Karttunen, Oct 22 2014

Examples

			For example, if n = 4, the power of 2 that divides 1*4*6*4*1 is 5.

References

  • I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, Wiley, 1991, pages 182, 183, 187 (Ex. 34).

Links

Crossrefs

Row sums of triangular table A065040.
Row 1 of array A249421.
Cf. A000295 (a(2^k-2)), A000337 (a(2^k)), A005803 (a(2^k-3)), A036799 (a(2^k+1)), A109363 (a(2^k-4)).
Cf. A000178, A000788, A001142, A002109, A007318, A007814, A174605, A249152, A249150, A249151, A249154, A249343, A249345, A249346, A249347.

Programs

  • Haskell
    a187059 = a007814 . a001142  -- Reinhard Zumkeller, Mar 16 2015
  • Mathematica
    a[n_] := Sum[IntegerExponent[Binomial[n, k], 2], {k, 0, n}]; Array[a, 100, 0]
A187059[n_] := Sum[#*((#+1)*2^k - n - 1) & [Floor[n/2^k]], {k, Floor[Log2[n]]}];
Array[A187059, 100, 0] (* Paolo Xausa, Feb 11 2025 *)
2*Accumulate[#] - Range[Length[#]]*# & [DigitCount[Range[0, 99], 2, 1]] (* Paolo Xausa, Feb 11 2025 *)
  • PARI
    a(n)=sum(k=0,n,valuation(binomial(n,k),2))
  • PARI
    \\ Much faster version, based on code for A065040 by Charles R Greathouse IV which if reduced even further gives the formula a(n) = 2*A000788(n) - A249154(n):
A065040(m,k) = (hammingweight(k)+hammingweight(m-k)-hammingweight(m));
A187059(n) = sum(k=0, n, A065040(n, k));
for(n=0, 4095, write("b187059.txt", n, " ", A187059(n)));
\\ Antti Karttunen, Oct 25 2014
  • Python
    def A187059(n): return (n+1)*n.bit_count()+sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1)) # Chai Wah Wu, Nov 11 2024

Formula

a(2^k-1) = 0 (19th century); a(2^k) = (k-1)*2^k+1 for k >= 1. (Use de Polignac.)
a(n) = Sum_{i=0..n} A065040(n,i) [where the entries of triangular table A065040(m,k) give the exponent of the maximal power of 2 dividing binomial coefficient A007318(m,k)].
a(n) = A007814(A001142(n)). - Jason Kimberley, Nov 02 2011
a(n) = A249152(n) - A174605(n). [Exponent of 2 in the n-th hyperfactorial minus exponent of 2 in the n-th superfactorial. Cf. for example Lagarias & Mehta paper or Peter Luschny's formula for A001142.] - Antti Karttunen, Oct 25 2014
a(n) = 2*A000788(n) - A249154(n). - Antti Karttunen, Nov 02 2014
a(n) = Sum_{i=1..n} (2*i-n-1)*v_2(i), where v_2(i) = A007814(i) is the exponent of the highest power of 2 dividing i. - Ridouane Oudra, Jun 02 2022
a(n) = Sum_{k=1..floor(log_2(n))} t*((t+1)*2^k - n - 1), where t = floor(n/(2^k)). - Paolo Xausa, Feb 11 2025, derived from Ridouane Oudra's formula above.

Extensions

Name clarified by Antti Karttunen, Oct 22 2014

A052515 Number of ordered pairs of complementary subsets of an n-set with both subsets of cardinality at least 2.

Original entry on oeis.org

0, 0, 0, 0, 6, 20, 50, 112, 238, 492, 1002, 2024, 4070, 8164, 16354, 32736, 65502, 131036, 262106, 524248, 1048534, 2097108, 4194258, 8388560, 16777166, 33554380, 67108810, 134217672, 268435398, 536870852, 1073741762
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

a(n) is the number of binary sequences of length n having at least two 0's and at least two 1's. a(4)=6 because there are six binary sequences of length four that have two or more 0's and two or more 1's: 0011, 0101, 0110, 1100, 1010, 1001. - Geoffrey Critzer, Feb 11 2009
For n>3, a(n) is also the sum of those terms from the n-th row of Pascal's triangle which also occur in A006987: 6, 10+10, 15+20+15, 21+35+35+21,... - Douglas Latimer, Apr 02 2012
From Dennis P. Walsh, Apr 09 2013: (Start)
Column 2 of triangle A200091.
Number of doubly-surjective functions f:[n]->[2].
Number of ways to distribute n different toys to 2 children so that each child gets at least 2 toys. (End)
a(n) is the number of subsets of an n-set of cardinality k with 2 <= k <= n - 2. - Rick L. Shepherd, Dec 05 2014

Links

Programs

  • Magma
    m:=35; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (Exp(x)-1-x)^2 )); [0,0,0,0] cat [Factorial(n+3)*b[n]: n in [1..m-5]]; // G. C. Greubel, May 13 2019
  • Maple
    Pairs spec := [S,{S=Prod(B,B),B=Set(Z,2 <= card)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    Join[{0,0,0}, LinearRecurrence[{4,-5,2}, {0,6,20}, 35]] (* G. C. Greubel, May 13 2019 *)
With[{nn=30},CoefficientList[Series[(Exp[x]-x-1)^2,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, May 29 2023 *)
  • PARI
    concat([0,0,0,0],Vec((6-4*x)/(1-x)^2/(1-2*x)+O(x^35))) \\ Charles R Greathouse IV, Apr 03 2012
  • PARI
    x='x+O('x^35); concat([0,0,0,0],Vec(serlaplace((exp(x)-x-1)^2))) \\ Joerg Arndt, Apr 10 2013
  • Sage
    (2*x^4*(3-2*x)/((1-x)^2*(1-2*x))).series(x, 35).coefficients(x, sparse=False) # G. C. Greubel, May 13 2019

Formula

E.g.f.: (exp(x) - x - 1)^2. - Joerg Arndt, Apr 10 2013
n*a(n+2) - (1+3*n)*a(n+1) + 2(1+n)*a(n) = 0, with a(0) = .. = a(3) = 0, a(4) = 6.
For n>2, a(n) = 2^n - 2n - 2 = A005803(n) - 2 = A070313(n) - 1 = A071099(n) - A071099(n+1) + 1 = 2*A000247(n-1). - Ralf Stephan, Jan 11 2004
G.f.: 2*x^4*(3-2*x)/((1-x)^2*(1-2*x)). - Colin Barker, Feb 19 2012

Extensions

More terms from Ralf Stephan, Jan 11 2004
Definition corrected by Rainer Rosenthal, Feb 12 2010
Definition further clarified by Rick L. Shepherd, Dec 05 2014

A183892 T(n,k)=Number of arrangements of n+3 numbers in 0..k with each number being the sum mod (k+1) of three others.

Original entry on oeis.org

8, 1, 22, 16, 121, 52, 1, 292, 631, 114, 8, 241, 2574, 2059, 240, 1, 982, 8681, 13344, 6399, 494, 16, 1, 20152, 63645, 59518, 19483, 1004, 1, 492, 32821, 201154, 363749, 250124, 58807, 2026, 8, 241, 57614, 542557, 1466628, 1905097, 1024174, 176859
Offset: 1

Views

Author

R. H. Hardin Jan 07 2011

Keywords

Comments

Table starts
....8.......1.......16..........1...........8...........1...........16
...22.....121......292........241.........982...........1..........492
...52.....631.....2574.......8681.......20152.......32821........57614
..114....2059....13344......63645......201154......542557......1172880
..240....6399....59518.....363749.....1466628.....4978219.....13388190
..494...19483...250124....1905097.....9469778....38280865....122033732
.1004...58807..1024174....9681073....58702904...277198249...1030189438
.2026..176859..4143944...48678653...357576342..1964161189...8431580544
.4072..531103.16671934..243871365..2161032260.13808869207..68127437030
.8166.1593931.66886900.1220204441.13012476622.96809981737.547474985956

Examples

			Some solutions for n=3
..1....2....2....1....2....1....1....0....0....2....1....2....0....0....1....1
..2....0....1....1....2....2....0....2....1....2....2....0....2....0....0....2
..2....0....2....0....0....1....0....0....1....0....0....1....2....1....2....2
..2....2....0....1....1....1....2....1....1....2....1....2....0....1....2....1
..0....2....1....2....2....0....1....2....1....0....1....1....2....2....2....1
..0....2....1....1....2....2....0....0....0....2....0....2....0....1....1....2

Links

Crossrefs

Column 1 is A005803(n+3)
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