A143007 -id:A143007 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

			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

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Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), Article 13.5.4.
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CombOS - Combinatorial Object Server, Generate middle levels Gray codes
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Index entries for "core" sequences.

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).
Row sums of A059481, A008459, A152229, A158815, A205946.
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.
Cf. A000346, A001700, A001791, A008549, A032443, A073016, A097805, A126869.
Cf. A003418, A056040, A001316, A261130, A356637.

List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018

a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011

a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];

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

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 *)

A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */

A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014

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

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

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

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

A002117 Apéry's number or Apéry's constant zeta(3). Decimal expansion of zeta(3) = Sum_{m >= 1} 1/m^3.

Original entry on oeis.org

1, 2, 0, 2, 0, 5, 6, 9, 0, 3, 1, 5, 9, 5, 9, 4, 2, 8, 5, 3, 9, 9, 7, 3, 8, 1, 6, 1, 5, 1, 1, 4, 4, 9, 9, 9, 0, 7, 6, 4, 9, 8, 6, 2, 9, 2, 3, 4, 0, 4, 9, 8, 8, 8, 1, 7, 9, 2, 2, 7, 1, 5, 5, 5, 3, 4, 1, 8, 3, 8, 2, 0, 5, 7, 8, 6, 3, 1, 3, 0, 9, 0, 1, 8, 6, 4, 5, 5, 8, 7, 3, 6, 0, 9, 3, 3, 5, 2, 5, 8, 1, 4, 6, 1, 9, 9, 1, 5

Offset: 1

N. J. A. Sloane

Sometimes called Apéry's constant.
"A natural question is whether Zeta(3) is a rational multiple of Pi^3. This is not known, though in 1978 R. Apéry succeeded in proving that Zeta(3) is irrational. In Chapter 8 we pointed out that the probability that two random integers are relatively prime is 6/Pi^2, which is 1/Zeta(2). This generalizes to: The probability that k random integers are relatively prime is 1/Zeta(k) ... ." [Stan Wagon]
In 2001 Tanguy Rivoal showed that there are infinitely many odd (positive) integers at which zeta is irrational, including at least one value j in the range 5 <= j <= 21 (refined the same year by Zudilin to 5 <= j <= 11), at which zeta(j) is irrational. See the Rivoal link for further information and references.
The reciprocal of this constant is the probability that three integers chosen randomly using uniform distribution are relatively prime. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 13 2005
Also the value of zeta(1,2), the double zeta-function of arguments 1 and 2. - R. J. Mathar, Oct 10 2011
Also the length of minimal spanning tree for large complete graph with uniform random edge lengths between 0 and 1, cf. link to John Baez's comment. - M. F. Hasler, Sep 26 2017
Sum of the inverses of the cubes (A000578). - Michael B. Porter, Nov 27 2017
This number is the average value of sigma_2(n)/n^2 where sigma_2(n) is the sum of the squares of the divisors of n. - Dimitri Papadopoulos, Jan 07 2022

			1.2020569031595942853997...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 261.
S. R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, pp. 40-53, 500.
A. Fletcher, J. C. P. Miller, L. Rosenhead, and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 84.
R. William Gosper, Strip Mining in the Abandoned Orefields of Nineteenth Century Mathematics, Computers in Mathematics (Stanford CA, 1986); Lecture Notes in Pure and Appl. Math., Dekker, New York, 125 (1990), 261-284; MR 91h:11154.
Xavier Gourdon, Analyse, Les Maths en tête, Ellipses, 1994, Exemple 3, page 224.
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section F17, Series associated with the zeta-function, p. 391.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press; 6 edition (2008), pp. 47, 268-269.
Paul Levrie, The Ubiquitous Apéry Number, Math. Intelligencer, Vol. 45, No. 2, 2023, pp. 118-119.
A. A. Markoff, Mémoire sur la transformation de séries peu convergentes en séries très convergentes, Mém. de l'Acad. Imp. Sci. de St. Pétersbourg, XXXVII, 1890.
Paul J. Nahin, In Pursuit of Zeta-3, Princeton University Press, 2021.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Stan Wagon, Mathematica In Action, W. H. Freeman and Company, NY, 1991, page 354.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 33.
A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with Elementary Solutions, Dover (1987), Ex. 92-93.

Harry J. Smith, Table of n, a(n) for n = 1..20002
T. Amdeberhan, Faster and Faster convergent series for zeta(3), arXiv:math/9804126 [math.CO], 1998.
Kunihiro Aoki and Ryo Furue, A model for the size distribution of marine microplastics: a statistical mechanics approach, arXiv:2103.10221 [physics.ao-ph], 2021.
Peter Bala, New series for old functions.
Peter Bala, Some series for zeta(3), Nov 2023.
John Baez, Comments about zeta(3), Azimuth Project blog, August 2017.
R. Barbieri, J. A. Mignaco, and E. Remiddi, Electron form factors up to fourth order. I., Il Nuovo Cim. 11A (4) (1972) 824-864, table II (7), (9), (19).
F. Beukers, A Note on the Irrationality of zeta(2) and zeta(3), Bull. London Math. Soc., 11 (3) (1979): 268-272.
J. Borwein and D. Bradley, Empirically determined Apéry-like formulas for zeta(4n+3), arXiv:math/0505124 [math.CA], 2005.
Mainendra Kumar Dewangan and Subhra Datta, Effective permeability tensor of confined flows with wall grooves of arbitrary shape, J. of Fluid Mechanics (2020) Vol. 891.
Dr. Math, Probability of Random Numbers Being Coprime.
L. Euler, On the sums of series of reciprocals, arXiv:math/0506415 [math.HO], 2005-2008.
L. Euler, De summis serierum reciprocarum, E41.
X. Gourdon and P. Sebah, The Apery's constant: zeta(3).
Brady Haran and Tony Padilla, Apéry's constant (calculated with Twitter), Numberphile video (2017).
W. Janous, Around Apéry's constant, J. Inequ. Pure Appl. Math. 7(1) (2006), #35.
Yasuyuki Kachi and Pavlos Tzermias, Infinite products involving zeta(3) and Catalan's constant, Journal of Integer Sequences, 15 (2012), #12.9.4.
Masato Kobayashi, Integral representations for zeta(3) with the inverse sine function, arXiv:2108.01247 [math.NT], 2021.
M. Kondratiewa and S. Sadov, Markov's transformation of series and the WZ method, arXiv:math/0405592 [math.CA], 2004.
Tobias Kyrion, A closed-form expression for zeta(3), arXiv:2008.05573 [math.GM], 2020.
John Landen, Mathematical memoirs respecting a variety of subjects Vol. I, London, 1780.
F. M. S. Lima, Approximate expressions for mathematical constants from PSLQ algorithm: a simple approach and a case study, arXiv:0910.2684 [math.NT], 2009-2012.
Jonah Lissner, Theoretical Physics Utilizations Of Riemann Zeta Function Odd Positive Integer Three, ResearchGate (2024).
Junqi Liu, Jujian Zhang, and Lihong Zhi, A formal proof of the irrationality of ζ(3) in Lean 4, arXiv preprint (2025). arXiv:2503.07625 [math.NT]
C. Lupu and D. Orr, Series representations for the Apéry constant zeta(3) involving the values zeta(2n), Ramanujan J. 48(3) (2019), 477-494.
R. J. Mathar, Yet another table of integrals, arXiv:1207.5845 [math.CA], 2012-2014.
G. P. Michon, Roger Apéry, Numericana.
S. D. Miller, An Easier Way to Show zeta(3) is Irrational.
Michael Penn, Euler's harmonic number identity, YouTube video, 2020.
Simon Plouffe, Zeta(3) or Apéry's constant to 2000 places.
Simon Plouffe, Zeta(2) to Zeta(4096) to 2048 digits each (gzipped file).
A. van der Poorten, A Proof that Euler Missed.
Tanguy Rivoal, Irrationality of the zeta Function on Odd Integers [ps file].
Tanguy Rivoal, Irrationality of the zeta Function on Odd Integers [pdf file].
Ernst E. Scheufens, From Fourier series to rapidly convergent series for zeta(3), Mathematics Magazine, Vol. 84, No. 1 (2011), pp. 26-32.
G. Villemin's Almanach of Numbers, Constante d'Apéry (in French).
S. Wedeniwski, The value of zeta(3) to 1000000 places [Gutenberg Project Etext].
S. Wedeniwski, Plouffe's Inverter, Apery's constant to 128000026 decimal digits.
S. Wedeniwski, The value of zeta(3) to 1000000 decimal digits.
Eric Weisstein's World of Mathematics, Apéry's Constant.
Eric Weisstein's World of Mathematics, Relatively Prime.
Wikipedia, Riemann zeta function.
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.
J. W. Wrench, Jr., Letter to N. J. A. Sloane, Feb 04 1971.
Wenzhe Yang, Apéry's irrationality proof, mirror symmetry and Beukers' modular forms, arXiv:1911.02608 [math.NT], 2019.
Wadim Zudilin, An elementary proof of Apéry's theorem, arXiv:math/0202159 [math.NT], 2002.
Index entries for zeta function.

Cf. A013631, A013679, A013661, A013663, A013667, A013669, A013671, A013675, A013677, A059956 (6/Pi^2), A084225; A084226.
Cf. A197070: 3*zeta(3)/4; A233090: 5*zeta(3)/8; A233091: 7*zeta(3)/8.
Cf. A001008, A002805, A143003, A143007.
Cf. A000578 (cubes).
Cf. sums of inverses: A152623 (tetrahedral numbers), A175577 (octahedral numbers), A295421 (dodecahedral numbers), A175578 (icosahedral numbers).

L:=RiemannZeta(: Precision:=100); Evaluate(L,3); // G. C. Greubel, Aug 21 2018

# Calculates an approximation with n exact decimal places (small deviation
# in the last digits are possible). Goes back to ideas of A. A. Markoff 1890.
zeta3 := proc(n) local s, w, v, k; s := 0; w := -1; v := 4;
for k from 2 by 2 to 7*n/2 do
    w := -w*v/k;
    v := v + 8;
    s := s + 1/(w*k^3);
od; 20*s; evalf(%, n) end:
zeta3(10000); # Peter Luschny, Jun 10 2020

RealDigits[ N[ Zeta[3], 100] ] [ [1] ]
(* Second program (historical interest): *)
d[n_] := 34*n^3 + 51*n^2 + 27*n + 5; 6/Fold[Function[d[#2-1] - #2^6/#1], 5, Reverse[Range[100]]] // N[#, 108]& // RealDigits // First
(* Jean-François Alcover, Sep 19 2014, after Apéry's continued fraction *)

fpprec : 100$ ev(bfloat(zeta(3)))$ bfloat(%); /* Martin Ettl, Oct 21 2012 */

default(realprecision, 20080); x=zeta(3); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002117.txt", n, " ", d)); \\ Harry J. Smith, Apr 19 2009

from mpmath import mp, apery
mp.dps=109
print([int(z) for z in list(str(apery).replace('.', ''))[:-1]]) # Indranil Ghosh, Jul 08 2017

Lima gives an approximation to zeta(3) as (236*log(2)^3)/197 - 283/394*Pi*log(2)^2 + 11/394*Pi^2*log(2) + 209/394*log(sqrt(2) + 1)^3 - 5/197 + (93*Catalan*Pi)/197. - Jonathan Vos Post, Oct 14 2009 [Corrected by Wouter Meeussen, Apr 04 2010]
zeta(3) = 5/2*Integral_(x=0..2*log((1+sqrt(5))/2), x^2/(exp(x)-1)) + 10/3*(log((1+sqrt(5))/2))^3. - Seiichi Kirikami, Aug 12 2011
zeta(3) = -4/3*Integral_{x=0..1} log(x)/x*log(1+x) = Integral_{x=0..1} log(x)/x*log(1-x) = -4/7*Integral_{x=0..1} log(x)/x*log((1+x)/(1-x)) = 4*Integral_{x=0..1} 1/x*log(1+x)^2 = 1/2*Integral_{x=0..1} 1/x*log(1-x)^2 = -16/7*Integral_{x=0..Pi/2} x*log(2*cos(x)) = -4/Pi*Integral_{x=0..Pi/2} x^2*log(2*cos(x)). - Jean-François Alcover, Apr 02 2013, after R. J. Mathar
From Peter Bala, Dec 04 2013: (Start)
zeta(3) = (16/7)*Sum_{k even} (k^3 + k^5)/(k^2 - 1)^4.
zeta(3) - 1 = Sum_{k >= 1} 1/(k^3 + 4*k^7) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n - 1)^6/((2*n - 1)*(n^2 - n + 5) - ...))))) (continued fraction).
More generally, there is a sequence of polynomials P(n,x) (of degree 2*n) such that
zeta(3) - Sum_{k = 1..n} 1/k^3 = Sum_{k >= 1} 1/( k^3*P(n,k-1)*P(n,k) ) = 1/((2*n^2 + 2*n + 1) - 1^6/(3*(2*n^2 + 2*n + 3) - 2^6/(5*(2*n^2 + 2*n + 7) - 3^6/(7*(2*n^2 + 2*n + 13) - ...)))) (continued fraction). See A143003 and A143007 for details.
Series acceleration formulas:
zeta(3) = (5/2)*Sum_{n >= 1} (-1)^(n+1)/( n^3*binomial(2*n,n) )
= (5/2)*Sum_{n >= 1} P(n)/( (2*n(2*n - 1))^3*binomial(4*n,2*n) )
= (5/2)*Sum_{n >= 1} (-1)^(n+1)*Q(n)/( (3*n(3*n - 1)*(3*n - 2))^3*binomial(6*n,3*n) ), where P(n) = 24*n^3 + 4*n^2 - 6*n + 1 and Q(n) = 9477*n^6 - 11421*n^5 + 5265*n^4 - 1701*n^3 + 558*n^2 - 108*n + 8 (Bala, section 7). (End)
zeta(3) = Sum_{n >= 1} (A010052(n)/n^(3/2)) = Sum_{n >= 1} ( (floor(sqrt(n)) - floor(sqrt(n-1)))/n^(3/2) ). - Mikael Aaltonen, Feb 22 2015
zeta(3) = Product_{k>=1} 1/(1 - 1/prime(k)^3). - Vaclav Kotesovec, Apr 30 2020
zeta(3) = 4*(2*log(2) - 1 - 2*Sum_{k>=2} zeta(2*k+1)/2^(2*k+1)). - Jorge Coveiro, Jun 21 2020
zeta(3) = (4*zeta'''(1/2)*(zeta(1/2))^2-12*zeta(1/2)*zeta'(1/2)*zeta''(1/2)+8*(zeta'(1/2))^3-Pi^3*(zeta(1/2))^3)/(28*(zeta(1/2))^3). - Artur Jasinski, Jun 27 2020
zeta(3) = Sum_{k>=1} H(k)/(k+1)^2, where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - Amiram Eldar, Jul 31 2020
From Artur Jasinski, Sep 30 2020: (Start)
zeta(3) = (5/4)*Li_3(1/f^2) + Pi^2*log(f)/6 - 5*log(f)^3/6,
zeta(3) = (8/7)*Li_3(1/2) + (2/21)*Pi^2 log(2) - (4/21) log(2)^3, where f is golden ratio (A001622) and Li_3 is the polylogarithm function, formulas published by John Landen in 1780, p. 118. (End)
zeta(3) = (1/2)*Integral_{x=0..oo} x^2/(e^x-1) dx (Gourdon). - Bernard Schott, Apr 28 2021
From Peter Bala, Jan 18 2022: (Start)
zeta(3) = 1 + Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)) = 25/24 + (2!)^4*Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)*(4*n^4 + 2^4)) = 28333/27000 + (3!)^4*Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)*(4*n^4 + 2^4)*(4*n^4 + 3^4)). In general, for k >= 1, we have zeta(3) = r(k) + (k!)^4*Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)*...*(4*n^4 + k^4)), where r(k) is rational.
zeta(3) = (6/7) + (64/7)*Sum_{n >= 1} n/(4*n^2 - 1)^3.
More generally, for k >= 0, it appears that zeta(3) = a(k) + b(k)*Sum_{n >= 1} n/( (4*n^2 - 1)*(4*n^2 - 9)*...*(4*n^2 - (2*k+1)^2) )^3, where a(k) and b(k) are rational.
zeta(3) = (10/7) - (128/7)*Sum_{n >= 1} n/(4*n^2 - 1)^4.
More generally, for k >= 0, it appears that zeta(3) = c(k) + d(k)*Sum_{n >= 1} n/( (4*n^2 - 1)*(4*n^2 - 9)*...*(4*n^2 - (2*k+1)^2) )^4, where c(k) and d(k) are rational. [added Nov 27 2023: for the values of a(k), b(k), c(k) and d(k) see the Bala 2023 link, Sections 8 and 9.]
zeta(3) = 2/3 + (2^13)/(3*7)*Sum_{n >= 1} n^3/(4*n^2 - 1)^6. (End)
zeta(3) = -Psi(2)(1/2)/14 (the second derivative of digamma function evaluated at 1/2). - Artur Jasinski, Mar 18 2022
zeta(3) = -(8*Pi^2/9) * Sum_{k>=0} zeta(2*k)/((2*k+1)*(2*k+3)*4^k) = (2*Pi^2/9) * (log(2) + 2 * Sum_{k>=0} zeta(2*k)/((2*k+3)*4^k)) (Scheufens, 2011, Glasser Math. Comp. 22 1968). - Amiram Eldar, May 28 2022
zeta(3) = Sum_{k>=1} (30*k-11) / (4*(2k-1)*k^3*(binomial(2k,k))^2) (Gosper, 1986 and Richard K. Guy reference). - Bernard Schott, Jul 20 2022
zeta(3) = (4/3)*Integral_{x >= 1} x*log(x)*(1 + log(x))*log(1 + 1/x^x) dx = (2/3)*Integral_{x >= 1} x^2*log(x)^2*(1 + log(x))/(1 + x^x) dx. - Peter Bala, Nov 27 2023
zeta_3(n) = 1/180*(-360*n^3*f(-3, n/4) + Pi^3*(n^4 + 20*n^2 + 16))/(n*(n^2 + 4)), where f(-3, n) = Sum_{k>=1} 1/(k^3*(exp(Pi*k/n) - 1)). Will give at least 1 digit of precision/term, example: zeta_3(5) = 1.202056944732.... - Simon Plouffe, Dec 21 2023
zeta(3) = 1 + (1/2)*Sum_{n >= 1} (2*n + 1)/(n^3*(n + 1)^3) = 5/4 - (1/4)*Sum_{n >= 1} (2*n + 1)/(n^4*(n + 1)^4) = 147/120 + (2/15)*Sum_{n >= 1} (2*n + 1)/(n^5*(n + 1)^5) - (64/15)*Sum_{n >= 1} (n + 1)/(n^5*(n + 2)^5) = 19/16 + (128/21)*Sum_{n >= 1} (n + 1)/(n^6*(n + 2)^6) - (1/21)*Sum_{n >= 1} (2*n + 1)/(n^6*(n + 1)^6). - Peter Bala, Apr 15 2024
Equals 7*Pi^3/180 - 2*Sum_{k>=1} 1/(k^3*(exp(2*Pi*k) - 1)) [Grosswald] (see Finch). - Stefano Spezia, Nov 01 2024
Equals 10*Integral_{x=0..1/2} arcsinh(x)^2/x dx = -5*Integral_{x=0..2*log(phi)} x*log(2*sinh(x/2))dx [Munthe Hjortnaes] (see Finch). - Stefano Spezia, Nov 03 2024
Equals Li_3(1) = Integral_{x=0..1} Li_2(x)/x dx = Integral_{x=0..1} Integral_{y=0..1} Li_1(xy)/xy dydx = Integral_{x=0..1} Integral_{y=0..1} Integral_{z=0..1} Li_0(xyz)/xyz dzdydx (see Beukers), in general Integral_{x_1,...,x_k=0..1} Li_{3-k}(Product_{n=1..k} x_n)/(Product_{n=1..k} x_n) dx_k...dx_1 = zeta(3), for any k > 0. - Miko Labalan, Dec 23 2024
zeta(3) = (1/2)*Sum_{m >= 1}(Sum_{n >= 1} 1/(m*n*(m+n))). - Ricardo Bittencourt, Feb 24 2025
zeta(3) = Integral_{x=0..1} Integral_{y=0..1} Integral_{z=0..1} 1/(1 - x*y*z) dz dy dx. - Kritsada Moomuang, May 22 2025
zeta(3) = Sum_{i, j >= 1} 1/(i^2*j*binomial(i+j, i)) = Sum_{k >= 1} 1/(k + 1)^2 * Sum_{j = 1..k} 1/j = zeta(2, 1) (multiple zeta value due to Euler). - Peter Bala, Aug 05 2025

More terms from David W. Wilson
Additional comments from Robert G. Wilson v, Dec 08 2000
Quotation from Stan Wagon corrected by N. J. A. Sloane on Dec 24 2005. Thanks to Jose Brox for noticing this error.
Edited by M. F. Hasler, Sep 26 2017

A000522 Total number of ordered k-tuples (k=0..n) of distinct elements from an n-element set: a(n) = Sum_{k=0..n} n!/k!.

Original entry on oeis.org

1, 2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, 9864101, 108505112, 1302061345, 16926797486, 236975164805, 3554627472076, 56874039553217, 966858672404690, 17403456103284421, 330665665962404000, 6613313319248080001, 138879579704209680022, 3055350753492612960485, 70273067330330098091156

Offset: 0

N. J. A. Sloane

Total number of permutations of all subsets of an n-set.
Also the number of one-to-one sequences that can be formed from n distinct objects.
Old name "Total number of permutations of a set with n elements", or the same with the word "arrangements", both sound too much like A000142.
Related to number of operations of addition and multiplication to evaluate a determinant of order n by cofactor expansion - see A026243.
a(n) is also the number of paths (without loops) in the complete graph on n+2 vertices starting at one vertex v1 and ending at another v2. Example: when n=2 there are 5 paths in the complete graph with 4 vertices starting at the vertex 1 and ending at the vertex 2: (12),(132),(142),(1342),(1432) so a(2) = 5. - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001; comment corrected by Jonathan Coxhead, Mar 21 2003
Also row sums of Table A008279, which can be generated by taking the derivatives of x^k. For example, for y = 1*x^3, y' = 3x^2, y" = 6x, y'''= 6 so a(4) = 1 + 3 + 6 + 6 = 16. - Alford Arnold, Dec 15 1999
a(n) is the permanent of the n X n matrix with 2s on the diagonal and 1s elsewhere. - Yuval Dekel, Nov 01 2003
(A000166 + this_sequence)/2 = A009179, (A000166 - this_sequence)/2 = A009628.
Stirling transform of A006252(n-1) = [1,1,1,2,4,14,38,...] is a(n-1) = [1,2,5,16,65,...]. - Michael Somos, Mar 04 2004
Number of {12,12*,21*}- and {12,12*,2*1}-avoiding signed permutations in the hyperoctahedral group.
a(n) = b such that Integral_{x=0..1} x^n*exp(-x) dx = a-b*exp(-1). - Sébastien Dumortier, Mar 05 2005
a(n) is the number of permutations on [n+1] whose left-to-right record lows all occur at the start. Example: a(2) counts all permutations on [3] except 231 (the last entry is a record low but its predecessor is not). - David Callan, Jul 20 2005
a(n) is the number of permutations on [n+1] that avoid the (scattered) pattern 1-2-3|. The vertical bar means the "3" must occur at the end of the permutation. For example, 21354 is not counted by a(4): 234 is an offending subpermutation. - David Callan, Nov 02 2005
Number of deco polyominoes of height n+1 having no reentrant corners along the lower contour (i.e., no vertical step that is followed by a horizontal step). In other words, a(n)=A121579(n+1,0). A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: a(1)=2 because the only deco polyominoes of height 2 are the vertical and horizontal dominoes, having no reentrant corners along their lower contours. - Emeric Deutsch, Aug 16 2006
Unreduced numerators of partial sums of the Taylor series for e. - Jonathan Sondow, Aug 18 2006
a(n) is the number of permutations on [n+1] (written in one-line notation) for which the subsequence beginning at 1 is increasing. Example: a(2)=5 counts 123, 213, 231, 312, 321. - David Callan, Oct 06 2006
a(n) is the number of permutations (written in one-line notation) on the set [n + k], k >= 1, for which the subsequence beginning at 1,2,...,k is increasing. Example: n = 2, k = 2. a(2) = 5 counts 1234, 3124, 3412, 4123, 4312. - Peter Bala, Jul 29 2014
a(n) and (1,-2,3,-4,5,-6,7,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Nov 01 2007
Consider the subsets of the set {1,2,3,...,n} formed by the first n integers. E.g., for n = 3 we have {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}. Let the variable sbst denote a subset. For each subset sbst we determine its number of parts, that is nprts(sbst). The sum over all possible subsets is written as Sum_{sbst=subsets}. Then a(n) = Sum_{sbst=subsets} nprts(sbst)!. E.g., for n = 3 we have 1!+1!+1!+1!+2!+2!+2!+3!=16. - Thomas Wieder, Jun 17 2006
Equals row sums of triangle A158359(unsigned). - Gary W. Adamson, Mar 17 2009
Equals eigensequence of triangle A158821. - Gary W. Adamson, Mar 30 2009
For positive n, equals 1/BarnesG(n+1) times the determinant of the n X n matrix whose (i,j)-coefficient is the (i+j)th Bell number. - John M. Campbell, Oct 03 2011
a(n) is the number of n X n binary matrices with i) at most one 1 in each row and column and ii) the subset of rows that contain a 1 must also be the columns that contain a 1. Cf. A002720 where restriction ii is removed. - Geoffrey Critzer, Dec 20 2011
Number of restricted growth strings (RGS) [d(1),d(2),...,d(n)] such that d(k) <= k and d(k) <= 1 + (number of nonzero digits in prefix). The positions of nonzero digits determine the subset, and their values (decreased by 1) are the (left) inversion table (a rising factorial number) for the permutation, see example. - Joerg Arndt, Dec 09 2012
Number of a restricted growth strings (RGS) [d(0), d(1), d(2), ..., d(n)] where d(k) >= 0 and d(k) <= 1 + chg([d(0), d(1), d(2), ..., d(k-1)]) and chg(.) gives the number of changes of its argument. Replacing the function chg(.) by a function asc(.) that counts the ascents in the prefix gives A022493 (ascent sequences). - Joerg Arndt, May 10 2013
The sequence t(n) = number of i <= n such that floor(e*i!) is a square is mentioned in the abstract of Luca & Shparlinski. The values are t(n) = 0 for 0 <= n <= 2 and t(n) = 1 for at least 3 <= n <= 300. - R. J. Mathar, Jan 16 2014
a(n) = p(n,1) = q(n,1), where p and q are polynomials defined at A248664 and A248669. - Clark Kimberling, Oct 11 2014
a(n) is the number of ways at most n people can queue up at a (slow) ticket counter when one or more of the people may choose not to queue up. Note that there are C(n,k) sets of k people who quene up and k! ways to queue up. Since k can run from 0 to n, a(n) = Sum_{k=0..n} n!/(n-k)! = Sum_{k=0..n} n!/k!. For example, if n=3 and the people are A(dam), B(eth), and C(arl), a(3)=16 since there are 16 possible lineups: ABC, ACB, BAC, BCA, CAB, CBA, AB, BA, AC, CA, BC, CB, A, B, C, and empty queue. - Dennis P. Walsh, Oct 02 2015
As the row sums of A008279, Motzkin uses the abbreviated notation $n_<^\Sigma$ for a(n).
The piecewise polynomial function f defined by f(x) = a(n)*x^n/n! on each interval [ 1-1/a(n), 1-1/a(n+1) ) is continuous on [0,1) and lim_{x->1} f(x) = e. - Luc Rousseau, Oct 15 2019
a(n) is composite for 3 <= n <= 2015, but a(2016) is prime (or at least a strong pseudoprime): see Johansson link. - Robert Israel, Jul 27 2020 [a(2016) is prime, ECPP certificate generated with CM 0.4.3 and checked by factordb. - Jason H Parker, Jun 15 2025]
In general, sequences of the form a(0)=a, a(n) = n*a(n-1) + k, n>0, will have a closed form of n!*a + floor(n!*(e-1))*k. - Gary Detlefs, Oct 26 2020
From Peter Bala, Apr 03 2022: (Start)
a(2*n) is odd and a(2*n+1) is even. More generally, a(n+k) == a(n) (mod k) for all n and k. It follows that for each positive integer k, the sequence obtained by reducing a(n) modulo k is periodic, with the exact period dividing k. Various divisibility properties of the sequence follow from this; for example, a(5*n+2) == a(5*n+4) == 0 (mod 5), a(25*n+7) == a(25*n+19) == 0 (mod 25) and a(13*n+4) == a(13*n+10)== a(13*n+12) == 0 (mod 13). (End)
Number of possible ranking options on a typical ranked choice voting ballot with n candidates (allowing undervotes). - P. Christopher Staecker, May 05 2024
From Thomas Scheuerle, Dec 28 2024: (Start)
Number of decorated permutations of size n.
Number of Le-diagrams with bounding box semiperimeter n, for n > 0.
By counting over all k = 1..n and n > 0, the number of positroid cells for the totally nonnegative real Grassmannian Gr(k, n), equivalently the number of Grassmann necklaces of type (k, n). (End)

			G.f. = 1 + 2*x + 5*x^2 + 16*x^3 + 65*x^4 + 326*x^5 + 1957*x^6 + 13700*x^7 + ...
With two objects we can form 5 sequences: (), (a), (b), (a,b), (b,a), so a(2) = 5.
From _Joerg Arndt_, Dec 09 2012: (Start)
The 16 arrangements of the 3-set and their RGS (dots denote zeros) are
  [ #]       RGS        perm. of subset
  [ 1]    [ . . . ]      [  ]
  [ 2]    [ . . 1 ]      [ 3 ]
  [ 3]    [ . 1 . ]      [ 2 ]
  [ 4]    [ . 1 1 ]      [ 2 3 ]
  [ 5]    [ . 1 2 ]      [ 3 2 ]
  [ 6]    [ 1 . . ]      [ 1 ]
  [ 7]    [ 1 . 1 ]      [ 1 3 ]
  [ 8]    [ 1 . 2 ]      [ 3 1 ]
  [ 9]    [ 1 1 . ]      [ 1 2 ]
  [10]    [ 1 1 1 ]      [ 1 2 3 ]
  [11]    [ 1 1 2 ]      [ 1 3 2 ]
  [12]    [ 1 1 3 ]      [ 2 3 1 ]
  [13]    [ 1 2 . ]      [ 2 1 ]
  [14]    [ 1 2 1 ]      [ 2 1 3 ]
  [15]    [ 1 2 2 ]      [ 3 1 2 ]
  [16]    [ 1 2 3 ]      [ 3 2 1 ]
(End)

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J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 16.
D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.
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).

Alois P. Heinz, Table of n, a(n) for n = 0..450 (first 101 terms from T. D. Noe)
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Janusz Brzozowski and Marek Szykuła, Large Aperiodic Semigroups, arXiv preprint arXiv:1401.0157 [cs.FL], 2013.
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P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
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CombOS - Combinatorial Object Server, Generate partial permutations
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Jan Geuenich, Tilting modules for the Auslander algebra of K[x]/(xn), arXiv:1803.10707 [math.RT], 2018.
Henry W. Gould, Letter to N. J. A. Sloane, Nov 1973, and various attachments.
R. K. Guy, Letter to N. J. A. Sloane, 1977
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Eric Weisstein's World of Mathematics, Binomial Sums
Index entries for sequences related to logarithmic numbers
Index entries for related partition-counting sequences
Index entries for sequences related to factorial numbers

Cf. A000166, A002627, A006231, A064383, A064384, A008290, A010844, A010845, A014508, A038159, A054091, A058006, A072453, A072456, A073591, A082030, A095000, A095177, A108625, A121579, A124779, A142992, A143007, A158359, A158821, A195254, A222637-A222639, A038155, A000217.
Average of n-th row of triangle in A068424 [Corrected by N. J. A. Sloane, Feb 29 2024].
Row sums of A008279 and A094816.
First differences give A001339. Second differences give A001340.
Partial sums are in A001338, A002104.
A row of the array in A144502.
See also A370973, Nearest integer to e*n!.

import Data.List (subsequences, permutations)
a000522 = length . choices . enumFromTo 1 where
choices = concat . map permutations . subsequences
-- Reinhard Zumkeller, Feb 21 2012, Oct 25 2010

[1] cat [n eq 1 select (n+1) else n*Self(n-1)+1: n in [1..25]]; // Vincenzo Librandi, Feb 15 2015

a(n):= exp(1)*int(x^n*exp(-x)*Heaviside(x-1), x=0..infinity); # Karol A. Penson, Oct 01 2001
A000522 := n->add(n!/k!,k=0..n);
G(x):=exp(x)/(1-x): f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..20);
# Zerinvary Lajos, Apr 03 2009
G:=exp(z)/(1-z): Gser:=series(G,z=0,21):
for n from 0 to 20 do a(n):=n!*coeff(Gser,z,n): end do
# Paul Weisenhorn, May 30 2010
k := 1; series(hypergeom([1,k],[],x/(1-x))/(1-x), x=0, 20); # Mark van Hoeij, Nov 07 2011
# one more Maple program:
a:= proc(n) option remember;
      `if`(n<0, 0, 1+n*a(n-1))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, Sep 13 2019
seq(simplify(KummerU(-n, -n, 1)), n = 0..23); # Peter Luschny, May 10 2022

Table[FunctionExpand[Gamma[n + 1, 1]*E], {n, 0, 24}]
nn = 20; Accumulate[Table[1/k!, {k, 0, nn}]] Range[0, nn]! (* Jan Mangaldan, Apr 21 2013 *)
FoldList[#1*#2 + #2 &, 0, Range@ 23] + 1 (* or *)
f[n_] := Floor[E*n!]; f[0] = 1; Array[f, 20, 0] (* Robert G. Wilson v, Feb 13 2015 *)
RecurrenceTable[{a[n + 1] == (n + 1) a[n] + 1, a[0] == 1}, a, {n, 0, 12}] (* Emanuele Munarini, Apr 27 2017 *)
nxt[{n_,a_}]:={n+1,a(n+1)+1}; NestList[nxt,{0,1},30][[All,2]] (* Harvey P. Dale, Jan 29 2023 *)

a(n) := if n=0 then 1 else n*a(n-1)+1; makelist(a(n),n,0,12); /* Emanuele Munarini, Apr 27 2017 */

{a(n) = local(A); if( n<0, 0, A = vector(n+1); A[1]=1; for(k=1, n, A[k+1] = k*A[k] + 1); A[n+1])}; /* Michael Somos, Jul 01 2004 */

{a(n) = if( n<0, 0, n! * polcoeff( exp( x +x * O(x^n)) / (1 - x), n))}; /* Michael Somos, Mar 06 2004 */

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1/(1-x)^2+x^2*deriv(A)/(1-x)); polcoeff(A, n) \\ Paul D. Hanna, Sep 03 2008

{a(n)=local(X=x+x*O(x^n));polcoeff(sum(m=0,n,(m+2)^m*x^m/(1+(m+1)*X)^(m+1)),n)} /* Paul D. Hanna */

a(n)=sum(k=0,n,binomial(n,k)*k!); \\ Joerg Arndt, Dec 14 2014

# program adapted from Alois P. Heinz's Maple code in A022493
@CachedFunction
def b(n, i, t):
    if n <= 1:
        return 1
    return sum(b(n - 1, j, t + (j == i)) for j in range(t + 2))
def a(n):
    return b(n, 0, 0)
v000522 = [a(n) for n in range(33)]
print(v000522)
# Joerg Arndt, May 11 2013

a(n) = n*a(n-1) + 1, a(0) = 1.
a(n) = A007526(n-1) + 1.
a(n) = A061354(n)*A093101(n).
a(n) = n! * Sum_{k=0..n} 1/k! = n! * (e - Sum_{k>=n+1} 1/k!). - Michael Somos, Mar 26 1999
a(0) = 1; for n > 0, a(n) = floor(e*n!).
E.g.f.: exp(x)/(1-x).
a(n) = 1 + Sum_{n>=k>=j>=0} (k-j+1)*k!/j! = a(n-1) + A001339(n-1) = A007526(n) + 1. Binomial transformation of n!, i.e., A000142. - Henry Bottomley, Jun 04 2001
a(n) = floor(2/(n+1))*A009578(n+1)-1. - Emeric Deutsch, Oct 24 2001
Integral representation as n-th moment of a nonnegative function on a positive half-axis: a(n) = e*Integral_{x>=0} x^n*e^(-x)*Heaviside(x-1) dx. - Karol A. Penson, Oct 01 2001
Formula, in Mathematica notation: Special values of Laguerre polynomials, a(n)=(-1)^n*n!*LaguerreL[n, -1-n, 1], n=1, 2, ... . This relation cannot be checked by Maple, as it appears that Maple does not incorporate Laguerre polynomials with second index equal to negative integers. It does check with Mathematica. - Karol A. Penson and Pawel Blasiak ( blasiak(AT)lptl.jussieu.fr), Feb 13 2004
G.f.: Sum_{k>=0} k!*x^k/(1-x)^(k+1). a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*k^(n-k)*(k+1)^k. - Vladeta Jovovic, Aug 18 2002
a(n) = Sum_{k=0..n} A008290(n, k)*2^k. - Philippe Deléham, Dec 12 2003
a(n) = Sum_{k=0..n} A046716(n, k). - Philippe Deléham, Jun 12 2004
a(n) = e*Gamma(n+1,1) where Gamma(z,t) = Integral_{x>=t} e^(-x)*x^(z-1) dx is incomplete gamma function. - Michael Somos, Jul 01 2004
a(n) = Sum_{k=0..n} P(n, k). - Ross La Haye, Aug 28 2005
Given g.f. A(x), then g.f. A059115 = A(x/(1-x)). - Michael Somos, Aug 03 2006
a(n) = 1 + n + n*(n-1) + n*(n-1)*(n-2) + ... + n!. - Jonathan Sondow, Aug 18 2006
a(n) = Sum_{k=0..n} binomial(n,k) * k!; interpretation: for all k-subsets (sum), choose a subset (binomial(n,k)), and permutation of subset (k!). - Joerg Arndt, Dec 09 2012
a(n) = Integral_{x>=0} (x+1)^n*e^(-x) dx. - Gerald McGarvey, Oct 19 2006
a(n) = Sum_{k=0..n} A094816(n, k), n>=0 (row sums of Poisson-Charlier coefficient matrix). - N. J. A. Sloane, Nov 10 2007
From Tom Copeland, Nov 01 2007, Dec 10 2007: (Start)
Act on 1/(1-x) with 1/(1-xDx) = Sum_{j>=0} (xDx)^j = Sum_{j>=0} x^j*D^j*x^j = Sum_{j>=0} j!*x^j*L(j,-:xD:,0) where Lag(n,x,0) are the Laguerre polynomials of order 0, D the derivative w.r.t. x and (:xD:)^j = x^j*D^j. Truncating the operator series at the j = n term gives an o.g.f. for a(0) through a(n) consistent with Jovovic's.
These results and those of Penson and Blasiak, Arnold, Bottomley and Deleham are related by the operator A094587 (the reverse of A008279), which is the umbral equivalent of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j)*j!*x^(n-j) = Sum_{j=0..n} (n!/j!)*x^j. Umbral substitution of b(.) for x and then letting b(n)=1 for all n connects the results. See A132013 (the inverse of A094587) for a connection between these operations and 1/(1-xDx).
(End)
From Peter Bala, Jul 15 2008: (Start)
a(n) = n!*e - 1/(n + 1/(n+1 + 2/(n+2 + 3/(n+3 + ...)))).
Asymptotic result (Ramanujan): n!*e - a(n) ~ 1/n - 1/n^3 + 1/n^4 + 2/n^5 - 9/n^6 + ..., where the sequence [1,0,-1,1,2,-9,...] = [(-1)^k*A000587(k)], for k>=1.
a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). For fixed k, define the derived sequence a_k(n) = (a(n+k)-a(k))/n, n = 1,2,3,... . Then a_k(n) is also a difference divisibility sequence.
For example, the derived sequence a_0(n) is just a(n-1). The set of integer sequences satisfying the difference divisibility property forms a ring with term-wise operations of addition and multiplication.
Recurrence relations: a(0) = 1, a(n) = (n-1)*(a(n-1) + a(n-2)) + 2, for n >= 1. a(0) = 1, a(1) = 2, D-finite with recurrence: a(n) = (n+1)*a(n-1) - (n-1)*a(n-2) for n >= 2. The sequence b(n) := n! satisfies the latter recurrence with the initial conditions b(0) = 1, b(1) = 1. This leads to the finite continued fraction expansion a(n)/n! = 1/(1-1/(2-1/(3-2/(4-...-(n-1)/(n+1))))), n >= 2.
Limit_{n->oo} a(n)/n! = e = 1/(1-1/(2-1/(3-2/(4-...-n/((n+2)-...))))). This is the particular case m = 0 of the general result m!/e - d_m = (-1)^(m+1) *(1/(m+2 -1/(m+3 -2/(m+4 -3/(m+5 -...))))), where d_m denotes the m-th derangement number A000166(m).
For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to A001339 (r=1), A082030 (r=2), A095000 (r=3) and A095177 (r=4).
For the corresponding results for the constants log(2), zeta(2) and zeta(3) refer to A142992, A108625 and A143007 respectively.
(End)
G.f. satisfies: A(x) = 1/(1-x)^2 + x^2*A'(x)/(1-x). - Paul D. Hanna, Sep 03 2008
From Paul Barry, Nov 27 2009: (Start)
G.f.: 1/(1-2*x-x^2/(1-4*x-4*x^2/(1-6*x-9*x^2/(1-8*x-16*x^2/(1-10*x-25*x^2/(1-... (continued fraction);
G.f.: 1/(1-x-x/(1-x/(1-x-2*x/(1-2*x/(1-x-3*x/(1-3*x/(1-x-4*x/(1-4*x/(1-x-5*x/(1-5*x/(1-... (continued fraction).
(End)
O.g.f.: Sum_{n>=0} (n+2)^n*x^n/(1 + (n+1)*x)^(n+1). - Paul D. Hanna, Sep 19 2011
G.f. hypergeom([1,k],[],x/(1-x))/(1-x), for k=1,2,...,9 is the generating function for A000522, A001339, A082030, A095000, A095177, A096307, A096341, A095722, and A095740. - Mark van Hoeij, Nov 07 2011
G.f.: 1/U(0) where U(k) = 1 - x - x*(k+1)/(1 - x*(k+1)/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 14 2012
E.g.f.: 1/U(0) where U(k) = 1 - x/(1 - 1/(1 + (k+1)/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 16 2012
G.f.: 1/(1-x)/Q(0), where Q(k) = 1 - x/(1-x)*(k+1)/(1 - x/(1-x)*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
G.f.: 2/(1-x)/G(0), where G(k) = 1 + 1/(1 - x*(2*k+2)/(x*(2*k+3) - 1 + x*(2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: (B(x)+ 1)/(2-2*x) = Q(0)/(2-2*x), where B(x) be g.f. A006183, Q(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + (1-x)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 08 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 30 2013
E.g.f.: e^x/(1-x) = (1 - 12*x/(Q(0) + 6*x - 3*x^2))/(1-x), where Q(k) = 2*(4*k+1)*(32*k^2 + 16*k + x^2 - 6) - x^4*(4*k-1)*(4*k+7)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
G.f.: conjecture: T(0)/(1-2*x), where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
0 = a(n)*(+a(n+1) - 3*a(n+2) + a(n+3)) + a(n+1)*(+a(n+1) - a(n+3)) + a(n+2)*(+a(n+2)) for all n>=0. - Michael Somos, Jul 04 2014
From Peter Bala, Jul 29 2014: (Start)
a(n) = F(n), where the function F(x) := Integral_{0..infinity} e^(-u)*(1 + u)^x du smoothly interpolates this sequence to all real values of x. Note that F(-1) = G and for n = 2,3,... we have F(-n) = (-1)^n/(n-1)! *( A058006(n-2) - G ), where G = 0.5963473623... denotes Gompertz's constant - see A073003.
a(n) = n!*e - e*( Sum_{k >= 0} (-1)^k/((n + k + 1)*k!) ).
(End)
a(n) = hypergeometric_U(1, n+2, 1). - Peter Luschny, Nov 26 2014
a(n) ~ exp(1-n)*n^(n-1/2)*sqrt(2*Pi). - Vladimir Reshetnikov, Oct 27 2015
a(n) = A038155(n+2)/A000217(n+1). - Anton Zakharov, Sep 08 2016
a(n) = round(exp(1)*n!), n > 1 - Simon Plouffe, Jul 28 2020
a(n) = KummerU(-n, -n, 1). - Peter Luschny, May 10 2022
a(n) = (e/(2*Pi))*Integral_{x=-oo..oo} (n+1+i*x)!/(1+i*x) dx. - David Ulgenes, Apr 18 2023
Sum_{i=0..n} (-1)^(n-i) * binomial(n, i) * a(i) = n!. - Werner Schulte, Apr 03 2024

Additional comments from Michael Somos

A000172 The Franel number a(n) = Sum_{k = 0..n} binomial(n,k)^3.

Original entry on oeis.org

1, 2, 10, 56, 346, 2252, 15184, 104960, 739162, 5280932, 38165260, 278415920, 2046924400, 15148345760, 112738423360, 843126957056, 6332299624282, 47737325577620, 361077477684436, 2739270870994736, 20836827035351596, 158883473753259752, 1214171997616258240

Offset: 0

N. J. A. Sloane

Cusick gives a general method of deriving recurrences for the r-th order Franel numbers (this is the sequence of third-order Franel numbers), with floor((r+3)/2) terms.
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
An identity of V. Strehl states that a(n) = Sum_{k = 0..n} C(n,k)^2 * binomial(2*k,n). Zhi-Wei Sun conjectured that for every n = 2,3,... the polynomial f_n(x) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*k,n) * x^(n-k) is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Conjecture: a(n) == 2 (mod n^3) iff n is prime. - Gary Detlefs, Mar 22 2013
a(p) == 2 (mod p^3) for any prime p since p | C(p,k) for all k = 1,...,p-1. - Zhi-Wei Sun, Aug 14 2013
a(n) is the maximal number of totally mixed Nash equilibria in games of 3 players, each with n+1 pure options. - Raimundas Vidunas, Jan 22 2014
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Diagonal of rational functions 1/(1 - x*y - y*z - x*z - 2*x*y*z), 1/(1 - x - y - z + 4*x*y*z), 1/(1 + y + z + x*y + y*z + x*z + 2*x*y*z), 1/(1 + x + y + z + 2*(x*y + y*z + x*z) + 4*x*y*z). - Gheorghe Coserea, Jul 04 2018
a(n) is the constant term in the expansion of ((1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 27 2019
Diagonal of rational function 1 / ((1-x)*(1-y)*(1-z) - x*y*z). - Seiichi Manyama, Jul 11 2020
Named after the Swiss mathematician Jérôme Franel (1859-1939). - Amiram Eldar, Jun 15 2021
It appears that a(n) is equal to the coefficient of (x*y*z)^n in the expansion of (1 + x + y - z)^n * (1 + x - y + z)^n * (1 - x + y + z)^n. Cf. A036917. - Peter Bala, Sep 20 2021

			O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 56*x^3 + 346*x^4 + 2252*x^5 + ...
O.g.f.: A(x) = 1/(1-2*x) + 3!*x^2/(1-2*x)^4 + (6!/2!^3)*x^4/(1-2*x)^7 + (9!/3!^3)*x^6/(1-2*x)^10 + (12!/4!^3)*x^8/(1-2*x)^13 + ... - _Paul D. Hanna_, Oct 30 2010
Let g.f. A(x) = Sum_{n >= 0} a(n)*x^n/n!^3, then
A(x) = 1 + 2*x + 10*x^2/2!^3 + 56*x^3/3!^3 + 346*x^4/4!^3 + ... where
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 + ...]^2. - _Paul D. Hanna_

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H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 56.
Murray Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 193.
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).

Indranil Ghosh, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe)
Boris Adamczewski, Jason P. Bell, and Eric Delaygue, Algebraic independence of G-functions and congruences à la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
Prarit Agarwal and June Nahmgoong, Singlets in the tensor product of an arbitrary number of Adjoint representations of SU(3), arXiv:2001.10826 [math.RT], 2020.
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P. Barrucand, Problem 75-4, A Combinatorial Identity, SIAM Rev., 17 (1975), 168. [Annotated scanned copy of statement of problem]
Arnaud Beauville, Les familles stables de courbes sur P_1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, Vol.. 294 (May 24 1982), pp. 657-660.
David Callan, A combinatorial interpretation for the identity Sum_{k=0..n} binomial(n,k) Sum_{j=0..k} binomial(k,j)^3= Sum_{k=0..n} binomial(n,k)^2 binomial(2k,k), arXiv:0712.3946 [math.CO], 2007.
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Marc Chamberland and Armin Straub, Apéry Limits: Experiments and Proofs, arXiv:2011.03400 [math.NT], 2020.
Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023.
M. Coster, Email, Nov 1990
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Eric Delaygue, Arithmetic properties of Apéry-like numbers, arXiv preprint arXiv:1310.4131 [math.NT], 2013.
Robert W. Donley Jr, Directed path enumeration for semi-magic squares of size three, arXiv:2107.09463 [math.CO], 2021.
Tomislav Došlic and Darko Veljan, Logarithmic behavior of some combinatorial sequences, Discrete Math., Vol. 308, No. 11 (2008), pp. 2182--2212. MR2404544 (2009j:05019) - From _N. J. A. Sloane_, May 01 2012
Carsten Elsner, On recurrence formulae for sums involving binomial coefficients, Fib. Q., Vol. 43, No. 1 (2005), pp. 31-45.
Jeff D. Farmer and Steven C. Leth, An asymptotic formula for powers of binomial coefficients, Math. Gaz., Vol. 89, No. 516 (2005), pp. 385-391.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See A p. 2.
Darij Grinberg, Introduction to Modern Algebra (UMN Spring 2019 Math 4281 Notes), University of Minnesota (2019).
S. Hassani, J.-M. Maillard, and N. Zenine, On the diagonals of rational functions: the minimal number of variables (unabridged version), arXiv:2502.05543 [math-ph], 2025. See pp. 34-35.
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Nick Hobson, Python program for this sequence.
Lawrence Hollom, Julien Portier, and Victor Souza, Double-jump phase transition for the reverse Littlewood-Offord problem, arXiv:2503.24202 [math.CO], 2025. See p. 7.
Bradley Klee, Checking Weierstrass data, 2023.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 282.
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Guo-Shuai Mao, Proof of some congruence conjectures of Z.-H. Sun involving Apéry-like numbers, arXiv:2111.08778 [math.NT], 2021.
Guo-Shuai Mao and Yan Liu, On a congruence conjecture of Z.-W. Sun involving Franel numbers, arXiv:2111.08775 [math.NT], 2021.
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Cf. A002893, A052144, A005260, A096191, A033581, A189791. Second row of array A094424.
Cf. A181543, A006480, A141057, A000578, A007318.
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.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Column k=3 of A372307.

a000172 = sum . map a000578 . a007318_row
-- Reinhard Zumkeller, Jan 06 2013

A000172 := proc(n)
    add(binomial(n,k)^3,k=0..n) ;
end proc:
seq(A000172(n),n=0..10) ; # R. J. Mathar, Jul 26 2014
A000172_list := proc(len) series(hypergeom([], [1, 1], x)^2, x, len);
seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
A000172_list(21); # Peter Luschny, May 31 2017

Table[Sum[Binomial[n,k]^3,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Aug 24 2011 *)
Table[ HypergeometricPFQ[{-n, -n, -n}, {1, 1}, -1], {n, 0, 20}]  (* Jean-François Alcover, Jul 16 2012, after symbolic sum *)
a[n_] := Sum[ Binomial[2k, n]*Binomial[2k, k]*Binomial[2(n-k), n-k], {k, 0, n}]/2^n; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 20 2013, after Zhi-Wei Sun *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/3, 2/3, 1, 27 x^2 / (1 - 2 x)^3] / (1 - 2 x), {x, 0, n}]; (* Michael Somos, Jul 16 2014 *)

{a(n)=polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)/(1-2*x+x*O(x^n))^(3*m+1)),n)} \\ Paul D. Hanna, Oct 30 2010

{a(n)=n!^3*polcoeff(sum(m=0,n,x^m/m!^3+x*O(x^n))^2,n)} \\ Paul D. Hanna, Jan 19 2011

A000172(n)={sum(k=0,(n-1)\2,binomial(n,k)^3)*2+if(!bittest(n,0),binomial(n,n\2)^3)} \\ M. F. Hasler, Sep 21 2015

def A000172():
    x, y, n = 1, 2, 1
    while True:
        yield x
        n += 1
        x, y = y, (8*(n-1)^2*x + (7*n^2-7*n + 2)*y) // n^2
a = A000172()
[next(a) for i in range(21)]   # Peter Luschny, Oct 12 2013

A002893(n) = Sum_{m = 0..n} binomial(n, m)*a(m) [Barrucand].
Sum_{k = 0..n} C(n, k)^3 = (-1)^n*Integral_{x = 0..infinity} L_k(x)^3 exp(-x) dx. - from Askey's book, p. 43.
D-finite with recurrence (n + 1)^2*a(n+1) = (7*n^2 + 7*n + 2)*a(n) + 8*n^2*a(n-1) [Franel]. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jan 31 2001
a(n) ~ 2*3^(-1/2)*Pi^-1*n^-1*2^(3*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
O.g.f.: A(x) = Sum_{n >= 0} (3*n)!/n!^3 * x^(2*n)/(1 - 2*x)^(3*n+1). - Paul D. Hanna, Oct 30 2010
G.f.: hypergeom([1/3, 2/3], [1], 27 x^2 / (1 - 2x)^3) / (1 - 2x). - Michael Somos, Dec 17 2010
G.f.: Sum_{n >= 0} a(n)*x^n/n!^3 = [ Sum_{n >= 0} x^n/n!^3 ]^2. - Paul D. Hanna, Jan 19 2011
G.f.: A(x) = 1/(1-2*x)*(1+6*(x^2)/(G(0)-6*x^2)),
with G(k) = 3*(x^2)*(3*k+1)*(3*k+2) + ((1-2*x)^3)*((k+1)^2) - 3*(x^2)*((1-2*x)^3)*((k+1)^2)*(3*k+4)*(3*k+5)/G(k+1) ;  (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
In 2011 Zhi-Wei Sun found the formula Sum_{k = 0..n} C(2*k,n)*C(2*k,k)*C(2*(n-k),n-k) = (2^n)*a(n) and proved it via the Zeilberger algorithm. - Zhi-Wei Sun, Mar 20 2013
0 = a(n)*(a(n+1)*(-2048*a(n+2) - 3392*a(n+3) + 768*a(n+4)) + a(n+2)*(-1280*a(n+2) - 2912*a(n+3) + 744*a(n+4)) + a(n+3)*(+288*a(n+3) - 96*a(n+4))) + a(n+1)*(a(n+1)*(-704*a(n+2) - 1232*a(n+3) + 288*a(n+4)) + a(n+2)*(-560*a(n+2) - 1372*a(n+3) + 364*a(n+4)) + a(n+3)*(+154*a(n+3) - 53*a(n+4))) + a(n+2)*(a(n+2)*(+24*a(n+2) + 70*a(n+3) - 20*a(n+4)) + a(n+3)*(-11*a(n+3) + 4*a(n+4))) for all n in Z. - Michael Somos, Jul 16 2014
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^3*C(n,k)^3 = C(n,r)^3*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^2. - Peter Luschny, May 31 2017
From Gheorghe Coserea, Jul 04 2018: (Start)
a(n) = Sum_{k=0..floor(n/2)} (n+k)!/(k!^3*(n-2*k)!) * 2^(n-2*k).
G.f. y=A(x) satisfies: 0 = x*(x + 1)*(8*x - 1)*y'' + (24*x^2 + 14*x - 1)*y' + 2*(4*x + 1)*y. (End)
a(n) = [x^n] (1 - x^2)^n*P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 56. - Peter Bala, Mar 24 2022
a(n) = (2^n/(4*Pi^2)) * Integral_{x,y=0..2*Pi} (1+cos(x)+cos(y)+cos(x+y))^n dx dy = (8^n/(Pi^2)) * Integral_{x,y=0..Pi} (cos(x)*cos(y)*cos(x+y))^n dx dy (Pla, 1995). - Amiram Eldar, Jul 16 2022
a(n) = Sum_{k = 0..n} m^(n-k)*binomial(n,k)*binomial(n+2*k,n)*binomial(2*k,k) at m = -4. Cf. A081798 (m = 1), A006480 (m = 0), A124435 (m = -1), A318109 (m = -2) and A318108 (m = -3). - Peter Bala, Mar 16 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=2*(1 + 4*x)*T(x) + (-1 + 14*x + 24*x^2)*T'(x) + x*(1 + x)*(-1 + 8*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (4/243)*(1 - 8*x + 240*x^2 - 464*x^3 + 16*x^4);
g3 = -(8/19683)*(1 - 12*x - 480*x^2 + 3080*x^3 - 12072*x^4 + 4128*x^5 +
    64*x^6);
which determine an elliptic surface with four singular fibers. (End)
From Peter Bala, Oct 31 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^2 * binomial(n-1, k). Cf. A361716.
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n + 1], [1, 1], -1). (End)

A008288 Square array of Delannoy numbers D(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653

Offset: 0

N. J. A. Sloane

In the Formula section, some contributors use T(n,k) = D(n-k, k) (for 0 <= k <= n), which is the triangular version of the square array (D(n,k): n,k >= 0). Conversely, D(n,k) = T(n+k,k) for n,k >= 0. - Petros Hadjicostas, Aug 05 2020
Also called the tribonacci triangle [Alladi and Hoggatt (1977)]. - N. J. A. Sloane, Mar 23 2014
D(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), (1,1). - Joerg Arndt, Jul 01 2011 [Corrected by N. J. A. Sloane, May 30 2020]
Or, triangle read by rows of coefficients of polynomials P[n](x) defined by P[0] = 1, P[1] = x+1; for n >= 2, P[n] = (x+1)*P[n-1] + x*P[n-2].
D(n, k) is the number of k-matchings of a comb-like graph with n+k teeth. Example: D(1, 3) = 7 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has seven 3-matchings: four triples of three teeth and the three triples {Aa, Bb, CD}, {Aa, Dd, BC}, {Cc, Dd, AB}. Also D(3, 1)=7, the 1-matchings of the same graph being the seven edges: {AB}, {BC}, {CD}, {Aa}, {Bb}, {Cc}, {Dd}. - Emeric Deutsch, Jul 01 2002
Sum of n-th antidiagonal of the array D is A000129(n+1). - Reinhard Zumkeller, Dec 03 2004 [Edited by Petros Hadjicostas, Aug 05 2020 so that the counting of antidiagonals of D starts at n = 0. That is, the sum of the terms in the n-th row of the triangles T is  A000129(n+1).]
The A-sequence for this Riordan type triangle (see one of Paul Barry's comments under Formula) is A112478 and the Z-sequence the trivial: {1, 0, 0, 0, ...}. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. This leads to the recurrence for the triangle given below. - Wolfdieter Lang, Jan 21 2008
The triangle or chess sums, see A180662 for their definitions, link the Delannoy numbers with twelve different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 and Kn15 have been added. It is remarkable that all knight sums are related to the tribonacci numbers, that is, A000073 and A001590, but none of the others. - Johannes W. Meijer, Sep 22 2010
This sequence, A008288, is jointly generated with A035607 as an array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Row n, for n > 0, of Roger L. Bagula's triangle in the Example section shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the numerator of the n-th convergent of the continued fraction [k, k, k, ...], where k = sqrt(x) + 1/sqrt(x); see A230000. - Clark Kimberling, Nov 13 2013
In an n-dimensional hypercube lattice, D(n,k) gives the number of nodes situated at a Minkowski (Manhattan) distance of k from a given node. In cellular automata theory, the cells at Manhattan distance k are called the von Neumann neighborhood of radius k. For k=1, see A005843. - Dmitry Zaitsev, Dec 10 2015
These numbers appear as the coefficients of series relating spherical and bispherical harmonics, in the solutions of Laplace's equation in 3D. [Majic 2019, Eq. 22] - Matt Majic, Nov 24 2019
From Peter Bala, Feb 19 2020: (Start)
The following remarks assume an offset of 1 in the row and column indices of the triangle.
The sequence of row polynomials T(n,x), beginning with T(1,x) = x, T(2,x) = x + x^2, T(3,x) = x + 3*x^2 + x^3, ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd(T(n,x), T(m,x)) = T(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the sequence (T(n,x): n >= 1) is a divisibility sequence in the polynomial ring Z[x]; that is, if n divides m then T(n,x) divides T(m,x) in Z[x].
Let S(x) = 1 + 2*x + 6*x^2 + 22*x^3 + ... denote the o.g.f. for the large Schröder numbers A006318. The power series (x*S(x))^n, n = 2, 3, 4, ..., can be expressed as a linear combination with polynomial coefficients of S(x) and 1: (x*S(x))^n = T(n-1,-x) - T(n,-x)*S(x). The result can be extended to negative integer n if we define T(0,x) = 0 and T(-n,x) = (-1)^(n+1) * T(n,x)/x^n. Cf. A115139.
[In the previous two paragraphs, D(n,x) was replaced with T(n,x) because the contributor is referring to the rows of the triangle T(n,k), not the rows of the array D(n,k). - Petros Hadjicostas, Aug 05 2020] (End)
Named after the French amateur mathematician Henri-Auguste Delannoy (1833-1915). - Amiram Eldar, Apr 15 2021
D(i,j) = D(j,i). With this and Dmitry Zaitsev's Dec 10 2015 comment, D(i,j) can be considered the number of points at L1 distance <= i in Z^j or the number of points at L1 distance <= j in Z^i from any given point. The rows and columns of D(i,j) are the crystal ball sequences on cubic lattices. See the first example below. The n-th term in the k-th crystal ball sequence can be considered the number of points at distance <= n from any point in a k-dimensional cubic lattice, or the number of points at distance <= k from any point in an n-dimensional cubic lattice. - Shel Kaphan, Jan 01 2023 and Jan 07 2023
Dimensions of hom spaces Hom(R^{(i)}, R^{(j)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023

			The square array D(i,j) (i >= 0, j >= 0) begins:
  1, 1,  1,   1,   1,   1,    1,    1,    1,    1, ... = A000012
  1, 3,  5,   7,   9,  11,   13,   15,   17,   19, ... = A005408
  1, 5, 13,  25,  41,  61,   85,  113,  145,  181, ... = A001844
  1, 7, 25,  63, 129, 231,  377,  575,  833, 1159, ... = A001845
  1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
  ...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section:  D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
   0,   0,   0,   0,   1,   0,  11,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   1,   0,   3,   0,  13,   0,  63,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   0,   0,   1,   0,  11,   0, ...
   [Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
   1;
   1,  1;
   1,  3,   1;
   1,  5,   5,   1;
   1,  7,  13,   7,    1;
   1,  9,  25,  25,    9,    1;
   1, 11,  41,  63,   41,   11,    1;
   1, 13,  61, 129,  129,   61,   13,   1;
   1, 15,  85, 231,  321,  231,   85,  15,   1;
   1, 17, 113, 377,  681,  681,  377, 113,  17,  1;
   1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
   ... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   1,  0;
   1,  1,  0;
   1,  3,  1,  0;
   1,  5,  5,  1,  0;
   1,  7, 13,  7,  1,  0;
   1,  9, 25, 25,  9,  1, 0;
   1, 11, 41, 63, 41, 11, 1, 0;
   ...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   0, 1;
   0, 1,  1;
   0, 1,  3,  1;
   0, 1,  5,  5,  1;
   0, 1,  7, 13,  7,  1;
   0, 1,  9, 25, 25,  9,  1;
   0, 1, 11, 41, 63, 41, 11, 1;
   ... (End)

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Sums of antidiagonals: A000129 (Pell numbers).
Main diagonal: A001850 (central Delannoy numbers), which has further information and references.
A002002, A026002, and A190666 are +-k-diagonals for k=1, 2, 3 resp. - Shel Kaphan, Jan 01 2023
Rows 0..10: A000012, A005408, A001844, A001845, A001846, A001847, A001848, A001849, A008417, A008419, A008421.
See also A027618.
Cf. A059446.
Has same main diagonal as A064861. Different from A100936.
Cf. A101164, A101167, A128966, A131887, A131935.
Cf. A035607, A108625, A142979, A142992, A143007.
Read mod small primes: A211312, A211313, A211314, A211315.
Triangle sums (see the comments): A000129 (Row1); A056594 (Row2); A000073 (Kn11 & Kn21); A089068 (Kn12 & Kn22); A180668 (Kn13 & Kn23); A180669 (Kn14 & Kn24); A180670 (Kn15 & Kn25); A099463 (Kn3 & Kn4); A116404 (Fi1 & Fi2); A006498 (Ca1 & Ca2); A006498(3*n) (Ca3 & Ca4); A079972 (Gi1 & Gi2); A079972(4*n) (Gi3 & Gi4); A079973(3*n) (Ze1 & Ze2); A079973(2*n) (Ze3 & Ze4).
Cf. A102413, A128966. (D(n,1)) = A005843. Cf. A115139.

a008288 n k = a008288_tabl !! n !! k
a008288_row n = a008288_tabl !! n
a008288_tabl = map fst $ iterate
    (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                       zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Jul 21 2013

A008288 := proc(n, k) option remember; if k = 0 then 1 elif n=k then 1 else procname(n-1, k-1) + procname(n-2, k-1) + procname(n-1, k) end if; end proc: seq(seq(A008288(n,k),k=0..n), n=0..10); # triangular indices n and k
P[0]:=1; P[1]:=x+1; for n from 2 to 12 do P[n]:=expand((x+1)*P[n-1]+x*P[n-2]); lprint(P[n]); lprint(seriestolist(series(P[n],x,200))); end do:

(* Next, A008288 jointly generated with A035607 *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A008288 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A035607 *)
(* Clark Kimberling, Mar 09 2012 *)
d[n_, k_] := Binomial[n+k, k]*Hypergeometric2F1[-k, -n, -n-k, -1]; A008288 = Flatten[Table[d[n-k, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Apr 05 2012, after 3rd formula *)

from functools import cache
@cache
def delannoy_row(n: int) -> list[int]:
    if n == 0: return [1]
    if n == 1: return [1, 1]
    rov = delannoy_row(n - 2)
    row = delannoy_row(n - 1) + [1]
    for k in range(n - 1, 0, -1):
        row[k] += row[k - 1] + rov[k - 1]
    return row
for n in range(10): print(delannoy_row(n))  # Peter Luschny, Jul 30 2023

for k in range(8):  # seen as an array, read row by row
    a = lambda n: hypergeometric([-n, -k], [1], 2)
    print([simplify(a(n)) for n in range(11)]) # Peter Luschny, Nov 19 2014

D(n, 0) = 1 = D(0, n) for n >= 0; D(n, k) = D(n, k-1) + D(n-1, k-1) + D(n-1, k).
Bivariate o.g.f.: Sum_{n >= 0, k >= 0} D(n, k)*x^n*y^k = 1/(1 - x - y - x*y).
D(n, k) = Sum_{d = 0..min(n,k)} binomial(k, d)*binomial(n+k-d, k) = Sum_{d=0..min(n,k)} 2^d*binomial(n, d)*binomial(k, d). [Edited by Petros Hadjicostas, Aug 05 2020]
Seen as a triangle read by rows: T(n, 0) = T(n, n) = 1 for n >= 0 and T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-1, k), 0 < k < n and n > 1. - Reinhard Zumkeller, Dec 03 2004
Read as a number triangle, this is the Riordan array (1/(1-x), x(1+x)/(1-x)) with T(n, k) = Sum_{j=0..n-k} C(n-k, j) * C(k, j) * 2^j. - Paul Barry, Jul 18 2005
T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k). - Paul Barry, May 21 2006
Let y^k(n) be the number of Khalimsky-continuous functions f from [0,n-1] to Z such that f(0) = 0 and f(n-1) = k. Then y^k(n) = D(i,j) for i = (1/2)*(n-1-k) and j = (1/2)*(n-1+k) where n-1+k belongs to 2Z. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007
Recurrence for triangle from A-sequence (see the Wolfdieter Lang comment above): T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j), n >= 1, k >= 1. [For k > n, the sum is empty, in which case T(n,k) = 0.]
From Peter Bala, Jul 17 2008: (Start)
The n-th row of the square array is the crystal ball sequence for the product lattice A_1 x ... x A_1 (n copies). A035607 is the table of the associated coordination sequences for these lattices.
The polynomial p_n(x) := Sum {k = 0..n} 2^k * C(n,k) * C(x,k) = Sum_{k = 0..n} C(n,k) * C(x+k,n), whose values [p_n(0), p_n(1), p_n(2), ... ] give the n-th row of the square array, is the Ehrhart polynomial of the n-dimensional cross polytope (the hyperoctahedron) [Bump et al. (2000), Theorem 6].
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The reciprocity law p_n(m) = p_m(n) reflects the symmetry of the table.
The polynomial p_n(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1) - x*f(x-1) = (2*n+1)*f(x), normalized so that f(0) = 1.
These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis [Bump et al. (2000), Theorem 4]. The o.g.f. for the p_n(x) is (1 + t)^x/(1 - t)^(x + 1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ... .
The square array of Delannoy numbers has a close connection with the constant log(2). The entries in the n-th row of the array occur in the series acceleration formula log(2) = (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n) + (-1)^n * Sum_{k>=1} (-1)^(k+1)/(k*D(n,k-1)*D(n,k)). [T(n,k) was replaced with D(n,k) in the formula to agree with the beginning of the paragraph. - Petros Hadjicostas, Aug 05 2020]
For example, the fourth row of the table (n = 3) gives the series log(2) = 1 - 1/2 + 1/3 - 1/(1*1*7) + 1/(2*7*25) - 1/(3*25*63) + 1/(4*63*129) - ... . See A142979 for further details.
Also the main diagonal entries (the central Delannoy numbers) give the series acceleration formula Sum_{n>=1} 1/(n*D(n-1,n-1)*D(n,n)) = (1/2)*log(2), a result due to Burnside. [T(n,n) was replaced here with D(n,n) to agree with the previous paragraphs. - Petros Hadjicostas, Aug 05 2020]
Similar relations hold between log(2) and the crystal ball sequences of the C_n lattices A142992. For corresponding results for the constants zeta(2) and zeta(3), involving the crystal ball sequences for root lattices of type A_n and A_n x A_n, see A108625 and A143007 respectively. (End)
From Peter Bala, Oct 28 2008: (Start)
Hilbert transform of Pascal's triangle A007318 (see A145905 for the definition of this term).
D(n+a,n) = P_n(a,0;3) for all integer a such that a >= -n, where P_n(a,0;x) is the Jacobi polynomial with parameters (a,0) [Hetyei]. The related formula A(n,k) = P_k(0,n-k;3) defines the table of asymmetric Delannoy numbers, essentially A049600. (End)
Seen as a triangle read by rows: T(n, k) = Hyper2F1([k-n, -k], [1], 2). - Peter Luschny, Aug 02 2014, Oct 13 2024.
From Peter Bala, Jun 25 2015: (Start)
O.g.f. for triangle T(n,k): A(z,t) = 1/(1 - (1 + t)*z - t*z^2) = 1 + (1 + t)*z + (1 + 3*t + t^2)*z^2 + (1 + 5*t + 5*t^2 + t^3)*z^3 + ....
1 + z*d/dz(A(z,t))/A(z,t) is the o.g.f. for A102413. (End)
E.g.f. for the n-th subdiagonal of T(n,k), n >= 0, equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 4*x + 4*x^2/2) = 1 + 5*x + 13*x^2/2! + 25*x^3/3! + 41*x^4/4! + 61*x^5/5! + .... - Peter Bala, Mar 05 2017 [The n-th subdiagonal of triangle T(n,k) is the n-th row of array D(n,k).]
Let a_i(n) be multiplicative with a_i(p^e) = D(i, e), p prime and e >= 0, then Sum_{n > 0} a_i(n)/n^s = (zeta(s))^(2*i+1)/(zeta(2*s))^i for i >= 0. - Werner Schulte, Feb 14 2018
Seen as a triangle read by rows: T(n,k) = Sum_{i=0..k} binomial(n-i, i) * binomial(n-2*i, k-i) for 0 <= k <= n. - Werner Schulte, Jan 09 2019
Univariate generating function: Sum_{k >= 0} D(n,k)*z^k = (1 + z)^n/(1 - z)^(n+1). [Dziemianczuk (2013), Eq. 5.3] - Matt Majic, Nov 24 2019
(n+1)*D(n+1,k) = (2*k+1)*D(n,k) + n*D(n-1,k). [Majic (2019), Eq. 22] - Matt Majic, Nov 24 2019
For i, j >= 1, D(i,j) = D(i,j-1) + 2*Sum_{k=0..i-1} D(k,j-1), or, because D(i,j) = D(j,i), D(i,j) = D(i-1,j) + 2*Sum_{k=0..j-1} D(i-1,k). - Shel Kaphan, Jan 01 2023
Sum_{k=0..n} T(n,k)^2 = A026933(n). - R. J. Mathar, Nov 07 2023
Let S(x) = (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x) denote the g.f. of the sequence of large Schröder numbers A006318. Read as a lower triangular array, the signed n-th row polynomial R(n, -x) = 1/sqrt(1 - 6*x + x^2) *( 1/S(x)^(n+1) + (x*S(x))^(n+1) ). For example, R(4, -x) = 1 - 7*x + 13*x^2 - 7*x^3 + x^4 = 1/sqrt(1 - 6*x + x^2) * ( 1/S(x)^5 + (x*S(x))^5 ). Cf. A102413. - Peter Bala, Aug 01 2024

Expanded description from Clark Kimberling, Jun 15 1997
Additional references from Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 28 2001
Changed the notation to make the formulas more precise. - N. J. A. Sloane, Jul 01 2002

A005259 Apery (Apéry) numbers: Sum_{k=0..n} (binomial(n,k)*binomial(n+k,k))^2.

Original entry on oeis.org

1, 5, 73, 1445, 33001, 819005, 21460825, 584307365, 16367912425, 468690849005, 13657436403073, 403676083788125, 12073365010564729, 364713572395983725, 11111571997143198073, 341034504521827105445, 10534522198396293262825, 327259338516161442321485

Offset: 0

Simon Plouffe, N. J. A. Sloane, May 20 1991

Conjecture: For each n = 1,2,3,... the Apéry polynomial A_n(x) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
The expansions of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + 7038*x^4 + 144550*x^5 + ... both appear to have integer coefficients. See A267220. - Peter Bala, Jan 12 2016
Diagonal of the rational function R(x, y, z, w) = 1 / (1 - (w*x*y*z + w*x*y + w*z + x*y + x*z + y + z)); also diagonal of rational function H(x, y, z, w) = 1/(1 - w*(1+x)*(1+y)*(1+z)*(x*y*z + y*z + y + z + 1)). - Gheorghe Coserea, Jun 26 2018
Named after the French mathematician Roger Apéry (1916-1994). - Amiram Eldar, Jun 10 2021

			G.f. = 1 + 5*x + 73*x^2 + 1445*x^3 + 33001*x^4 + 819005*x^5 + 21460825*x^6 + ...
a(2) = (binomial(2,0) * binomial(2+0,0))^2 + (binomial(2,1) * binomial(2+1,1))^2 + (binomial(2,2) * binomial(2+2,2))^2 = (1*1)^2 + (2*3)^2 + (1*6)^2 = 1 + 36 + 36 = 73. - _Michael B. Porter_, Jul 14 2016

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Apéry's number or Apéry's constant zeta(3) is A002117. - N. J. A. Sloane, Jul 11 2023
Cf. A002736, A005258, A005429, A005430, A059415, A059416, A063007, A000172.
Cf. A006221, A000578, A006353.
Related to diagonal of rational functions: A268545-A268555.
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.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
Cf. A092826 (prime terms).

List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k)^2)); # Muniru A Asiru, Sep 28 2018

a005259 n = a005259_list !! n
a005259_list = 1 : 5 : zipWith div (zipWith (-)
   (tail $ zipWith (*) a006221_list a005259_list)
   (zipWith (*) (tail a000578_list) a005259_list)) (drop 2 a000578_list)
-- Reinhard Zumkeller, Mar 13 2014

[&+[Binomial(n, k) ^2 *Binomial(n+k, k)^2: k in [0..n]]:n in  [0..17]]; // Marius A. Burtea, Jan 20 2020

a := proc(n) option remember; if n=0 then 1 elif n=1 then 5 else (n^(-3))* ( (34*(n-1)^3 + 51*(n-1)^2 + 27*(n-1) +5)*a((n-1)) - (n-1)^3*a((n-1)-1)); fi; end;
# Alternative:
a := n -> hypergeom([-n, -n, 1+n, 1+n], [1, 1, 1], 1):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Jan 19 2020

Table[HypergeometricPFQ[{-n, -n, n+1, n+1}, {1,1,1}, 1],{n,0,13}] (* Jean-François Alcover, Apr 01 2011 *)
Table[Sum[(Binomial[n,k]Binomial[n+k,k])^2,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 15 2011 *)
a[ n_] := SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ 1 / (1 - t (1 + x ) (1 + y ) (1 + z ) (x y z + (y + 1) (z + 1))), {t, 0, n}], {x, 0, n}], {y, 0, n}], {z, 0, n}]; (* Michael Somos, May 14 2016 *)

a(n)=sum(k=0,n,(binomial(n,k)*binomial(n+k,k))^2) \\ Charles R Greathouse IV, Nov 20 2012

def A005259(n):
    m, g = 1, 0
    for k in range(n+1):
        g += m
        m *= ((n+k+1)*(n-k))**2
        m //=(k+1)**4
    return g # Chai Wah Wu, Oct 02 2022

D-finite with recurrence (n+1)^3*a(n+1) = (34*n^3 + 51*n^2 + 27*n + 5)*a(n) - n^3*a(n-1), n >= 1.
Representation as a special value of the hypergeometric function 4F3, in Maple notation: a(n)=hypergeom([n+1, n+1, -n, -n], [1, 1, 1], 1), n=0, 1, ... - Karol A. Penson Jul 24 2002
a(n) = Sum_{k >= 0} A063007(n, k)*A000172(k). A000172 = Franel numbers. - Philippe Deléham, Aug 14 2003
G.f.: (-1/2)*(3*x - 3 + (x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2 - 7*x + 1)*(x+1)^(-3)*(x^2 - 34*x + 1)^(1/2)+(1/2)*(x^3 + 30*x^2 - 24*x + 1)*(x+1)^(-3))^2. - Mark van Hoeij, Oct 29 2011
Let g(x, y) = 4*cos(2*x) + 8*sin(y)*cos(x) + 5 and let P(n,z) denote the Legendre polynomial of degree n. Then G. A. Edgar posted a conjecture of Alexandru Lupas that a(n) equals the double integral 1/(4*Pi^2)*int {y = -Pi..Pi} int {x = -Pi..Pi} P(n,g(x,y)) dx dy. (Added Jan 07 2015: Answered affirmatively in Math Overflow question 178790) - Peter Bala, Mar 04 2012; edited by G. A. Edgar, Dec 10 2016
a(n) ~ (1+sqrt(2))^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 01 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * C(n+k,k)^2. - Joerg Arndt, May 11 2013
0 = (-x^2+34*x^3-x^4)*y''' + (-3*x+153*x^2-6*x^3)*y'' + (-1+112*x-7*x^2)*y' + (5-x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Peter Bala, Jan 18 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j) * C(n,k)^2 * C(n+k,k)^2 * C(n,j) * C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} C(n,k) * C(n+k,k) * C(k,j)^3 (see Koepf, p. 55).
a(n) = Sum_{0 <= j, k <= n} C(n,k)^2 * C(n,j)^2 * C(3*n-j-k,2*n) (see Koepf, p. 119).
Diagonal coefficients of the rational function 1/((1 - x - y)*(1 - z - t) - x*y*z*t) (Straub, 2014). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 22 2020
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k). - Peter Bala, Jul 18 2024
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,k)^2 * C(n,j)^2 * C(k+j,k), see Labelle et al. link. - Max Alekseyev, Mar 12 2025

a(n) = A(2*n,n,2*n,n) in the notation of Straub, equation 8. It follows from Straub, Theorem 1.2, that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and all positive integers n and k.

More generally, for positive integers r and s the sequence {A143007(r*n, s*n) : n >= 0} satisfies the same supercongruences. For cases, see A005259(r = s = 1), A363865 (r = 3, s = 1) and A363866 (r = 3, s = 2).

a(n) = A(3*n,n,3*n,n) in the notation of Straub, equation 8. It follows from Straub, Theorem 1.2, that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and all positive integers n and k.

More generally, for positive integers r and s the sequence {A143007(r*n, s*n) : n >= 0} satisfies the above supercongruences. For other cases, see A005259 (r = s = 1), A363864 (r = 2, s = 1) and A363866 (r = 3, s = 2).

a(n) = A(3*n,2*n,3*n,2*n) in the notation of Straub, equation 8. It follows from Straub, Theorem 1.2, that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and all positive integers n and k.

More generally, for positive integers r and s, the sequence {A143007(r*n, s*n) : n >= 0} satisfies the above supercongruences. For other cases, see A005259 (r = s = 1), A363864 (r = 2, s = 1) and A363865 (r = 3, s = 1).

The sequence of Apéry numbers A005259 forms the main diagonal of A143007, i.e., A005259(n) = A143007(n, n). The Apéry numbers satisfy the supercongruences A005259(n*p^r) == A005259(n^p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

More generally, for positive integers r and s, the sequence defined by a(r,s;n) = A143007(r*n - 1, s*n - 1) may also satisfy the same supercongruences. This is the case r = 2, s = 1. Compare with the comments in A363864.

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A363864 a(n) = A143007(2*n,n).

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A363865 a(n) = A143007(3*n,n).

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A363866 a(n) = A143007(3*n,2*n).

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A364244 a(n) = A143007(2*n-1, n-1) for n >= 1.

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

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A002117 Apéry's number or Apéry's constant zeta(3). Decimal expansion of zeta(3) = Sum_{m >= 1} 1/m^3.

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A363866 a(n) = A143007(3n,2n).

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