A002894 -id:A002894 - 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, Riordan-Bernstein Polynomials, Hankel Transforms and Somos Sequences, Journal of Integer Sequences, Vol. 15 (2012), Article 12.8.2.
Paul Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), Article 13.5.1.
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

A006480 De Bruijn's S(3,n): (3n)!/(n!)^3.

Original entry on oeis.org

1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000

Offset: 0

N. J. A. Sloane

Number of paths of length 3n in an n X n X n grid from (0,0,0) to (n,n,n), using steps (0,0,1), (0,1,0), and (1,0,0).
Appears in Ramanujan's theory of elliptic functions of signature 3.
S(s,n) = Sum_{k=0..2n} (-1)^(k+n) * binomial(2n, k)^s. The formula S(3,n) = (3n)!/(n!)^3 is due to Dixon (according to W. N. Bailey 1935). - Charles R Greathouse IV, Dec 28 2011
a(n) is the number of ballot results that end in a 3-way tie when 3n voters each cast two votes for two out of three candidates vying for 2 slots on a county board; in such a tie, each of the three candidates receives 2n votes. Note there are C(3n,2n) ways to choose the voters who cast a vote for the youngest candidate. The n voters who did note vote for the youngest candidate voted for the two older candidates. Then there are C(2n,n) ways to choose the other n voters who voted for both the youngest and the second youngest candidate. The remaining voters vote for the oldest candidate. Hence there are C(3n,2n)*C(2n,n)=(3n)!/(n!)^3 ballot results. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X+Y+1/(X*Y))^(3*n). - Mark van Hoeij, May 07 2013
For n > 2 a(n) is divisible by (n+2)*(n+1)^2, a(n) = (n+1)^2*(n+2)*A161581(n). - Alexander Adamchuk, Dec 27 2013
a(n) is the number of permutations of the multiset {1^n, 2^n, 3^n}, the number of ternary words of length 3*n with n of each letters. - Joerg Arndt, Feb 28 2016
Diagonal of the rational function 1/(1 - x - y - z). - Gheorghe Coserea, Jul 06 2016
At least two families of elliptic curves, x = 2*H1 = (p^2+q^2)*(1-q) and x = 2*H2 = p^2+q^2-3*p^2*q+q^3 (0Bradley Klee, Feb 25 2018
The ordinary generating function also determines periods along a family of tetrahedral-symmetric sphere curves ("du troisième ordre"). Compare links to Goursat "Étude des surfaces..." and "Proof Certificate". - Bradley Klee, Sep 28 2018

			G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...

L. A. Aizenberg and A. P. Yuzhakov, "Integral representations and residues in multidimensional complex analysis", American Mathematical Society, 1983, p. 194.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 174.
N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
George E. Andrews, The well-poised thread: An Organized Chronicle of Some Amazing Summations and their Implications, Ramanujan J., 1 (1997), 7-23; see Section 8.
A. Bostan, S. Boukraa, J.-M. Maillard, and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, Journal of Physics A: Mathematical and Theoretical, Vol. 48, No. 50 (2015), 504001; arXiv preprint, arXiv:1507.03227 [math-ph], 2015.
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
Harlan J. Brothers, Pascal's Prism: Supplementary Material.
Robert M. Dickau, 3-dimensional shortest-path diagrams.
Henry W. Gould, Tables of Combinatorial Identities, Edited by J. Quaintance.
Édouard Goursat, Étude des surfaces qui admettent tous les plans de symétrie d'un polyèdre régulier, Annales scientifiques de l'École Normale Supérieure, Série 3 : Volume 4 (1887), 165-166.
Bob Hinman, Letter to N. J. A. Sloane, Aug. 1980.
Brad Klee, Geometric G.F. for Ramanujan Periods, seqfans mailing list, 2017.
Bradley Klee, Proof Certificate.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024.
Gilbert Labelle and Annie Lacasse, Closed paths whose steps are roots of unity, in FPSAC 2011, Reykjavík, Iceland DMTCS proc. AO, 2011, pp. 599-610.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, Sums of powers of Catalan triangle numbers, Discrete Mathematics, Vol. 340, No. 10 (2017), pp. 2388-2397; arXiv preprint, arXiv:1602.04347 [math.NT], 2016.
Jovan Mikić, A Method For Examining Divisibility Properties Of Some Binomial Sums, J. Int. Seq., Vol. 21 (2018), Article 18.8.7.
Michaël Moortgat, The Tamari order for D^3 and derivability in semi-associative Lambek-Grishin Calculus, 15th Workshop: Computational Logic and Applications (CLA 2020).
Trey Peck, Letter to N. J. A. Sloane, Aug. 1980.
Karol A. Penson and Allan I. Solomon, Coherent states from combinatorial sequences, in: E. Kapuscik and A. Horzela (eds.), Quantum theory and symmetries, World Scientific, 2002, pp. 527-530; arXiv preprint, arXiv:quant-ph/0111151, 2001.
Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, A K Peters, 1996, p. 22.
Srinivasa Ramanujan, Modular Equations and Approximations to Pi, Quarterly Journal of Mathematics, XLV (1914), 350-372.
Bruno Salvy, GFUN and the AGM.
Renzo Sprugnoli, Riordan array proofs of identities in Gould's book, 2006.
Dennis P. Walsh, The probability of a run-off election when three equally-favored candidates vie for two slots.
Jacques-Arthur Weil, Supplementary Material for the Paper "Diagonals of rational functions and selected differential Galois groups".
Eric Weisstein's World of Mathematics, Binomial Sums.
Wikipedia, Dixon's identity.

Cf. A000984, A001459, A008977, A050983, A050984, A091683, A161581, A181545.
Row 3 of A187783.
Related to diagonal of rational functions: A268545-A268555. Elliptic Integrals: A002894, A113424, A000897. Factors: A005809, A000984. Integrals: A007004, A024486. Sphere Curves: A318245, A318495.

List([0..20],n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018

[Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011

seq((3*n)!/(n!)^3, n=0..16); # Zerinvary Lajos, Jun 28 2007

Sum [ (-1)^(k+n) Binomial[ 2n, k ]^3, {k, 0, 2n} ]
a[ n_] := If[ n < 0, 0, (-1)^n HypergeometricPFQ[ {-2 n, -2 n, -2 n}, {1, 1}, 1]]; (* Michael Somos, Oct 22 2014 *)
Table[Multinomial[n, n, n], {n, 0, 100}] (* Emanuele Munarini, Oct 25 2016 *)
CoefficientList[Series[Hypergeometric2F1[1/3,2/3,1,27*x],{x,0,5}],x] (* Bradley Klee, Feb 28 2018 *)
Table[(3n)!/(n!)^3,{n,0,20}] (* Harvey P. Dale, Mar 09 2025 *)

makelist(multinomial_coeff(n,n,n),n,0,24); /* Emanuele Munarini, Oct 25 2016 */

{a(n) = if( n<0, 0, (3*n)! / n!^3)}; /* Michael Somos, Dec 03 2002 */

{a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while( m<=n, m*=3; A = subst( (1 + 2*x) * subst(A, x, (x/3)^3), x, serreverse(x * (1 + x + x^2) / (1 + 2*x)^3 / 3 + O(x^m)))); polcoeff(A, n))}; /* Michael Somos, Dec 03 2002 */

from math import factorial
def A006480(n): return factorial(3*n)//factorial(n)**3 # Chai Wah Wu, Oct 04 2022

Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 1/2 * sqrt(3) * 27^n / (Pi*n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Karol A. Penson, Nov 21 2001: (Start)
O.g.f.: hypergeom([1/3, 2/3], [1], 27*x).
E.g.f.: hypergeom([1/3, 2/3], [1, 1], 27*x).
Integral representation as n-th moment of a positive function on [0, 27]:
a(n) = int( x^n*(-1/24*(3*sqrt(3)*hypergeom([2/3, 2/3], [4/3], 1/27*x)* Gamma(2/3)^6*x^(1/3) - 8*hypergeom([1/3, 1/3], [2/3], 1/27*x)*Pi^3)/Pi^3 /x^(2/3)/Gamma(2/3)^3), x=0..27). This representation is unique. (End)
a(n) = Sum_{k=-n..n} (-1)^k*binomial(2*n, n+k)^3. - Benoit Cloitre, Mar 02 2005
a(n) = C(2n,n)*C(3n,n) = A104684(2n,n). - Paul Barry, Mar 14 2006
G.f. satisfies: A(x^3) = A( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x). - Paul D. Hanna, Oct 29 2010
D-finite with recurrence: n^2*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = (n+1)^2*(n+2)*A161581(n) for n>2. - Alexander Adamchuk, Dec 27 2013
0 = a(n)^2*(472392*a(n+1)^2 - 83106*a(n+1)*a(n+2) + 3600*a(n+2)^2) + a(n)*a(n+1)*(-8748*a(n+1)^2 + 1953*a(n+1)*a(n+2) - 120*a(n+2)^2) + a(n+1)^2*(36*a(n+1)^2 - 12*a(n+1)*a(n+2) + a(n+2)^2) for all n in Z. - Michael Somos, Oct 22 2014
0 = x*(27*x-1)*y'' + (54*x-1)*y' + 6*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
From Peter Bala, Jul 15 2016: (Start)
a(n) = 3*binomial(2*n - 1,n)*binomial(3*n - 1,n) = 3*[x^n] 1/(1 - x)^n * [x^n] 1/(1 - x)^(2*n) for n >= 1.
a(n) = binomial(2*n,n)*binomial(3*n,n) = ([x^n](1 + x)^(2*n)) *([x^n](1 + x)^(3*n)) = [x^n](F(x)^(6*n)), where F(x) = 1 + x + 2*x^2 + 14*x^3 + 127*x^4 + 1364*x^5 + 16219*x^6 + ... appears to have integer coefficients. Cf. A002894.
This sequence occurs as the right-hand side of several binomial sums:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)^3 = a(n) (Dixon's identity).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n,n - k)*binomial(3*n + k,k) = a(n) (Gould, Vol. 4, 6.86)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = a(n).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n + k,k)*binomial(3*n,n - k) = a(n).
Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = a(n).
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(4*n - k,2*n) = a(n) (see Gould, Vol.5, 9.23).
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = a(n) (Sprugnoli, Section 2.9, Table 10, p. 123). (End)
From Bradley Klee, Feb 28 2018: (Start)
a(n) = A005809(n)*A000984(n).
G.f.: F(x) = 1/(2*Pi) Integral_{z=0..2*Pi} 2F1(1/3,2/3; 1/2; 27*x*sin^2(z)) dz.
With G(x) = x*2F1(1/3,2/3; 2; 27*x): F(x) = d/dx G(x). (Cf. A007004) (End)
F(x)*G(1/27-x) + F(1/27-x)*G(x) = 1/(4*Pi*sqrt(3)). - Bradley Klee, Sep 29 2018
Sum_{n>=0} 1/a(n) = A091683. - Amiram Eldar, Nov 15 2020
From Peter Bala, Sep 20 2021: (Start)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)^2 * binomial(k,n). Cf. A001459.
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(3*n,2*n)*C(2*n,n) and apply Mestrovic, equation 39, p. 12). (End)
a(n) = 6*A060542(n). - R. J. Mathar, Jun 21 2023
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)^3 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^3 * binomial(2*n, n+k)^3 = x*(x + n)*(x + 2*n)*a(n) (x arbitrary). Compare with Dixon's identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^3 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Aug 14 2023: (Start)
a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
Row 1 of A364509. (End)
From Peter Bala, Oct 10 2024: (Start)
The following hold for n >= 1:
a(n) = Sum_{k = 0.. 2*n} (-1)^(n+k) * k/n * binomial(2*n, k)^3 = 3/2 * Sum_{k = 0.. 2*n} (-1)^(n+k) * (k/n)^2 * binomial(2*n, k)^3.
a(n) = 3/2 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)^2*binomial(2*n-1, k).
a(n) = 3 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)*binomial(2*n-1, k)^2. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(2*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

a(14)-a(16) from Eric W. Weisstein

Terms a(17) and beyond from T. D. Noe, Jun 29 2008

A049541 Decimal expansion of 1/Pi.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Dec 11 1999

The ratio of the volume of a regular octahedron to the volume of the circumscribed sphere (which has circumradius a*sqrt(2)/2 = a*A010503, where a is the octahedron's edge length; see MathWorld link). For similar ratios for other Platonic solids, see A165922, A165952, A165953 and A165954. - Rick L. Shepherd, Oct 01 2009

Corresponds to a gauge point marked "M" on slide rule calculating devices in the 20th century. The Pickworth reference notes its use in calculating the area of the curved surface of a cylinder. - Peter Munn, Aug 14 2020

			0.3183098861837906715377675267450287240689192914809128974953...

J.-P. Delahaye, Pi - die Story (German translation), Birkhäuser, 1999 Baasel, p. 245. French original: Le fascinant nombre Pi, Pour la Science, Paris, 1997.
C. N. Pickworth, The Slide Rule, 24th Ed., Pitman, London, 1945, p. 53, Gauge Points.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 27.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Mohammad K. Azarian, An Expression for Pi, Problem #870, College Mathematics Journal, Vol. 39, No. 1, January 2008, p. 66. Solution appeared in Vol. 40, No. 1, January 2009, pp. 62-64.
J. Bohr, Ramanujan's Method of Approximating Pi.
J. Borwein, Ramanujan's Sum.
Heng Huat Chan, Shaun Cooper, and Wen-Chin Liaw, The Rogers-Ramanujan continued fraction and a quintic iteration for 1/Pi, Proc. Amer. Math. Soc. 135 (2007), 3417-3424.
D. V. Chudnovsky and G. V. Chudnovsky, The computation of classical constants, Proc. Nati. Acad. Sci. USA, Vol. 86, pp. 8178-8182, November 1989.
J. Guillera, A New Method to Obtain Series for 1/Pi and 1/Pi^2, Experimental Mathematics, Volume 15, Issue 1, 2006.
R. Matsumoto, Ramanujan Type Series. [Broken link]
A. S. Nimbran, Deriving Forsyth-Glaisher type series for 1/π and Catalan's constant by an elementary method, The Mathematics Student, Indian Math. Soc., Vol. 84, Nos. 1-2, Jan.-June (2015), 69-86. [Broken link]
Eric W. Weisstein, Octahedron.
Index entries for transcendental numbers.

Cf. A000796, A165922, A165952, A165953, A165954, A063723, A010503. - Rick L. Shepherd, Oct 01 2009

Cf. A088538 (4/Pi).

MATLAB
```
1/pi % Altug Alkan, Apr 10 2016
```

R:= RealField(100); 1/Pi(R); // G. C. Greubel, Aug 21 2018

Digits:=100: evalf(1/Pi); # Wesley Ivan Hurt, Aug 29 2016

RealDigits[N[1/Pi, 10, 100]][[1]] (* Vladimir Joseph Stephan Orlovsky, Jun 18 2009 *)

1/Pi \\ Charles R Greathouse IV, Jun 16 2011

Equals (1/(12-16*A002162))*Sum_{n>=0} A002894(n)*H(n)/(A001025(n) * A016754(n-1)), where H(n) denotes the n-th harmonic number. - John M. Campbell, Aug 28 2016
1/Pi = Sum_{m>=0} binomial(2*m, m)^3 * (42*m+5)/(2^(12*m+4)), Ramanujan, from the J.-P. Delahaye reference. - Wolfdieter Lang, Sep 18 2018; corrected by Bernard Schott, Mar 26 2020
1/Pi = 12*Sum_{n >= 0} (-1)^n*((6*n)!/(n!^3*(3*n)!))*(13591409 + 545140134*n)/640320^(3*n + 3/2) [Chudnovsky]. - Sanjar Abrarov, Mar 31 2020
1/Pi = (sqrt(8)/9801) * Sum_{n >= 0} ((4*n)!/((n!)^4)) * (26390*n + 1103)/(396^(4*n)) [Ramanujan, 1914]. - Bernard Schott, Mar 26 2020
Equal Sum_{k>=2} tan(Pi/2^k)/2^k. - Amiram Eldar, Aug 05 2020
Floor((3/8)*Sum_{n>=1} sigma[3](n)*n/exp(Pi*n/(10^((1/5)*k+(1/5))))) mod 10, will give the k-th digit of 1/Pi. - Simon Plouffe, Dec 19 2023

A002057 Fourth convolution of Catalan numbers: a(n) = 4binomial(2n+3,n)/(n+4).

Original entry on oeis.org

1, 4, 14, 48, 165, 572, 2002, 7072, 25194, 90440, 326876, 1188640, 4345965, 15967980, 58929450, 218349120, 811985790, 3029594040, 11338026180, 42550029600, 160094486370, 603784920024, 2282138106804, 8643460269248, 32798844771700, 124680849918352

Offset: 0

N. J. A. Sloane

a(n) is sum of the (flattened) list obtained by the iteration of: replace each integer k with the list 0,...,k+1 on the starting value 0. Length of this list is Catalan(n) or A000108. - Wouter Meeussen, Nov 11 2001
a(n-2) is the number of n-th generation vertices in the tree of sequences with unit increase labeled by 3 (cf. Zoran Sunic reference). - Benoit Cloitre, Oct 07 2003
Number of standard tableaux of shape (n+2,n-1). - Emeric Deutsch, May 30 2004
a(n) = CatalanNumber(n+3) - 2*CatalanNumber(n+2). Proof. From its definition as a convolution of Catalan numbers, a(n) counts lists of 4 Dyck paths of total size (semilength) = n. Connect the 4 paths by 3 upsteps (U) and append 3 downsteps (D). This is a reversible procedure. So a(n) is also the number of Dyck (n+3)-paths that end DDD (D for downstep). Let C(n) denote CatalanNumber(n) (A000108). Since C(n+3) is the total number of Dyck (n+3)-paths and C(n+2) is the number that end UD, we have (*) C(n+3) - C(n+2) is the number of Dyck (n+3)-paths that end DD. Also, (**) C(n+2) is the number of Dyck (n+3)-paths that end UDD (change the last D in a Dyck (n+2)-path to UDD). Subtracting (**) from (*) yields a(n) = C(n+3) - 2C(n+2) as claimed. - David Callan, Nov 21 2006
Convolution square of the Catalan sequence without one of the initial "1"'s: (1 + 4x + 14x^2 + 48x^3 + ...) = (1/x^2) * square(x + 2x^2 + 5x^3 + 14x^4 + ...)
a(n) is the number of binary trees with n+3 internal nodes in which both subtrees of the root are nonempty.  Cf. A068875 [Sedgewick and Flajolet]. - Geoffrey Critzer, Jan 05 2013
With offset 4, a(n) is the number of permutations on {1,2,...,n} that are 123-avoiding, i.e., do not contain a three-term monotone subsequence, for which the first ascent is at positions (4,5); for example, there are 48 123-avoiding permutations on n=7 for which the first ascent is at spots (4,5). See Connolly link. There it is shown in general that the k-th Catalan Convolution is the number of 123-avoiding permutations for which the first ascent is at (k, k+1). (For n=k, the first ascent is defined to be at positions (k,k+1) if the permutation is the decreasing permutation with no ascents.) - Anant Godbole, Jan 17 2014
With offset 4, a(n) is the number of permutations on {1,2,...,n} that are 123-avoiding and for which the integer n is in the 4th spot; see Connolly link. - Anant Godbole, Jan 17 2014
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that have exactly one east step below the subdiagonal y = x-1. Details can be found in Section 3.1 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that bounce off the diagonal y = x to the right exactly once but do not bounce off y = x to the left. Details can be found in Section 4.2 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that horizontally cross the diagonal y = x exactly once but do not cross the diagonal vertically. Details can be found in Section 4.3 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
Apparently also Young tableaux of (non-partition) shape [n+1, 1, 1, n+1], see example file. - Joerg Arndt, Dec 30 2023

			From _Peter Bala_, Apr 14 2017: (Start)
This sequence appears on the main diagonal of a generalized Catalan triangle. Construct a lower triangular array (T(n,k)), n,k >= 0 by placing the sequence [0,0,0,1,1,1,1,...] in the first column and then filling in the remaining entries in the array using the rule T(n,k) = T(n,k-1) + T(n-1,k). The resulting array begins
  n\k| 0 1  2  3  4   5   6   7  ...
  ---+-------------------------------
   0 | 0
   1 | 0 0
   2 | 0 0  0
   3 | 1 1  1  1
   4 | 1 2  3  4  4
   5 | 1 3  6 10 14  14
   6 | 1 4 10 20 34  48  48
   7 | 1 5 15 35 69 117 165 165
   ...
(see Tedford 2011; this is essentially the array C_4(n,k) in the notation of Lee and Oh). Compare with A279004. (End)

Pierre de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 11, coefficients of P_4(z).
C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., Vol. 14 (1922), pp. 55-62, 122-138 and 143-146.
Robert Sedgewick and Phillipe Flajolet, Analysis of Algorithms, Addison Wesley, 1996, page 225.
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).

T. D. Noe and Fung Lam, Table of n, a(n) for n = 0..1000 (first 100 terms were computed by T. D. Noe)
Anwar Al Ghabra, K. Gopala Krishna, Patrick Labelle, and Vasilisa Shramchenko, Enumeration of multi-rooted plane trees, arXiv:2301.09765 [math.CO], 2023.
Joerg Arndt, The a(3)=48 Young tableaux for shape [4,1,1,4].
Andrei Asinowski and Cyril Banderier, From geometry to generating functions: rectangulations and permutations, arXiv:2401.05558 [cs.DM], 2024. See page 2.
Jean-Luc Baril and Helmut Prodinger, Enumeration of partial Lukasiewicz paths, arXiv:2205.01383 [math.CO], 2022.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Julia E. Bergner, Cedric Harper, Ryan Keller, and Mathilde Rosi-Marshall, Action graphs, planar rooted forests, and self-convolutions of the Catalan numbers, arXiv:1807.03005 [math.CO], 2018.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, Bounce statistics for rational lattice paths, arXiv:1707.09918 [math.CO], 2017, p. 9.
A. Cayley, On the partitions of a polygon, Proc. London Math. Soc., 22 (1891), 237-262 = Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 93ff.
Giulio Cerbai, Anders Claesson, and Luca Ferrari, Stack sorting with restricted stacks, arXiv:1907.08142 [cs.DS], 2019.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014 (see page 4).
Samuel Connolly, Zachary Gabor, and Anant Godbole, The location of the first ascent in a 123-avoiding permutation, arXiv:1401.2691 [math.CO], 2014.
S. J. Cyvin, J. Brunvoll, E. Brendsdal, B. N. Cyvin, and E. K. Lloyd, Enumeration of polyene hydrocarbons: a complete mathematical solution, J. Chem. Inf. Comput. Sci., Vol. 35, No. 4 (1995) pp. 743-751.
S. J. Cyvin, J. Brunvoll, E. Brendsdal, B. N. Cyvin, and E. K. Lloyd, Enumeration of polyene hydrocarbons: a complete mathematical solution, J. Chem. Inf. Comput. Sci., Vol. 35, No. 4 (1995) pp. 743-751. [Annotated scanned copy]
Olivier Danvy, Summa Summarum: Moessner's Theorem without Dynamic Programming, arXiv:2412.03127 [cs.DM], 2024. See p. 31.
Harold M. Edwards, A Normal Form for Elliptic Curves, Bulletin of the A.M.S., Vol. 44, No. 3 (2007), pp. 393-422.
Richard K. Guy, Letter to N. J. A. Sloane, May 1990
Richard K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seq., Vol. 3 (2000), Article 00.1.6.
V. E. Hoggatt, Jr., 7-page typed letter to N. J. A. Sloane with suggestions for new sequences, circa 1977.
V. E. Hoggatt, Jr. and M. Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., Vol. 14, No. 5 (1976), pp. 395-405.
C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., Vol. 14 (1922), pp. 55-62, 122-138 and 143-146. [Annotated scanned copy]
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart., Vol. 38, No. 5 (2000), pp. 408-419. See Eq.(3).
Kyu-Hwan Lee and Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Nik Lygeros and Olivier Rozier, A new solution to the equation tau(rho) == 0 (mod p), J. Int. Seq., Vol. 13 (2010), Article 10.7.4.
Ran Pan and Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
John Riordan, Letter to N. J. A. Sloane, Nov 10 1970.
John Riordan, Letter of 04/11/74.
L. W. Shapiro, A Catalan triangle, Discrete Math., Vol. 14, No. 1 (1976), pp. 83-90.
L. W. Shapiro, A Catalan triangle, Discrete Math., Vol. 14, No. 1 (1976), pp. 83-90. [Annotated scanned copy]
Zoran Sunic, Self describing sequences and the Catalan family tree, Elect. J. Combin., Vol. 10 (2003), Article N5.
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2.
Steven J. Tedford, Combinatorial interpretations of convolutions of the Catalan numbers, Integers, Vol. 11 (2011), Article A3.
Wen-Jin Woan, Lou Shapiro, and D. G. Rogers, The Catalan numbers, the Lebesgue integral and 4^{n-2}, Amer. Math. Monthly, Vol. 104, No. 10 (1997), pp. 926-931.

T(n, n+4) for n=0, 1, 2, ..., array T as in A047072. Also a diagonal of A059365 and of A009766.
Cf. A001003.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A000108, A000245, A000344, A003517, A000588, A003518, A003519, A001392, A001622.
Cf. A145596 (row sums), A279004.

List([0..25],n->4*Binomial(2*n+3,n)/(n+4)); # Muniru A Asiru, Mar 05 2018

[4*Binomial(2*n+3,n)/(n+4): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016

a := n -> 32*4^n*GAMMA(5/2+n)*(1+n)/(sqrt(Pi)*GAMMA(5+n)):
seq(a(n),n=0..23); # Peter Luschny, Dec 14 2015
A002057List := proc(m) local A, P, n; A := [1]; P := [1,1,1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-1]]);
A := [op(A), P[-1]] od; A end: A002057List(27); # Peter Luschny, Mar 26 2022

Table[Plus@@Flatten[Nest[ #/.a_Integer:> Range[0, a+1]&, {0}, n]], {n, 0, 10}]
Table[4 Binomial[2n+3,n]/(n+4),{n,0,30}] (* or *) CoefficientList[ Series[ (1-Sqrt[1-4 x]+2 x (-2+Sqrt[1-4 x]+x))/(2 x^4),{x,0,30}],x] (* Harvey P. Dale, May 05 2011 *)

{a(n) = if( n<0, 0, n+=2; 2*binomial(2*n, n-2) / n)}; /* Michael Somos, Jul 31 2005 */

x='x+O('x^100); Vec((1-(1-4*x)^(1/2)+2*x*(-2+(1-4*x)^(1/2)+x))/(2*x^4)) \\ Altug Alkan, Dec 14 2015

[2*(n+1)*catalan_number(n+2)/(n+4) for n in (0..30)] # G. C. Greubel, May 27 2022

a(n) = A033184(n+4, 4) = 4*binomial(2*n+3, n)/(n+4) = 2*(n+1)*A000108(n+2)/(n+4).
G.f.: c(x)^4 with c(x) g.f. of A000108 (Catalan).
Row sums of A145596. Column 4 of A033184. By specializing the identities for the row polynomials given in A145596 we obtain the results a(n) = Sum_{k = 0..n} (-1)^k*binomial(n+1,k+1)*a(k)*4^(n-k) and a(n) = Sum_{k = 0..floor(n/2)} binomial(n+1,2*k+1) * Catalan(k+1) * 2^(n-2*k). From the latter identity we can derive the congruences a(2n+1) == 0 (mod 4) and a(2n) == Catalan(n+1) (mod 4). It follows that a(n) is odd if and only if n = (2^m - 4) for some m >= 2. - Peter Bala, Oct 14 2008
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=3, a(n-3) = (-1)^(n-3) * coeff(charpoly(A,x), x^3). - Milan Janjic, Jul 08 2010
G.f.: (1-sqrt(1-4*x) + 2*x*(-2+sqrt(1-4*x) + x))/(2*x^4). - Harvey P. Dale, May 05 2011
a(n+1) = A214292(2*n+4,n). - Reinhard Zumkeller, Jul 12 2012
D-finite with recurrence: (n+4)a(n) = 8*(2*n-1)*a(n-3) - 20*(n+1)*a(n-2) + 4*(2*n+5)*a(n-1). - Fung Lam, Jan 29 2014
D-finite with recurrence: (n+4)*a(n) - 2*(3*n+7)*a(n-1) + 4*(2*n+1)*a(n-2) = 0. - R. J. Mathar, Jun 03 2014
Asymptotics: a(n) ~ 4^(n+3)/sqrt(4*Pi*n^3). - Fung Lam, Mar 31 2014
a(n) = 32*4^n*Gamma(5/2+n)*(1+n)/(sqrt(Pi)*Gamma(5+n)). - Peter Luschny, Dec 14 2015
a(n) = C(n+1) - 2*C(n) where C is Catalan number A000108. Yuchun Ji, Oct 18 2017 [Note: Offset is off by 2]
E.g.f.: d/dx ( 2*exp(2*x)*BesselI(2,2*x)/x ). - Ilya Gutkovskiy, Nov 01 2017
From Bradley Klee, Mar 05 2018: (Start)
With F(x) = 16/(1+sqrt(1-4*x))^4 g.f. of A002057, xi(x) = F(x/4)*(x/4)^2, K(16*x) = 2F1(1/2,1/2;1;16*x) g.f. of A002894, q(x) g.f. of A005797, and q'(x) g.f. of A274344:
  K(x) = (1+sqrt(xi(x)))*K(xi(x)).
  2*K(1-x) = (1+sqrt(xi(x)))*K(1-xi(x)).
  q(x) = sqrt(q(xi(16*x)/16)) = q'(xi(16*x)/16)/sqrt(xi(16*x)/16). (End)
From Amiram Eldar, Jan 02 2022: (Start)
Sum_{n>=0} 1/a(n) = 5/4 + Pi/(18*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 183*log(phi)/(25*sqrt(5)) - 77/100, where phi is the golden ratio (A001622). (End)
a(n) = Integral_{x=0..4} x^n*W(x) dx where W(x) = -x^(3/2)*(1 - x/2)*sqrt(4 - x)/Pi, defined on the open interval (0,4). - Karol A. Penson, Nov 13 2022

A008977 a(n) = (4*n)!/(n!)^4.

Original entry on oeis.org

1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600, 53644737765488792839237440000, 12309355935372581458927646400000

Offset: 0

N. J. A. Sloane

Number of paths of length 4*n in an n X n X n X n grid from (0,0,0,0) to (n,n,n,n).
a(n) occurs in Ramanujan's formula 1/Pi = (sqrt(8)/9801) * Sum_{n>=0} (4*n)!/(n!)^4 * (1103 + 26390*n)/396^(4*n). - Susanne Wienand, Jan 05 2013
a(n) is the number of ballot results that lead to a 4-way tie when 4*n voters each cast three votes for three out of four candidates vying for 3 slots on a county commission; each of these ballot results give 3*n votes to each of the four candidates. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X + Y + Z + 1/(X*Y*Z))^(4*n). - Mark van Hoeij, May 07 2013
In Narumiya and Shiga on page 158 the g.f. is given as a hypergeometric function. - Michael Somos, Aug 12 2014
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w+x+y+z)). - Gheorghe Coserea, Jul 15 2016

			a(13)=52!/(13!)^4=53644737765488792839237440000 is the number of ways of dealing the four hands in Bridge or Whist. - _Henry Bottomley_, Oct 06 2000
a(1)=24 since, in a 4-voter 3-vote election that ends in a four-way tie for candidates A, B, C, and D, there are 4! ways to arrange the needed vote sets {A,B,C}, {A,B,D}, {A,C,D}, and {B,C,D} among the 4 voters. - _Dennis P. Walsh_, May 02 2013
G.f. = 1 + 24*x + 2520*x^2 + 369600*x^3 + 63063000*x^4 + 11732745024*x^5 + ...

T. D. Noe, Table of n, a(n) for n=0..100
R. M. Dickau, Paths through a 4-D lattice.
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. 41.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Michaël Moortgat, The Tamari order for D^3 and derivability in semi-associative Lambek-Grishin Calculus, 15th Workshop: Computational Logic and Applications (CLA 2020).
N. Narumiya and H. Shiga, The mirror map for a family of K3 surfaces induced from the simplest 3-dimensional reflexive polytope, Proceedings on Moonshine and related topics (Montréal, QC, 1999), 139-161, CRM Proc. Lecture Notes, 30, Amer. Math. Soc., Providence, RI, 2001. MR1877764 (2002m:14030).

Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.
Row 4 of A187783.
Related to diagonal of rational functions: A268545-A268555.

[Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Maple
```
A008977 := n->(4*n)!/(n!)^4;
```

Table[(4n)!/(n!)^4,{n,0,16}] (* Harvey P. Dale, Oct 24 2011 *)
a[ n_] := If[ n < 0, 0, (4 n)! / n!^4]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/4, 2/4, 3/4}, {1, 1}, 256 x], {x, 0, n}]; (* Michael Somos, Aug 12 2014 *)

A008977(n):=(4*n)!/(n!)^4$ makelist(A008977(n),n,0,20); /* Martin Ettl, Nov 15 2012 */

a(n) = (4*n)!/n!^4; \\ Gheorghe Coserea, Jul 15 2016

from math import factorial
def A008977(n): return factorial(n<<2)//factorial(n)**4 # Chai Wah Wu, Mar 15 2023

a(n) = A139541(n)*(A001316(n)/A049606(n))^3. - Reinhard Zumkeller, Apr 28 2008
Self-convolution of A178529, where A178529(n) = (4^n/n!^2) * Product_{k=0..n-1} (8*k + 1)*(8*k + 3).
G.f.: hypergeom([1/8, 3/8], [1], 256*x)^2. - Mark van Hoeij, Nov 16 2011
a(n) ~ 2^(8*n - 1/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Mar 07 2014
G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(24*n)), where F(x) = 1 + x + 29*x^2 + 2246*x^3 + 239500*x^4 + 30318701*x^5 + 4271201506*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008978, A008979, A186420 and A188662. (End)
0 = (x^2-256*x^3)*y''' + (3*x-1152*x^2)*y'' + (1-816*x)*y' - 24*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..3*n} (-1)^(n+k)*binomial(4*n,n + k)* binomial(n + k,k)^4.
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(4*n,k)*binomial(n + k,k)^4. (End)
E.g.f.: 3F3(1/4,1/2,3/4; 1,1,1; 256*x). - Ilya Gutkovskiy, Jan 23 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(4*n). (End)
D-finite with recurrence n^3*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Aug 01 2022
a(n) = 24*A082368(n). - R. J. Mathar, Jun 21 2023

A093388 (n+1)^2a(n+1) = (17n^2+17n+6)a(n) - 72n^2a(n-1).

Original entry on oeis.org

1, 6, 42, 312, 2394, 18756, 149136, 1199232, 9729882, 79527084, 654089292, 5408896752, 44941609584, 375002110944, 3141107339328, 26402533581312, 222635989516122, 1882882811380284, 15967419789558804, 135752058036988848, 1156869080242393644

Offset: 0

Matthijs Coster, Apr 29 2004

nonn

This is the Taylor expansion of a special point on a curve described by Beauville.

This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017

			A(x) = 1 + 6*x + 42*x^2 + 312*x^3 + 2394*x^4 + 18756*x^5 + ... is the g.f.

Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.

Seiichi Manyama, Table of n, a(n) for n = 0..1050 (terms 0..200 from Vincenzo Librandi)
Arnaud Beauville, Les familles stables de courbes sur P_1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Some Arithmetic Properties of Short Random Walk Integrals, May 2011.
Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023.
Matthijs Coster, Sequences
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See F p. 2.
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Bradley Klee, Checking Weierstrass data, 2023.
Robert S. Maier, On Rationally Parametrized Modular Equations, arXiv:math/0611041 [math.NT], 2006.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012. - From _N. J. A. Sloane_, Dec 16 2012
Zhi-Hong Sun, Congruences for Apéry-like numbers, arXiv:1803.10051 [math.NT], 2018.
Zhi-Hong Sun, New congruences involving Apéry-like numbers, arXiv:2004.07172 [math.NT], 2020.
H. Verrill, Some congruences related to modular forms, Section 2.2.
D. Zagier, Integral solutions of Apery-like recurrence equations. See line F in sporadic solutions table of page 5.

This is the seventh sequence in the family beginning A002894, A006077, A081085, A005258, A000172, A002893.
Cf. A091401.
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.

f:=proc(n) option remember; local m; if n=0 then RETURN(1); fi; if n=1 then RETURN(6); fi; m:=n-1; ((17*m^2+17*m+6)*f(n-1)-72*m^2*f(n-2))/n^2; end;

Table[(-1)^n*Sum[Binomial[n,k]*(-8)^k*Sum[Binomial[n-k,j]^3,{j,0,n-k}],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

a(n)=(-1)^n*sum(k=0,n,binomial(n,k)*(-8)^k*sum(j=0,n-k,binomial(n-k,j)^3));

seq(N) = {
  my(a = vector(N)); a[1] = 6; a[2] = 42;
  for (n=3, N, a[n] = ((17*n^2 - 17*n + 6)*a[n-1] - 72*(n-1)^2*a[n-2])/n^2);
  concat(1,a);
};
seq(20)  \\ Gheorghe Coserea, Aug 26 2016

a(n) = (-1)^n * Sum_{k=0..n} binomial(n, k) * (-8)^k * Sum_{j=0..n-k} binomial(n-k, j)^3. - Helena Verrill (verrill(AT)math.lsu.edu), Aug 09 2004
G.f.: hypergeom([1/3, 2/3],[1],x^2*(8*x-1)/(2*x-1/3)^3)/(1-6*x). - Mark van Hoeij, Oct 25 2011
a(n) ~ 3^(2*n+3/2)/(Pi*n). - Vaclav Kotesovec, Oct 14 2012
G.f. A(x) satisfies: 0 = x*(x+8)*(x+9)*y'' + (3*x^2 + 34*x + 72)*y' + (x+6)*y, where y(x) = A(-x/72). - Gheorghe Coserea, Aug 26 2016
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=6*(-1 + 12*x)*T(x) + (1 - 34*x + 216*x^2)*T'(x) + x*(-1 + 8*x)*(-1 + 9*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 12*(-1 + 6*x)*(-1 + 18*x - 84*x^2 + 24*x^3);
g3 = -8*(1 - 12*x + 24*x^2)*(-1 + 24*x - 192*x^2 + 504*x^3 + 72*x^4);
which determine an elliptic surface with four singular fibers.  (End)

A036917 G.f.: (4/Pi^2)EllipticK(4x^(1/2))^2.

Original entry on oeis.org

1, 8, 88, 1088, 14296, 195008, 2728384, 38879744, 561787864, 8206324928, 120929313088, 1794924383744, 26802975999424, 402298219288064, 6064992788397568, 91786654611673088, 1393772628452578264, 21227503080738294464, 324160111169327247424

Offset: 0

N. J. A. Sloane

			G.f. = 1 + 8*x + 88*x^2 +  1088*x^3 + 14296*x^5 + 195008*x^5 + ... - _Michael Somos_, May 29 2023

M. Petkovsek et al., "A=B", Peters, p. ix of second printing.

Reinhard Zumkeller, Table of n, a(n) for n = 0..500
B. Adamczewski, J. P. Bell, and E. Delaygue, Algebraic independence of G-functions and congruences "a la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
E. Delaygue, Arithmetic properties of Apery-like numbers, arXiv preprint arXiv:1310.4131 [math.NT], 2013.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Ji-Cai Liu and He-Xia Ni, Supercongruences for Almkvist--Zudilin sequences, arXiv:2004.07652 [math.NT], 2020. See Vn.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
Zhi-Hong Sun, New congruences involving Apéry-like numbers, arXiv:2004.07172 [math.NT], 2020. See Vn.
Zhi-Hong Sun, Congruences for two types of Apery-like sequences, arXiv:2005.02081 [math.NT], 2020.

Cf. A002894, A036915, A057703.
Cf. A007318, A036916, A036829.
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.)

a036917 n = sum $ map
   (\k -> (a007318 (2*n-2*k) (n-k))^2 * (a007318 (2*k) k)^2) [0..n]
-- Reinhard Zumkeller, May 24 2012

a[n_] := (16 (n - 1/2)(2*n^2 - 2*n + 1)a[n - 1] - 256(n - 1)^3 a[n - 2])/n^3; a[0] = 1; a[1] = 8; Array[a, 19, 0] (* Or *)
f[n_] := Sum[(Binomial[2 (n - k), n - k] Binomial[2 k, k])^2, {k, 0, n}]; Array[f, 19, 0] (* Or *)
lmt = 20; Take[ 4^Range[0, 2 lmt]*CoefficientList[ Series[(4/Pi^2) EllipticK[4 x^(1/2)]^2, {x, 0, lmt}], x^(1/2)], lmt] (* Robert G. Wilson v *)
a[n_] := HypergeometricPFQ[{1/2, 1/2, -n, -n}, {1, 1/2-n, 1/2-n}, 1] * 4^n * (2n-1)!!^2 / n!^2 (* Vladimir Reshetnikov, Mar 08 2014 *)
a[ n_] := SeriesCoefficient[ EllipticTheta[3, 0, EllipticNomeQ[16*x]]^4, {x, 0, n}]; (* Michael Somos, May 30 2023 *)

for(n=0,25, print1(sum(k=0,n, (binomial(2*n-2*k,n-k) *binomial(2*k,k))^2), ", ")) \\ G. C. Greubel, Oct 24 2017

a(n) = if(n<0, 0, polcoeff(agm(1, sqrt(1 - 16*x + x*O(x^n)))^-2, n)); /* Michael Somos, May 29 2023 */

a(n) = (16*(n-1/2)*(2*n^2-2*n+1)*a(n-1)-256*(n-1)^3*a(n-2))/n^3.
a(n) = Sum_{k=0..n} (C(2 * (n-k), n-k) * C(2 * k, k))^2. [corrected by Tito Piezas III, Oct 19 2010]
a(n) = hypergeom([1/2, 1/2, -n, -n], [1, 1/2-n, 1/2-n], 1) * 4^n * (2n-1)!!^2 / n!^2. - Vladimir Reshetnikov, Mar 08 2014
a(n) ~ 2^(4*n+1) * log(n) / (n*Pi^2) * (1 + (4*log(2) + gamma)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Nov 28 2015
G.f. y=A(x) satisfies: 0 = x^2*(16*x - 1)^2*y''' + 3*x*(16*x - 1)*(32*x - 1)*y'' + (1792*x^2 - 112*x + 1)*y' + 8*(32*x - 1)*y. - Gheorghe Coserea, Jul 03 2018
G.f.: 1 / AGM(1, sqrt(1 - 16*x))^2. - Vaclav Kotesovec, Oct 01 2019
It appears that a(n) is equal to the coefficient of (x*y*z*t)^n in the expansion of (1+x+y+z-t)^n * (1+x+y-z+t)^n * (1+x-y+z+t)^n * (1-x+y+z+t)^n. Cf. A000172. - Peter Bala, Sep 21 2021
G.f. y = A(x) satisfies 0 = x*(1 - 16*x)*(2*y''*y - y'*y') + 2*(1 - 32*x)*y*y' - 16*y*y. - Michael Somos, May 29 2023
Expansion of theta_3(0, q)^4 in powers of m/16 where the modulus m = k^2. - Michael Somos, May 30 2023
From Paul D. Hanna, Mar 25 2024: (Start)
G.f. ( Sum_{n>=0} binomial(2*n,n)^2 * x^n )^2.
G.f. Sum_{n>=0} binomial(2*n,n)^3 * x^n * (1 - 16*x)^n. (End)

Replaced complicated definition via a formula with simple generating function provided by Vladeta Jovovic, Dec 01 2003. Thanks to Paul D. Hanna for suggesting this. - N. J. A. Sloane, Mar 25 2024

A002897 a(n) = binomial(2n,n)^3.

Original entry on oeis.org

1, 8, 216, 8000, 343000, 16003008, 788889024, 40424237568, 2131746903000, 114933031928000, 6306605327953216, 351047164190381568, 19774031697705428416, 1125058699232216000000, 64561313052442296000000

Offset: 0

N. J. A. Sloane, Simon Plouffe

Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*z + x + y + z)). - Gheorghe Coserea, Jul 14 2016

Conjecture: The g.f. is also the diagonal of the rational function 1/(1 - (x + y)*(1 - 4*z*t) - z - t) = 1/det(I - M*diag(x, y, z, t)), I the 4 x 4 unit matrix and M the 4 x 4 matrix [1, 1, 1, 1; 1, 1, 1, 1; 1, 1, 1, -1; 1 , 1, -1, 1]. If true, then a(n) = [(x*y*z)^n] (1 + x + y + z)^(2*n)*(1 + x + y - z)^n*(1 + x - y + z)^n. - Peter Bala, Apr 10 2022

S. Ramanujan, Modular Equations and Approximations to pi, pp. 23-39 of Collected Papers of Srinivasa Ramanujan, Ed. G. H. Hardy et al., AMS Chelsea 2000. See page 36, equation (25).
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..100
David H. Bailey, Jonathan M. Borwein, David Broadhurst and M. L. Glasser, Elliptic integral evaluations of Bessel moments, arXiv:0801.0891 [hep-th], 2008.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Yen Lee Loh, A general method for calculating lattice Green functions on the branch cut, arXiv:1706.03083 [math-ph], 2017.
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Cf. A000897, A002894, A006480, A008977, A108625, A176285, A183204, A186420, A188662.

Related to diagonal of rational functions: A268545-A268555.

[Binomial(2*n, n)^3: n in [0..20]]; // Vincenzo Librandi, Nov 18 2011

a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/2, 1/2, 1/2}, {1, 1}, 64x], {x, 0, n}];
Table[Binomial[2n,n]^3,{n,0,20}] (* Harvey P. Dale, Dec 06 2017 *)

{a(n) = binomial(2*n, n)^3}; /* Michael Somos, Jan 31 2007 */

[binomial(2*n, n)**3 for n in range(21)] # Zerinvary Lajos, Apr 21 2009

Expansion of (K(k)/(Pi/2))^2 in powers of (kk'/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at modulus k. - Michael Somos, Jan 31 2007
G.f.: F(1/2, 1/2, 1/2; 1, 1; 64x) where F() is a hypergeometric function. - Michael Somos, Jan 31 2007
G.f.: hypergeom([1/4,1/4],[1],64*x)^2. - Mark van Hoeij, Nov 17 2011
D-finite with recurrence n^3*a(n) - 8*(2*n - 1)^3*a(n-1) = 0. - R. J. Mathar, Mar 08 2013
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)^3 = ( [x^n](1 + x)^(2*n) )^3 = [x^n](F(x)^(8*n)), where F(x) = 1 + x + 6*x^2 + 111*x^3 + 2806*x^4 + 84456*x^5 + 2832589*x^6 + 102290342*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002894, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 64^n/(Pi*n)^(3/2). - Ilya Gutkovskiy, Jul 13 2016
0 = (-x^2 + 64*x^3)*y''' + (-3*x + 288*x^2)*y'' + (-1 + 208*x)*y' + 8*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = Sum_{k = 0..n} (2*n + k)!/(k!^3*(n - k)!^2). Cf. A001850(n) = Sum_{k = 0..n} (n + k)!/(k!^2*(n - k)!). - Peter Bala, Jul 27 2016
It appears that a(n) is the coefficient of (x*y*z)^(2*n) in the expansion of (1 + x*y + x*z - y*z)^(2*n) * (1 + x*y - x*z + y*z)^(2*n) * (1 - x*y + x*z + y*z)^(2*n).  Cf. A000172. - Peter Bala, Sep 21 2021
From Peter Bala, Sep 24 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k,n).
a(n) = the coefficient of (x*y*z*t^2)^n in the expansion of 1/(1 - x - y)*(1 - z - t) - x*y*z*t) (a(n) = A(n,n,n,2*n) in the notation of Straub, Theorem 1.2). (End)
a(n) = (8/5) * Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k-1,n) for n >= 1. - Peter Bala, Jul 09 2024
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(2*n, k). Cf. A183204. - Peter Bala, Oct 12 2024
From Peter Bala, Oct 16 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(2*n+k, k)*A108625(n, k) = 8 * Sum_{k = 0..n} (-1)^(n+k+1) * binomial(n-1, k)*binomial(2*n+k-1, k)*A108625(n, k) = (8/5) * Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(2*n+k-1, k)*A108625(n, k) for n >= 1. Cf. A176285. (End)

A002898 Number of n-step closed paths on hexagonal lattice.

Original entry on oeis.org

1, 0, 6, 12, 90, 360, 2040, 10080, 54810, 290640, 1588356, 8676360, 47977776, 266378112, 1488801600, 8355739392, 47104393050, 266482019232, 1512589408044, 8610448069080, 49144928795820, 281164160225520, 1612061452900080, 9261029179733760, 53299490722049520

Offset: 0

N. J. A. Sloane

Also, number of closed paths of length n on the honeycomb tiling.
The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.
From David Callan, Aug 25 2009: (Start)
a(n) = number of 2 X n matrices, entries from {1,2,3}, second row a (multiset) permutation of the first, with no constant columns. For example, a(2)=6 counts the matrices
   1 2   1 3   2 1   2 3   3 1   3 2
   2 1   3 1   1 2   3 2   1 3   2 3. (End)
Also, a(n) is the constant coefficient in the expansion of (x + 1/x + y + 1/y + x/y + y/x)^n. - Christopher J. Smyth, Sep 25 2018
a(n) is the constant term in the expansion of (-2 + (1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 27 2019
a(n) is the number of paths from (0,0,0) to (n,n,n) using the six permutations of (0,1,2) as steps, i.e., the steps (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - William J. Wang, Dec 07 2020

			O.g.f.: 1 + 6*x^2 + 12*x^3 + 90*x^4 + 360*x^5 + 2040*x^6 + ...
O.g.f.: 1 + 6*x^2*(1+2*x) + 90*x^4*(1+2*x)^2 + 1680*x^6*(1+2*x)^3 + 34650*x^8*(1+2*x)^4 + ... + A006480(n)*x^(2*n)*(1+2*x)^n + .... - _Paul D. Hanna_, Feb 26 2012

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Cyril Banderier, Analytic combinatorics of random walks and planar maps, PhD Thesis, 2001 (>6 Mb).
Volker Braun, Philip Candelas, and Xenia de la Ossa, Two One-Parameter Special Geometries, arXiv preprint arXiv:1512.08367 [hep-th], 2015.
Gunther Cornelissen, David Hokken, and Berend Ringeling, The asymptotic Mahler measure of Gaussian periods, arXiv:2507.09303 [math.NT], 2025. See p. 40.
Cyril Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Davidson Noby Joseph and Igor Boettcher, Walking on Archimedean Lattices: Insights from Bloch Band Theory, arXiv:2507.12662 [cond-mat.stat-mech], 2025. See p. 18.
Leonard F. Klosinski, Gerald L. Alexanderson and Loren C. Larson, Solution to 1995 Putnam problem A-6, Am. Math. Monthly, 1996, p. 674.
Gilbert Labelle and Annie Lacasse, Closed paths whose steps are roots of unity, in FPSAC 2011, Reykjavík, Iceland DMTCS proc. AO, 2011, 599-610.
Yen Lee Loh, A general method for calculating lattice Green functions on the branch cut, arXiv:1706.03083 [math-ph], 2017.
Gabriele Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2
Grzegorz Siudem and Agata Fronczak, Bell polynomials in the series expansions of the Ising model, arXiv:2007.16132 [math-ph], 2020.

Cf. A000172, A006480, A337905-A337907, A094060, A002894 (returns square lattice), A002893 (honeycomb net).

a:= proc(n) option remember; `if`(n<3, [1, 0, 6][n+1], ((n-1)*
      n*a(n-1) +24*(n-1)^2*a(n-2) +36*(n-1)*(n-2)*a(n-3))/n^2)
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Dec 08 2020

a[n_] := Sum[(-2)^(n-i)*Binomial[i, j]^3*Binomial[n, i], {i, 0, n}, {j, 0, i}]; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 21 2011, after Vasu Tewari *)

{a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3*x^(2*m)*(1+2*x+x*O(x^n))^m),n)} /* Paul D. Hanna, Feb 26 2012 */

D-finite with recurrence a(0) = 1, a(1) = 0, a(2) = 6, 36*(n+2)*(n+1)*a(n) +24*(n+2)^2*a(n+1) +(n+3)*(n+2)*a(n+2) -(n+3)^2*a(n+3) = 0.
E.g.f.: (BesselI(0,2*x))^3 + 2*Sum_{k>=1} (BesselI(k,2*x))^3. - Karol A. Penson Aug 18 2006
a(n) = Sum_{i=0..n} (-2)^(n-i)*binomial(n, i)*(Sum_{j=0..i} binomial(i, j)^3). - Vasu Tewari (vasu(AT)math.ubc.ca), Aug 04 2010
O.g.f.: (4/Pi)*EllipticK( 8*sqrt(z^3*(1+3*z))/(1-12*z^2+sqrt((1-6*z)*(1+2*z)^3)) ) / sqrt(2 - 24*z^2 + 2*sqrt((1-6*z)*(1+2*z)^3)). - Sergey Perepechko, Feb 08 2011
O.g.f.: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)*(1+2*x)^n. - Paul D. Hanna, Feb 26 2012
a(n) ~ sqrt(3)*6^n/(2*Pi*n). - Vaclav Kotesovec, Aug 13 2013
O.g.f.: 2F1(1/3,2/3; 1; 27*x^2*(1+2*x)). - R. J. Mathar, Sep 29 2020

More terms from David Bloom, Mar 1997

Formula and further terms from Cyril Banderier, Oct 12 2000

A000897 a(n) = (4n)! / ((2n)!*n!^2).

Original entry on oeis.org

1, 12, 420, 18480, 900900, 46558512, 2498640144, 137680171200, 7735904619300, 441233078286000, 25467973278667920, 1484298740174927040, 87202550985276963600, 5157850293780050462400, 306839461354466267304000, 18344908596179023234548480

Offset: 0

N. J. A. Sloane

Appears in Ramanujan's theory of elliptic functions of signature 4.
H. A. Verrill proves that a(n) = Sum_{p + q + r = 3n} w^(p-q) * {(3n)!/(p! q! r!)}^2, with p, q, r >= 0 and w = primitive 3rd root of unity.
The family of elliptic curves "x=2*H1=p^2+q^2-(1/4)*q^4, 0sqrt(-1)*q" to H1 produces "x=2*H2=p^2-q^2-(1/4)*q^4, 0Bradley Klee, Feb 25 2018
Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z)). - Gheorghe Coserea, Aug 09 2018

			G.f.: 1 + 12*x + 420*x^2 + 18480*x^3 + 900900*x^4 + 46558512*x^5 + 2498640144*x^6 + ...

E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
H. J. Brothers, Pascal's Prism: Supplementary Material
B. Klee, Geometric G.F. for Ramanujan Periods, seqfan mailing list, 2017.
S. Ramanujan, Modular Equations and Approximations to Pi, Quarterly Journal of Mathematics, XLV (1914), 350-372.
L. C. Shen, A note on Ramanujan's identities involving the hypergeometric function 2F1(1/6,5/6;1;z), The Ramanujan Journal, 30.2 (2013), 211-222.
H. A. Verrill, Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations, arXiv:math/0407327v1 [math.CO], 2004.

Cf. A002897, A008977, A186420, A188662. Elliptic Integrals: A002894, A113424, A006480. Factors: A005809, A005810, A000984, A001448.

a:=n->Sum([0..3*n],k->(-1)^k*Binomial(3*n,k)*Binomial(6*n-k,3*n)*
Binomial(2*k,k));;
A000897:=List([0..14],n->a(n)); # Muniru A Asiru, Feb 11 2018

seq((4*n)!/(n!)^4/binomial(2*n,n), n=0..14); # Zerinvary Lajos, Jun 28 2007

Table[(4n)!/((2n)! n!^2), {n, 0, 30}] (* Stefan Steinerberger, Apr 14 2006 *)
a[ n_] := Binomial[ 4 n, 2 n] Binomial[ 2 n, n]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 3/4, 1, 64 x], {x, 0, n}]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 4 n}, (-1)^n m! SeriesCoefficient[ BesselI[ 0, 2 x] BesselJ[ 0, 2 x], {x, 0, m}]]]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := 64^n Pochhammer[1/4, n] Pochhammer[3/4, n] / n!^2; (* Michael Somos, Aug 12 2014 *)

{a(n) = if( n<0, 0, (4*n)! / ((2*n)! * n!^2))}; /* Michael Somos, Oct 31 2005 */

E.g.f.: Sum_{k>=0} (-1)^k * a(k) * x^(4*k) / (4*k)! = BesselI(0, 2x) * BesselJ(0, 2x).
G.f.: F(1/4, 3/4; 1; 64*x). - Michael Somos, Oct 31 2005
a(n) = A008977(n)/A000984(n) - Zerinvary Lajos, Jun 28 2007
Sum_{k>=0} a(k) * x^(3k)/(3k)!^2 = f(x)*f(x*w)*f(x/w) where f(x) = BesselI(0, 2*sqrt(x)) and w = primitive 3rd root of unity. - Michael Somos, Jul 25 2007
In general, for (BesselI(b, 2x))*(BesselJ(b, 2x))=((x^(2*b))/((GAMMA(b+1))^2)*(1-(x^4)/(Q(0)+(x^4))); Q(k)=(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)-(x^4)+(x^4)*(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)/Q(k+1)) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
D-finite with recurrence 0 = a(n)*4*(4*n + 1)*(4*n + 3) - a(n+1)*(n + 1)^2 for all n in Z. - Michael Somos, Aug 12 2014
0 = a(n)*(-4026531840*a(n+2) +2005401600*a(n+3) -103896576*a(n+4) +1251948*a(n+5)) + a(n+1)*(+41418752*a(n+2) -30435328*a(n+3) +1863228*a(n+4) -24604*a(n+5)) + a(n+2)*(-16896*a(n+2) +75608*a(n+3) -6740*a(n+4) +105*a(n+5)) for all n in Z. - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(3*n,n)*binomial(4*n,n) = A005809(n)*A005810(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(12*n)), where F(x) = 1 + x + 6*x^2 + 105*x^3 + 2448*x^4 + 67043*x^5 + 2028307*x^6 + ... appears to have integer coefficients. Cf. A002894, A002897, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 2^(6*n-1/2)/(Pi*n). - Ilya Gutkovskiy, Jul 12 2016
G.f.: 2*EllipticK(sqrt((sqrt(1-64*x)-1)/(2*sqrt(1-64*x))))/(Pi*(1-64*x)^(1/4)) where EllipticK is the complete elliptic integral of the first kind (in Maple's notation). - Robert Israel, Jul 12 2016
a(n) = Sum_{k = 0..3*n} (-1)^k*C(3*n,k)*C(6*n-k,3*n)*C(2*k,k). - Peter Bala, Feb 10 2018
From Bradley Klee, Feb 27 2018: (Start)
a(n) = A000984(n)*A001448(n).
G.f.: (1/(sqrt(2)*Pi))*Integral_{q=-oo..oo} 1/sqrt(q^2+(1/4)*q^4+(1-64*x)) dq.
G.f.: (1/(2*Pi))*Integral_{phi=0..2*Pi} 1/sqrt(1-64*x*sin^4(phi)) dphi. (End)
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..2*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (4*n)!/((2*n)!*n!^2);
(3/2)*Sum_{k = 0..n} 2*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (4*n)!/((2*n)!*n!^2).
Cf. A001451. (End)
a(n) = (4^n/n!^2)*Product_{k = 0..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(3*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

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

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A006480 De Bruijn's S(3,n): (3n)!/(n!)^3.

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A049541 Decimal expansion of 1/Pi.

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A002057 Fourth convolution of Catalan numbers: a(n) = 4*binomial(2*n+3,n)/(n+4).

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A008977 a(n) = (4*n)!/(n!)^4.

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

A002057 Fourth convolution of Catalan numbers: a(n) = 4binomial(2n+3,n)/(n+4).

A093388 (n+1)^2a(n+1) = (17n^2+17n+6)a(n) - 72n^2a(n-1).

A036917 G.f.: (4/Pi^2)EllipticK(4x^(1/2))^2.