A245086 -id:A245086 - cp's OEIS frontend

A002894 a(n) = binomial(2n, n)^2.

1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000

Offset: 0

N. J. A. Sloane

a(n) is the number of monotonic paths (only moving N and E) in the lattice [0..2n] X [0..2n] that contain the points (0,0), (n,n) and (2n,2n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Expansion of K(k) / (Pi/2) in powers of m/16 = (k/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at k. - Michael Somos, Mar 04 2003
Square lattice walks that start and end at origin after 2n steps. - Gareth McCaughan and Michael Somos, Jun 12 2004
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic) then a(n)=E[(tr(A^k))^{2n}] for any k > 4. - Andrew V. Sutherland, Apr 01 2008
From R. H. Hardin, Feb 03 2016 and R. J. Mathar, Feb 18 2016: (Start)
Also, number of 2 X (2n) arrays of permutations of 2n copies of 0 or 1 with row sums equal.
For example, some solutions for n=3:
  0 1 0 1 0 1   0 1 0 1 1 0   0 0 1 0 1 1   1 1 1 0 0 0
  1 0 0 0 1 1   1 1 0 1 0 0   0 0 0 1 1 1   0 0 1 1 0 1
There is a simple combinatorial argument to show that this is a(n): We have 2n copies of 0's and 1's and need equal row sums. Therefore there must be n 1's in each of the two rows. Otherwise there are no constraints, so there are C(2n,n) ways of placing the 1's in the first row and independently C(2n,n) ways of placing the 1's in the second. The product is clearly C(2n,n)^2. (End)
Also the even part of the bisection of A241530. One half of the odd part is given in A000894. - Wolfdieter Lang, Sep 06 2016
From Peter Bala, Jan 26 2018: (Start)
Let S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} be a set of four column vectors. Then a(n) equals the number of 3 X k arrays whose columns belong to the set S and whose row sums are all equal to n (apply Eger, Theorem 3). An example is given below. Equivalently, a(n) equals the number of lattice paths from (0,0,0) to (n,n,n) using steps (1,0,0), (0,1,0), (1,0,1) and (0,1,1).
The o.g.f. for the sequence equals the diagonal of the rational function 1/(1 - (x + y + x*z + y*z)).
Row sums of A069466. (End)
Also, the constant term in the expansion of (x + 1/x + y + 1/y)^(2n). - Christopher J. Smyth, Sep 26 2018
Number of ways to place 2n^2 nonattacking pawns on a 2n x 2n board. - Tricia Muldoon Brown, Dec 12 2018
For n>0, a(n) is the number of Littlewood polynomials of degree 4n-1 that have a closed Lill path. A polynomial p(x) has a closed Lill path if and only if p(x) is divisible by x^(2)+1. - Raul Prisacariu, Aug 28 2024

			G.f. = 1 + 4*x + 36*x^2 + 400*x^3 + 4900*x^4 + 63504*x^5 + 853776*x^6 + ... - _Michael Somos_, Aug 06 2014
From _Peter Bala_, Jan 26 2018: (Start)
a(2) = 36: The thirty six 3 x k arrays with columns belonging to the set of column vectors S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} and having all row sums equal to 2 are the 6 distinct arrays obtained by permuting the columns of
  /1 1 0 0\
  |0 0 1 1|,
  \0 0 1 1/
the 6 distinct arrays obtained by permuting the columns of
  /0 0 1 1\
  |1 1 0 0|
  \0 0 1 1/
and the 24 arrays obtained by permuting the columns of
  /1 0 1 0\
  |0 1 0 1|. (End)
  \0 0 1 1/

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 591,828.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
Leonard Lipshitz and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.
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, Table of n, a(n) for n = 0..100
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
R. Bacher, Meander algebras, Institut Fourier, 1999.
E. Barcucci, A. Frosini and S. Rinaldi, On directed-convex polyominoes in a rectangle, Discr. Math., 298 (2005). 62-78.
Arnaud Beauville, Les familles stables de courbes elliptiques sur P^1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982, 657-660. MR0664643 (83h:14008)
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
Tricia Muldoon Brown, The Problem of Pawns, The Electronic Journal of Combinatorics (2019) Vol. 26, Issue 3, #P3.21. Also arXiv:1811.09606, [math.CO], 2018.
John Maxwell Campbell, New series involving harmonic numbers and squared central binomial coefficients, Rocky Mountain J. Math., 49 (2019), 2513-2544.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
Murray Elder, Cogrowth, 2011.
M. Elder, A. Rechnitzer, E. J. Janse van Rensburg, and T. Wong, The cogrowth series for BS(N,N) is D-finite, arXiv:1309.4184 [math.GR], 2013.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, 2009; see page 90.
Florian Fürnsinn and Sergey Yurkevich, Algebraicity of hypergeometric functions with arbitrary parameters, arXiv:2308.12855 [math.CA], 2023.
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.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2010.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024. See p. 14.
L. Lipshitz and A. J. van der Poorten, Rational functions, diagonals, automata and arithmetic
Raul Prisacariu, Littlewood Polynomials of Degree n with Closed Lill Paths
Eric M. Rains, High powers of random elements of compact Lie groups, Probability Theory and Related Fields 107 (1997), 219-241.
Grzegorz Siudem and Agata Fronczak, Bell polynomials in the series expansions of the Ising model, arXiv:2007.16132 [math-ph], 2020.
Eric Weisstein's World of Mathematics, Lattice Path.
D. Zagier, Integral solutions of Apéry-like recurrence equations. See line G in sporadic solutions table of page 5.

Row sums of A069466.
Row 2 of A268367 (even terms).
Equals 4*A060150.
Cf. A000984, A000515, A010370, A054474 (INVERTi transform), A172390, A000897, A002897, A006480, A008977, A186420, A188662, A000894, A241530, A002898 (walks hex lattice).

[Binomial(2*n, n)^2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Maple
```
A002894 := n-> binomial(2*n,n)^2.
```

CoefficientList[Series[Hypergeometric2F1[1/2, 1/2, 1, 16x], {x, 0, 20}], x]
Table[Binomial[2n,n]^2,{n,0,20}] (* Harvey P. Dale, Jul 06 2011 *)
a[ n_] := SeriesCoefficient[ EllipticK[16 x] / (Pi/2), {x, 0, n}]; (* Michael Somos, Aug 06 2014 *)
a[n_] := 16^n HypergeometricPFQ[{1/2, -2 n, 2 n + 1}, {1, 1}, 1];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)

PARI
```
{a(n) = binomial(2*n, n)^2};
```

{a(n) = if( n<0, 0, polcoeff( polcoeff( polcoeff( 1 / (1 - x * (y + z + 1/y + 1/z)) + x * O(x^(2*n)), 2*n), 0), 0))}; /* Michael Somos, Jun 12 2004 */

[binomial(2*n, n)**2 for n in range(17)]  # Zerinvary Lajos, Apr 21 2009

D-finite with recurrence: (n+1)^2*a(n+1) = 4*(2*n + 1)^2*a(n). - Matthijs Coster, Apr 28 2004
a(n) ~ Pi^(-1)*n^(-1)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
G.f.: F(1/2, 1/2; 1; 16*x) = 1 / AGM(1, (1 - 16*x)^(1/2)) = K(4*sqrt(x)) / (Pi/2), where AGM(x, y) is the arithmetic-geometric mean of Gauss and Legendre. - Michael Somos, Mar 04 2003
G.f.: 2*EllipticK(4*sqrt(x))/Pi, using Maple's convention for elliptic integrals.
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2x)^2.
a(n) = A000984(n)^2. - Jonathan Vos Post, Jun 17 2007
E.g.f.: (BesselI(0, 2*x))^2 = 1+2*x^2/(U(0)-2*x^2); U(k) = 2*x^2*(2*k+1)+(k+1)^3-2*x^2*(2*k+3)*(k+1)^3/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
In generally, for (BesselI(b, 2x))^2=((x^(2*b))/(GAMMA(b+1))^2)*(1+(2*x^2)*(2*b+1)/(Q(0)-(2*x^2)*(2*b+1)); Q(k)=(2*x^2)*(2*k+2*b+1)+(k+1)*(k+b+1)*(k+2*b+1)-(2*x^2)*(k+1)*(k+b+1)*(k+2*b+1)*(2*k+2*b+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 4*(2*k+1)^2*x*(1+4*x)^2/(4*(2*k+1)^2*x*(1+4*x)^2 + (k+1)^2*(1+4*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
0 = +a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for all n in Z. - Michael Somos, Aug 06 2014
Integral representation as the n-th moment of a positive function W(x) on (0,16), in Maple notation, W(x) = EllipticK(sqrt(1-x/16))/(2*Pi^2*sqrt(x)); a(n) = Integral_{x=0..16} x^n*W(x) dx, n>=0. The function W(x) is singular at x=0 and W(16) = 1/(16*Pi). This representation is unique since W(x) is the solution of the Hausdorff moment problem. - Stanley Smith and Karol A. Penson, Jun 19 2015
a(n) ~ 16^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)^2/((4*n+1)* Pi). - Peter Luschny, Oct 14 2015
a(n) = binomial(2*n,n)*binomial(2*n,n) = ( [x^n](1 + x)^(2*n) ) *( [x^n](1 + x)^(2*n) ) = [x^n](F(x)^(4*n)), where F(x) = 1 + x + x^2 + 4*x^3 + 20*x^4 + 120*x^5 + 798*x^6 + 5697*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) = Sum_{k = 0..n} binomial(2*n + k,k)*binomial(n,k)^2. Cf. A005258(n) = Sum_{k = 0..n} binomial(n + k,k)*binomial(n,k)^2. - Peter Bala, Jul 27 2016
a(n) = A241530(2*n), n >= 0. - Wolfdieter Lang, Sep 06 2016
E.g.f.: 2F2(1/2,1/2; 1,1; 16*x). - Ilya Gutkovskiy, Jan 23 2018
a(n) = 16^n*hypergeom([1/2, -2*n, 2*n + 1], [1, 1], 1). - Peter Luschny, Mar 14 2018
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-4)^(n-k) = a(n). - Peter Bala, Mar 16 2018
a(n) = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A245086(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)). - Peter Bala, Mar 23 2022
a(n) = Sum_{k=0..n} multinomial(2n [k k (n-k) (n-k)]), which is another way to count random walks on Z^2, with steps of (0,+-1) or (+-1,0), that return to the point of origin after 2n steps (not necessarily for the first time), as is C(2n,n)^2. - Shel Kaphan, Jan 12 2023
0 = a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for n>=0. - Michael Somos, May 30 2023
From Peter Bala, Sep 12 2023: (Start)
Right-hand side of the binomial coefficient identities
1) Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k,n) = a(n).
2) 2*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k-1,n)*C(2*n+k-1,n) = a(n) for n >= 1.
3) (4/3)*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k-1,n) = a(n) for n >= 1. (End)

Edited by N. J. A. Sloane, Feb 18 2016

A183204 Central terms of triangle A181544.

Original entry on oeis.org

1, 4, 48, 760, 13840, 273504, 5703096, 123519792, 2751843600, 62659854400, 1451780950048, 34116354472512, 811208174862904, 19481055861877120, 471822589361293680, 11511531876280913760, 282665135367572129040

Offset: 0

Paul D. Hanna, Dec 30 2010

The g.f. for row n of triangle A181544 is (1-x)^(3n+1)*Sum_{k>=0}C(n+k-1,k)^3*x^k.
This sequence is s_7 in Cooper's paper. - Jason Kimberley, Nov 06 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y*z + x*y + x*z + y + z)). - Gheorghe Coserea, Jul 14 2016
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017

			Triangle A181544 begins:
(1);
1, (4), 1;
1, 20, (48), 20, 1;
1, 54, 405, (760), 405, 54, 1;
1, 112, 1828, 8464, (13840), 8464, 1828, 112, 1; ...

Seiichi Manyama, Table of n, a(n) for n = 0..702 (terms 0..499 from Jason Kimberley)
S. Cooper, Sporadic sequences, modular forms and new series for 1/pi, Ramanujan J., December 2012, Volume 29, Issue 1, pp 163-183.
Shaun Cooper, Jesús Guillera, Armin Straub, and Wadim Zudilin, Crouching AGM, Hidden Modularity, arXiv:1604.01106 [math.NT], 5-April-2016.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See s7 p. 3.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5.
Robert Osburn, Armin Straub, and Wadim Zudilin, A modular supercongruence for 6F5: an Apéry-like story, arXiv:1701.04098 [math.NT], 2017.
Wadim Zudilin, A generating function of the squares of Legendre polynomials, preprint arXiv:1210.2493 [math.CA], 2012.

Cf. A108625, A181544, A183205, A245086.
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.)

P:=PolynomialRing(Integers()); C:=Binomial;
A183204:=func; // or directly:
A183204:=func;
[A183204(n):n in[0..16]]; // Jason Kimberley, Oct 29 2012

Table[Sum[Binomial[n,j]^2 * Binomial[2*j,n] * Binomial[j+n,j],{j,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 05 2015 *)

{a(n)=polcoeff((1-x)^(3*n+1)*sum(j=0, 2*n, binomial(n+j, j)^3*x^j), n)}

a(n) = [x^n] (1-x)^(3n+1) * Sum_{k>=0} C(n+k-1,k)^3*x^k.
a(n) = Sum_{j = 0..n} C(n,j)^2 * C(2*j,n) * C(j+n,j). [Formula of Wadim Zudilin provided by Jason Kimberley, Nov 06 2012]
1/Pi = sqrt(7) Sum_{n>=0} (-1)^n a(n) (11895n + 1286)/22^(3n+3). [Cooper, equation (41)] - Jason Kimberley, Nov 06 2012
G.f.: sqrt((1-13*x+(1-26*x-27*x^2)^(1/2))/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2)))*hypergeom([1/12,5/12],[1],13824*x^7/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2))^3)^2. - Mark van Hoeij, May 07 2013
a(n) ~ 3^(3*n+3/2) / (4 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Apr 05 2015
G.f. A(x) satisfies 1/(1+4*x)^2 * A( x/(1+4*x)^3 ) = 1/(1+2*x)^2 * A( x^2/(1+2*x)^3 ) [see Cooper, Guillera, Straub, Zudilin]. - Joerg Arndt, Apr 08 2016
a(n) = (-1)^n*binomial(3n+1,n)* 4F3({-n,n+1,n+1,n+1};{1,1,2(n+1)}; 1). - M. Lawrence Glasser, May 15 2016
Conjecture D-finite with recurrence: n^3*a(n) - (2*n-1)*(13*n^2-13*n+4)*a(n-1) - 3*(n-1)*(3*n-4)*(3*n-2)*a(n-2) = 0. - R. J. Mathar, May 15 2016
0 = (-x^2+26*x^3+27*x^4)*y''' + (-3*x+117*x^2+162*x^3)*y'' + (-1+86*x+186*x^2)*y' + (4+24*x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Jeremy Tan, Mar 14 2024: (Start)
The conjectured D-finite recurrence can be proved by Zeilberger's algorithm.
a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,n) * binomial(2*n-k,n) = [(w*x*y*z)^n] ((w+y)*(x+z)*(y+z)*(w+x+y+z))^n. (End)
a(n) = Sum_{0 <= j, k <= n} binomial(n, k)^2 * binomial(n, j)^2 * binomial(k+j, n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(n, k). - Peter Bala, Jul 08 2024
From Peter Bala, Sep 18 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n+k, k)^3*binomial(3*n+1, n-k). Cf A245086.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*A143007(n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A275652 a(n) = binomial(3n,3n/2)binomial(2n,n)binomial(5n/2,n/2)/binomial(n,n/2).

Original entry on oeis.org

1, 10, 300, 11440, 485100, 21841260, 1022041020, 49128552000, 2408829328620, 119918393838100, 6042249840712800, 307438844121252480, 15770112362658517500, 814459593645444166560, 42308586942403276440000, 2208850973597860123741440, 115825519836558228435979500

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)* binomial(3*n+k,3*n-k)*binomial(2*k,k)*binomial(2*n-k,n) = binomial(3*n,3*n/2)*binomial(2*n,n)*binomial(5*n/2,n/2)/binomial(n,n/2).
We also have Sum_{k = 0..3*n} (-1)^k*binomial(3*n+k,3*n-k)* binomial(2*k,k)*binomial(2*n-k,n) = binomial(3*n,3*n/2)*binomial(2*n,n)* binomial(5*n/2,n/2) /binomial(n,n/2).
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n+k,2*n-k)* binomial(2*k,k)*binomial(2*n-k,n) = binomial(2*n,n)^2 = A002894(n). See also A275653, A275654 and A275655.

Cf. A002894, A245086, A275653, A275654, A275655, A276098, A276100, A276101, A276102, A352651, A352652, A365025.

seq(simplify(factorial(3*n)*factorial(n/2)*factorial(5*n/2)/(factorial(n)^3*factorial(3*n/2)^2)), n = 0 .. 20);

Table[Binomial[3 n, 3 n/2] Binomial[2 n, n] Binomial[5 n/2, n/2] / Binomial[n, n/2], {n, 0, 16}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = sum(k = 0, n, binomial(2*n-k-1,n-k)*binomial(3*n,k)^2); \\ Michel Marcus, Apr 21 2022

from math import factorial
from sympy import factorial2
def A275652(n): return int(factorial(3*n)*factorial2(5*n)*factorial2(n)//factorial2(3*n)**2//factorial(n)**3) # Chai Wah Wu, Aug 08 2023

a(n) = (3*n)!*(5*n/2)!*(n/2)!/((3*n/2)!^2*n!^3).
Recurrence: a(n) = 5*(3*n - 1)*(3*n - 5)*(5*n - 2)*(5*n - 4)*(5*n - 6)*(5*n - 8)/(n^2*(n - 1)^2*(3*n - 2)*(3*n - 4)) * a(n-2).
a(n) = [x^n] G(x)^(5*n), where G(x) = 1 + 2*x + 12*x^2 + 184*x^3 + 3811*x^4 + 92796*x^5 + 2497358*x^6 + ... appears to have integer coefficients.
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^5, where F(x) = 1 + 2*x + 32*x^2 + 824*x^3 + 26291*x^4 + 947506*x^5 + 36934522*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(5/3)*5^(5*n/2)/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 22 2022: (Start)
For n >= 1, a(n) = (5/3)*binomial(m*n,2*n)*binomial(m*n/2,2*n)*binomial(2*n,n)^2/ binomial(m*n/2,n)^2 at m = -1. See A352651 for the case m = 1.
a(n) = Sum_{k = 0..n} binomial(2*n-k-1,n-k)*binomial(3*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(3*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A245086.
a(p) == a(1) (mod p^3) for prime p >= 5.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)
Row 1 of A365025. - Peter Bala, Aug 18 2023

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 4, -6, -20, 1, 10, 36, 0, 70, 1, 18, 300, 400, 90, -252, 1, 28, 1050, 11440, 4900, 0, 924, 1, 40, 2646, 77616, 485100, 63504, -1680, -3432, 1, 54, 5544, 316540, 6370650, 21841260, 853776, 0, 12870, 1, 70, 10296, 972400, 42031990, 554822268, 1022041020, 11778624, 34650, -48620

Offset: 0

Peter Bala, Jul 19 2023

The first row of the table is a signed version of the central binomial coefficients A000984. The central binomial coefficients satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p ^(3*r)) for all primes p >= 5 and all positive integers n and r (see Meštrović, equation 39). We conjecture that each row sequence of the table satisfies the same supercongruences.

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252    ...  (-1)^k*A000984(k)
  1 |  1   0     -6        0          90             0    ...  A245086
  2 |  1   4     36      400        4900         63504    ...  A002894
  3 |  1  10    300    11440      485100      21841260    ...  A275652
  4 |  1  18   1050    77616     6370650     554822268    ...  A275653
  5 |  1  28   2646   316540    42031990    5921058528    ...  A275654
  6 |  1  40   5544   972400   189290920   39089615040    ...  A275655
  7 |  1  54  10296  2484000   665091000  188907932304    ...  A364304
  8 |  1  70  17550  5567380  1960044750  732012601320    ...  A364305

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Wikipedia, Dixon's identity

Cf. A000984 (row 0 unsigned), A245086 (row 1), A002894 (row 2), A275652 (row 3), A275653 (row 4), A275654 (row 5), A275655 (row 6), A364304 (row 7), A364305 (row 8).

Cf. A364113, A364298

T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k, (1 + x)/(1 - x)), x, 11), x, k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);

T(n,k) = Sum_{i = 0..k} binomial(n*k, k-i)^2 * binomial((n-2)*k+i-1, i).
For n >= 2, T(n,k) = binomial((n-1)*k, k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 3)*k and b = -k.
For n >= 3, T(n,k) = ((n - 1)*k)! * ((n + 1)*k/2)! * ((n - 3)*k/2)! / ( ((n - 1)*k/2)!^2 * k!^2 *  ((n - 3)*k)! ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.

A336179 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Sum_{j=0..n} (-k)^j * binomial(n,j)^3.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, -1, -6, 1, 1, -2, -11, 0, 1, 1, -3, -14, 47, 90, 1, 1, -4, -15, 136, 241, 0, 1, 1, -5, -14, 261, 106, -2281, -1680, 1, 1, -6, -11, 416, -639, -8492, -3779, 0, 1, 1, -7, -6, 595, -2294, -17523, 35344, 104831, 34650, 1, 1, -8, 1, 792, -5135, -25624, 188049, 395008, -110207, 0, 1

Offset: 0

Seiichi Manyama, Jul 10 2020

Column k is the diagonal of the rational function 1 / (1 + y + z + x*y + y*z - k*z*x - (k-1)*x*y*z).

Column k is the diagonal of the rational function 1 / ((1-x)*(1-y)*(1-z) + k*x*y*z).

			Square array begins:
  1,  1,     1,     1,      1,      1, ...
  1,  0,    -1,    -2,     -3,     -4, ...
  1, -6,   -11,   -14,    -15,    -14, ...
  1,  0,    47,   136,    261,    416, ...
  1, 90,   241,   106,   -639,  -2294, ...
  1,  0, -2281, -8492, -17523, -25624, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-3 give: A000012, A245086, A336181, A336182.
Main diagonal gives A336180.
Cf. A307884, A336163.

Unprotect[Power]; 0^0 = 1; T[n_, k_] := Sum[(-k)^j * Binomial[n, j]^3, {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 11 2020 *)

A361716 a(n) = Sum_{k = 0..n-1} (-1)^kbinomial(n,k)^2binomial(n-1,k).

Original entry on oeis.org

0, 1, -3, -8, 45, 126, -840, -2400, 17325, 50050, -378378, -1100736, 8576568, 25069968, -199536480, -585307008, 4732755885, 13919870250, -113936715750, -335813478000, 2775498395670, 8194328596740, -68263497731520, -201822515032320

Offset: 0

Peter Bala, Mar 23 2023

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integers n and k.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Wikipedia, Dixon's identity.

Cf. A006480, A245086, A361710, A361711.

seq(add((-1)^k*binomial(n,k)^2*binomial(n-1,k), k = 0..n-1), n = 0..20);

A361716[n_]:=Sum[(-1)^k*Binomial[n,k]^2Binomial[n-1,k],{k,0,n-1}];Array[A361716,30,0] (* Paolo Xausa, Oct 06 2023 *)

a(n) = sum(k = 0, n-1, (-1)^k*binomial(n,k)^2*binomial(n-1,k)); \\ Michel Marcus, Mar 26 2023

from math import comb
def A361716(n): return (sum(comb(n,k)**3*k if k&1 else -comb(n,k)**3*k for k in range(n+1)))//(n if n&1 else -n) if n else 0 # Chai Wah Wu, Mar 27 2023

a(n) = Sum_{k = 0..n} (-1)^(n+k) * (k/n) * binomial(n,k)^3.
a(2*n) = (-1)^n * (1/2) * (3*n)!/n!^3 for n >= 1; a(2*n+1) = (-1)^n * (3*n+1)/(2*n+1) * (3n)!/n!^3.
a(2*n) = (1/2)*A245086(2*n) = (1/2)*(-1)^n*A006480(n) for n >= 1.
a(2*n+1) = A361710(2*n+1) = A361711(2*n+1).
a(n) = hypergeom([1 - n, - n, - n], [1, 1], 1) for n >= 1.
P-recursive: n^2*(n-1)*(6*n^2-20*n+17)*a(n) = -( 6*(3*n^2-6*n+2)*(n-1)*a(n-1) + (3*n-6)*(3*n-5)*(3*n-4)*(6*n^2-8*n+3)*a(n-2) ).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(3*n,n-k)*binomial(n+k,k)^2 for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n+k,n)*binomial(2*n-k-1,n). - Peter Bala, Sep 13 2023

A273630 a(n) = Sum_{k = 0..n} (-1)^kk^3binomial(n,k)^3.

Original entry on oeis.org

0, -1, 0, 162, 0, -11250, 0, 576240, 0, -25259850, 0, 1007242236, 0, -37685439792, 0, 1346871240000, 0, -46504059326010, 0, 1562983866658500, 0, -51407781284599740, 0, 1661123953798807680, 0, -52886433789393750000, 0, 1662782404368229351200

Offset: 0

Peter Bala, Jul 17 2016

Let d(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)^3. Clearly, by symmetry of the binomial coefficients we have d(2*n + 1) = 0. Dixon's identity is the result d(2*n) = (-1)^n*(3*n)!/n!^3. A generalization is: for r a nonnegative integer there holds Sum_{k = 0..n} (-1)^k*binomial(k,r)^3*binomial(n,k)^3 = (-1)^r*binomial(n,r)^3*d(n - r). This is the case r = 1. See A273631 (case r = 2) and A245086 (case r = 0).

Peter Bala, A generalization of Dixon's identity
J. Ward, 100 Years of Dixon's Identity, Irish Mathematical Society Bulletin 27, 46-54, 1991
Wikipedia, Dixon's identity

Cf. A006480, A245086, A273631.

[&+[(-1)^k*k^3 *Binomial(n, k)^3: k in [0..n]]: n in [0..70]]; // Vincenzo Librandi, Jul 23 2016

seq(add((-1)^k*k^3*binomial(n,k)^3, k = 0..n), n = 0..30);

Table[Sum[(-1)^k*k^3 Binomial[n, k]^3, {k, 0, n}], {n, 0, 27}] (* Michael De Vlieger, Jul 22 2016 *)

a(n) = sum(k=0, n, (-1)^k*k^3*binomial(n, k)^3) \\ Felix Fröhlich, Jul 22 2016

from math import factorial
def A273630(n): return (1 if (m:=n>>1)&1 else -1)*n**3*factorial(3*m)//factorial(m)**3 if n&1 else 0 # Chai Wah Wu, Oct 04 2022

a(2*n) = 0; a(2*n + 1) = (-1)^(n+1)*(2*n + 1)^3*(3*n)!/n!^3.
a(2*n + 1) = -(2*n + 1)^3*A245086(2*n) = (-1)^(n+1)* (2*n + 1)^3*A006480(n).
a(n) = Sum_{k = 1..n} (-1)^k*multinomial(n, 1, k - 1, n - k)^3.
Recurrence: a(n) = -3*n^3*(3*n - 5)*(3*n - 7)/((n - 1)^2*(n - 2)^3) * a(n-2).

A273628 a(n) = (7n)!/((5n)!*n!^2).

Original entry on oeis.org

1, 42, 6006, 1085280, 217567350, 46262007792, 10217700004512, 2317454130543552, 536022010184210550, 125863265857621191900, 29909151834298018538256, 7176685161839833601969280, 1735941935586019529116213920, 422752608090008019258722317800

Offset: 0

Peter Bala, Jul 15 2016

This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n + k,n)*binomial(4*n - k,n) = (-1)^m*a(m) for n = 2*m. For similar results see A001451, A006480 and A273629. Note the related sums:
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n + k,n)*binomial(4*n + k,n) = (-1)^n*(2*n)!*(4*n)!/(n!^3*(3*n)!) = (-1)^n*binomial(2*n,n)*binomial(4*n,n) = (-1)^n*A000984(n)*A005810(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = (3*n)!/n!^3 = A006480(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n + k,n)*binomial(4*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n + k,n)*binomial(4*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)*binomial(4*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n).

Cf. A000984, A001451, A005810, A006480, A008979, A245086, A273629.

[Factorial(7*n) div (Factorial(5*n)*Factorial(n)^2): n in [0..15]]; // Vincenzo Librandi, Jul 16 2016

Maple
```
seq((7*n)!/((5*n)!*n!^2), n = 0..20);
```

Table[(7 n)!/((5 n)! n!^2), {n, 0, 13}] (* or *)
Table[Binomial[7 n, n] Binomial[6 n, n], {n, 0, 13}] (* Michael De Vlieger, Jul 15 2016 *)

a(n) = (7*n)!/((5*n)!*n!^2) = binomial(7*n,2*n)*binomial(2*n,n).
a(n) = binomial(7*n,n)*binomial(6*n,n) = [x^n](1 + x)^(7*n) * [x^n](1 + x)^(6*n).
It appears that a(n) = [x^n] F(x)^(42*n), where F(x) = 1 + x + 30*x^2 + 2280*x^3 + 232715*x^4 + 27800465*x^5 + 3661895341*x^6 + ... has all integer coefficients. Cf. A273629 and A008979.
Recurrence: 5*n^2*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)*a(n) = 7*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n-1).
a(n) ~ 5^(-5*n-1/2)*7^(7*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 15 2016
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(6*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A273629 a(n) = (9n)!/((7n)!*n!^2).

Original entry on oeis.org

1, 72, 18360, 5920200, 2118223800, 803927196072, 316938365223480, 128313095514575400, 52976845635264939960, 22204947580777261872000, 9418997650746914743158360, 4034374193416822645489549632, 1741969558937890710303111545400

Offset: 0

Peter Bala, Jul 15 2016

This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(4*n + k,n)*binomial(5*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. For similar results see A001451, A006480 and A273628.
Note the related sums:
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(4*n - k,n)*binomial(5*n - k,n) = binomial(2*n,n)*binomial(4*n,n) = A000984(n)*A005810(n);
Sum_{k = 0..2*n} (-1)^k*binomial(n,k)*binomial(4*n + k,n)*binomial(5*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(n,k)*binomial(4*n - k,n)*binomial(5*n - k,n) = binomial(2*n,n) = A000984(n).
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(4*n + k,n)*binomial(5*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(4*n - k,n)*binomial(5*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n).

Cf. A000984, A001451, A004381, A005810, A006480, A008979, A169958, A245086, A273628.

[Factorial(9*n)/(Factorial(7*n)*Factorial(n)^2): n in [0..40]]; // Vincenzo Librandi, Jul 17 2016

Maple
```
seq((9*n)!/((7*n)!*n!^2), n = 0..20);
```

Table[Factorial[9 n] / (Factorial[7 n] Factorial[n]^2), {n, 0, 20}] (* Vincenzo Librandi, Jul 17 2016 *)

a(n) = (9*n)!/((7*n)!*n!^2) = binomial(9*n,2*n)* binomial(2*n,n).
a(n) = binomial(8*n,n)*binomial(9*n,n) = A004381(n)*A169958(n).
a(n) = [x^n](1 + x)^(8*n) * [x^n] (1 + x)^(9*n).
It appears that a(n) = [x^n] F(x)^(72*n), where F(x) = 1 + x + 56*x^2 + 7700*x^3 + 1422008*x^4 + 307144278*x^5 + 73118586828*x^6 + ... has all integer coefficients. Cf. A273628 and A008979.
Recurrence: 7*n^2*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n) = 9*(9*n - 1)*(9*n - 2)*(9*n - 3)*(9*n - 4)*(9*n - 5)*(9*n - 6)*(9*n - 7)*(9*n - 8)*a(n-1).
a(n) ~ 3^(18*n+1)*7^(-7*n-1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 15 2016
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(8*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A273631 a(n) = Sum_{k = 0..n} (-1)^kbinomial(k,2)^3binomial(n,k)^3.

Original entry on oeis.org

0, 0, 1, 0, -1296, 0, 303750, 0, -36879360, 0, 3157481250, 0, -217564322976, 0, 12926105848656, 0, -689598074880000, 0, 33901459248661290, 0, -1562983866658500000, 0, 68423756889802253940, 0, -2870422192164339671040, 0, 116191495035298068750000

Offset: 0

Peter Bala, Jul 17 2016

Let d(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)^3. Clearly, by symmetry of the binomial coefficients we have d(2*n + 1) = 0. Dixon's identity is the result d(2*n) = (-1)^n*(3*n)!/n!^3. A generalization is: for r a nonnegative integer there holds Sum_{k = 0..n} (-1)^k*binomial(k,r)^3*binomial(n,k)^3 = (-1)^r*binomial(n,r)^3*d(n - r). This is the case r = 2. See A273630 (case r = 1) and A245086 (case r = 0).

Peter Bala, A generalization of Dixon's identity
J. Ward, 100 Years of Dixon's Identity, Irish Mathematical Society Bulletin 27, 46-54, 1991
Wikipedia, Dixon's identity

Cf. A006480, A245086, A273630.

[&+[(-1)^k*Binomial(k,2)^3*Binomial(n,k)^3: k in [0..n]]: n in [0..70]]; // Vincenzo Librandi, Jul 23 2016

seq(add((-1)^k*binomial(k,2)^3*binomial(n,k)^3, k = 0..n), n = 0..30);

Table[Sum[(-1)^k*Binomial[k, 2]^3 Binomial[n, k]^3, {k, 0, n}], {n, 0, 27}] (* Michael De Vlieger, Jul 22 2016 *)

a(n) = sum(k=0, n, (-1)^k*binomial(k, 2)^3*binomial(n, k)^3) \\ Felix Fröhlich, Jul 22 2016

from math import factorial
def A273631(n): return 0 if n&1 or n == 0 else (-1 if (m:=n-1>>1)&1 else 1)*((m+1)*(n-1))**3*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022

a(0) = 0 and a(2*n + 2) = (-1)^n*binomial(2*n + 2,2)^3*(3*n)!/n!^3 for n >= 0. a(2*n + 1) = 0.
a(2*n + 2) = (-1)^n*(n + 1)^3*(2*n + 1)^3 * A006480(n) for n >= 0.
a(n) = Sum_{k = 2..n} (-1)^k*multinomial(n, 2, k - 2, n - k)^3.
Recurrence: a(n) = -3*n^3*(n - 1)^3*(3*n - 8)*(3*n - 10)/((n - 2)^5*(n - 3)^3) * a(n-2).

cp's OEIS Frontend

A002894 a(n) = binomial(2n, n)^2.

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A183204 Central terms of triangle A181544.

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A275652 a(n) = binomial(3*n,3*n/2)*binomial(2*n,n)*binomial(5*n/2,n/2)/binomial(n,n/2).

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A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

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A336179 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Sum_{j=0..n} (-k)^j * binomial(n,j)^3.

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A361716 a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)^2*binomial(n-1,k).

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A273630 a(n) = Sum_{k = 0..n} (-1)^k*k^3*binomial(n,k)^3.

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A275652 a(n) = binomial(3n,3n/2)binomial(2n,n)binomial(5n/2,n/2)/binomial(n,n/2).

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

A361716 a(n) = Sum_{k = 0..n-1} (-1)^kbinomial(n,k)^2binomial(n-1,k).

A273630 a(n) = Sum_{k = 0..n} (-1)^kk^3binomial(n,k)^3.

A273628 a(n) = (7n)!/((5n)!*n!^2).

A273629 a(n) = (9n)!/((7n)!*n!^2).

A273631 a(n) = Sum_{k = 0..n} (-1)^kbinomial(k,2)^3binomial(n,k)^3.