A188662 -id:A188662 - 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

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

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)

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

A008978 a(n) = (5*n)!/(n!)^5.

Original entry on oeis.org

1, 120, 113400, 168168000, 305540235000, 623360743125120, 1370874167589326400, 3177459078523411968000, 7656714453153197981835000, 19010638202652030712978200000, 48334775757901219912115629238400, 125285878026462826569986857692288000

Offset: 0

N. J. A. Sloane

Number of paths of length 5n in Z^5 from (0,0,0,0,0) to (n,n,n,n,n).

Vincenzo Librandi, Table of n, a(n) for n = 0..100
V. Batyrev, Review of "Mirror Symmetry and Algebraic Geometry", by D. A. Cox and S. Katz, Bull. Amer. Math. Soc., 37 (No. 4, 2000), 473-476.
R. M. Dickau, 5-D shortest path diagrams
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A006480, A008977, A002894, A002897, A186420, A188662, A001460.

Row 5 of A187783.

[Factorial(5*n)/Factorial(n)^5: n in [0..10]]; // Vincenzo Librandi, Mar 08 2014

Table[(5 n)!/(n)!^5, {n, 0, 20}] (* Vincenzo Librandi, Mar 08 2014 *)

a(n) = (5*n)!/(n!)^5; \\ Michel Marcus, Mar 08 2014

a(n) ~ 5^(5*n+1/2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 07 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n]( F(x)^(120*n) ), where F(x) = 1 + x + 353*x^2 + 318986*x^3 + 408941594*x^4 + 633438203535*x^5 + 1105336091531052*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. (End)
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(5*n,n + k)*binomial(n + k,k)^5.
a(n) = Sum_{k = 0..5*n} (-1)^(n+k)*binomial(5*n,k)*binomial(n + k,k)^5. (End)
From Ilya Gutkovskiy, Nov 23 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1,1,1; 3125*x).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1,1,1,1; 3125*x). (End)
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*u)^n] (1 + x + y + z + u )^(5*n). (End)
a(n) = 120*A322252(n). - R. J. Mathar, Jun 21 2023
a(n) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/n^4. - Neven Sajko, Jul 21 2023

A113424 a(n) = (6n)!/((3n)!(2n)!*n!).

Original entry on oeis.org

1, 60, 13860, 4084080, 1338557220, 465817912560, 168470811709200, 62588625639883200, 23717177328413240100, 9124964373613212524400, 3553261127084984957001360, 1397224499394244497967972800, 553883078634868423069470550800, 221068174083308549543680044926400

Offset: 0

Michael Somos, Oct 31 2005

Appears in Ramanujan's theory of elliptic functions of signature 6.
The family of elliptic curves "x=2*H=p^2+q^2-q^3, 0Bradley Klee, Feb 25 2018
The power series with coefficients a(n) * n! plays a central role in the Faber-Zagier relations on the moduli space of algebraic curves; see Pandharipande and Pixton, Section 0.2. - Harry Richman, Aug 19 2024

			G.f. = 1 + 60*x + 13860*x^2 + 4084080*x^3 + 1338557220*x^4 + ... - _Michael Somos_, Dec 02 2018

G. C. Greubel, Table of n, a(n) for n = 0..250
J. Cremona, Elliptic Curves over Q, LMFDB 2017.
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.
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 p. 23.
Bradley Klee, Geometric G.F. for Ramanujan Periods, seqfans mailing list, 2017.
Bradley Klee, On LMFDB period data, LMFDB-support mailing list, 2018.
Bradley Klee, Weierstrass Solution of Cubic Anharmonic Oscillation, Wolfram Demonstrations Project, 2018.
Romeo Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
R. Pandharipande and A. Pixton, Relations in the tautological ring of the moduli space of curves, arXiv:1301.4561 [math.AG], 2020.
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.

a(n) = A347304(6*n)
Elliptic Integrals: A002894, A006480, A000897. Factors: A005809, A066802.
Cf. A188662.

List([0..15],n->Factorial(6*n)/(Factorial(3*n)*Factorial(2*n)*Factorial(n))); # Muniru A Asiru, Apr 08 2018

a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/6, 5/6, 1, 432 x], {x, 0, n}];
Table[Multinomial[n, 2 n, 3 n], {n, 0, 15}] (* Vladimir Reshetnikov, Oct 12 2016 *)
a[ n_] := Multinomial[n, 2 n, 3 n]; (* Michael Somos, Dec 02 2018 *)

{a(n) = if( n<0, 0, (6*n)! / ((3*n)! * (2*n)! * n!))};

G.f.: hypergeometric2F1(1/6, 5/6; 1; 432 * x).
a(n) ~ 432^n/(2*Pi*n). - Ilya Gutkovskiy, Oct 13 2016
a(n) = A005809(n)*A066802(n). - Bradley Klee, Feb 25 2018
0 = a(n)*(-267483013447680*a(n+2) +25577192448000*a(n+3) -204669037440*a(n+4) +372142500*a(n+5)) +a(n+1)*(+408751349760*a(n+2) -57870650880*a(n+3) +546809652*a(n+4) -1088188*a(n+5)) +a(n+2)*(-17884800*a(n+2) +21466920*a(n+3) - 295844*a(n+4) +693*a(n+5)) for all n in Z. - Michael Somos, May 16 2018
From Peter Bala, Feb 28 2020: (Start)
a(n) = C(6*n,2*n)*C(4*n,n).
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k (apply Mestrovic, equation 39).
(-1)^n*a(n) = [x^(2*n)*y^(2*n)] ( (1 + x + y)*(1 - x + y) )^(4*n).
a(n) = [x^n] ( F(x) )^(60*n), where F(x) = 1 + x + 56*x^2 + 7355*x^3 + 1290319*x^4 + 264117464*x^5 + 59508459679*x^6 + ... appears to have integer coefficients. We conjecture that for k >= 1 the sequence defined by b_k(n) := [x^n] F(x)^(k*n) satisfies the above supercongruences for primes p >= 7. (End)
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..2*n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (6*n)!/((3*n)!*(2*n)!*n!);
Sum_{k = 0..3*n} 3*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (6*n)!/((3*n)!(2*n)!*n!).
Cf. A001451. (End)
From Peter Bala, Feb 26 2023: (Start)
a(n) = (4^n/n!^2) * Product_{k = n..3*n-1} 2*k + 1.
a(n) = (12^n/n!^2) * Product_{k = 0..n-1} (6*k + 1)*(6*k + 5). (End)
a(n) = 12*(6*n - 1)*(6*n - 5)*a(n-1)/n^2. - Neven Sajko, Jul 19 2023
From Karol A. Penson, Dec 26 2023: (Start)
a(n) = Integral_{x=0..432} x^n*W(x) dx, n>=0, where W(x) = sqrt(18)*MeijerG([[], [0, 0]], [[-1/6, -5/6], []], x/432)/(1296*Pi), where MeijerG is the Meijer G - function.
Apparently, W(x) cannot be represented by any other function. W(x) is positive on x = [0, 432], it diverges at x=0, and monotonically decreases for x>0. It appears that at x=432, W(x) tends to a constant value close to 0.000368414. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 432] is unique, as W(x) is the solution of the Hausdorff moment problem. (End)
W(x) can be represented in terms of two 2F1 hypergeometric functions, W(x) = hypergeom([1/6, 1/6], [1/3], x/432)/(6*sqrt(Pi)*Gamma(2/3)*Gamma(5/6)*x^(5/6)) - Gamma(2/3)*Gamma(5/6)*sqrt(3)*hypergeom([5/6, 5/6], [5/3], x/432)/(1152*Pi^(5/2)*x^(1/6)), x on (0, 432). - Karol A. Penson, May 16 2025

A008979 a(n) = (6n)!/(n!)^6.

Original entry on oeis.org

1, 720, 7484400, 137225088000, 3246670537110000, 88832646059788350720, 2670177736637149247308800, 85722533226982363751829504000, 2889253496242619386328267523990000, 101097362223624462291180422369532000000, 3644153415887633116359073848179365185734400, 134567406165969006655507763343147223231094784000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..100
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000897, A000984, A002894, A002897, A006480, A008977, A008978, A071549, A071550, A071551, A071552, A186420, A188662.

[Factorial(6*n)/Factorial(n)^6: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

seq( (6*n)!/(n!)^6, n=0..20); # G. C. Greubel, Feb 17 2020

Table[(6 n)!/(n)!^6, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)

vector(21, n, my(m=n-1); (6*m)!/(m!)^6 ) \\ G. C. Greubel, Feb 17 2020

[factorial(6*n)/factorial(n)^6 for n in (0..20)] # G. C. Greubel, Feb 17 2020

From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*
  binomial(5*n,n)*binomial(6*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) * ( [x^n](1 + x)^(6*n) ) = [x^n](F(x)^(720*n)), where F(x) = 1 + x + 4478*x^2 + 53085611*x^3 + 926072057094*x^4 + 19977558181209910*x^5 + 493286693783478576177*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A008978, A186420 and A188662. (End)
a(n) ~ 3^(6*n+1/2)*4^(3*n-1)/(Pi*n)^(5/2). - Ilya Gutkovskiy, Jul 12 2016
From Peter Bala, Feb 14 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*u*v)^n] (1 + x + y + z + u + v)^(6*n). (End)

A186420 a(n) = binomial(2n,n)^4.

Original entry on oeis.org

1, 16, 1296, 160000, 24010000, 4032758016, 728933458176, 138735983333376, 27435582641610000, 5588044012339360000, 1165183173971324375296, 247639903129149250277376, 53472066459540320483696896, 11701285507234585729600000000, 2589980371199606611713600000000

Offset: 0

Emanuele Munarini, Feb 21 2011

			G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,1},256x) where 4F3 is a hypergeometric series.

Seiichi Manyama, Table of n, a(n) for n = 0..417

Cf. binomial(2n,n)^k: A000984 (k=1), A002894 (k=2), A002897 (k=3), this sequence (k=4).

Cf. A000108, A000888, A186414, A186415, A186416, A186418, A186419, A000897, A006480, A008977, A188662.

Table[Binomial[2n,n]^4,{n,0,20}]
Table[Coefficient[Series[HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {1, 1, 1}, 256 x], {x, 0, n}], x, n], {n, 0, 14}] (* Michael De Vlieger, Jul 13 2016 *)

Maxima
```
makelist(binomial(2*n,n)^4,n,0,40);
```

a(n) = A000984(n)^4 = A002894(n)^2.
a(n) = binomial(2*n,n)^4 = ( [x^n](1 + x)^(2*n) )^4 = [x^n](F(x)^(16*n)), where F(x) = 1 + x + 25*x^2 + 1798*x^3 + 183442*x^4 + 22623769*x^5 + 3142959012*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 256^n/(Pi*n)^2. - Ilya Gutkovskiy, Jul 13 2016

A001451 a(n) = (5n)!/((3n)!n!n!).

Original entry on oeis.org

1, 20, 1260, 100100, 8817900, 823727520, 79919739900, 7962100660800, 808906548235500, 83426304143982800, 8707404737345073760, 917663774856743842200, 97491279924241456098300, 10427604345391237790688000, 1121786259855036145008408000

Offset: 0

N. J. A. Sloane

From Peter Bala, Jul 15 2016: (Start)
This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. Cf. A273628 and A273629.
Note the similar results:
Sum_{k = 0..n} (-1)^k*binomial(n,k)* binomial(2*n + k,n)*binomial(3*n + k,n) = (-1)^n*(3*n)!/n!^3 = (-1)^n*A006480(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n)^2 = A002894(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n). (End)
Choose three noncollinear step vectors to satisfy the zero sum, 3*v_1 + v_2 + v_3 = 0. Then a(n) is the number of loop plane walks of length 5*n which depart from and return to the origin. Equivalently, a(n) counts distinct permutations of a (5*n)-digit integer with digits 1,2,3 of multiplicity 3*n,n,n respectively. - Bradley Klee, Aug 12 2018

			G.f. = 1 + 20*x + 1260*x^2 + 100100*x^3 + 8817900*x^4 + 823727520*x^5 + ... - _Michael Somos_, Aug 12 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000897, A002894, A002897, A006480, A008977, A186420, A188662, A273628, A273629.

List([0..15],n->Factorial(5*n)/(Factorial(3*n)*Fact0rial(n)*Factorial(n))); # Muniru A Asiru, Aug 12 2018

[Factorial(5*n)/(Factorial(3*n)*Factorial(n)*Factorial(n)): n in [0..30]]; // Vincenzo Librandi, May 22 2011

Maple
```
f := n->(5*n)!/((3*n)!*n!*n!);
```

Table[(5*n)!/((3*n)!*n!*n!), {n, 0, 20}] (* Vincenzo Librandi, Sep 04 2012 *)

a(n) = binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n](F(x)^(20*n)), where F(x) = 1 + x + 12*x^2 + 390*x^3 + 16984*x^4 + 867042*x^5 + 48848541*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 3^(-3*n-1/2)*5^(5*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 13 2016
G.f.: G(x) = 4F3(1/5,2/5,3/5,4/5;1/3,2/3,1;(5^5/3^3)*x). Let G^(n)(x) = d^n/dx^n G(x), and c = {120, 15000*x-6, 45000*x^2-114*x, 25000*x^3-135*x^2, 3125*x^4-27*x^3}, then Sum_{n=0..4} c_n*G^(n)(x) = 0. - Bradley Klee, Aug 12 2018
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..3*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (5*n)!/((3*n)!*n!^2);
Sum_{k = 0..n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (5*n)!/((3*n)!*n!^2). Cf. A000897 and A113424. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(4*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 12 2024

cp's OEIS Frontend

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

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

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

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

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A008978 a(n) = (5*n)!/(n!)^5.

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A113424 a(n) = (6*n)!/((3*n)!*(2*n)!*n!).

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

A113424 a(n) = (6n)!/((3n)!(2n)!*n!).

A001451 a(n) = (5n)!/((3n)!n!n!).

A357569 a(n) = binomial(3n,n)^2 - 27binomial(2*n,n).