A001451 -id:A001451 - cp's OEIS frontend

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

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

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

A023982 Sum of exponents in prime-power factorization of multinomial coefficient M(5n;3n,n,n).

Original entry on oeis.org

0, 3, 6, 7, 9, 13, 13, 17, 17, 17, 19, 21, 20, 27, 27, 28, 26, 30, 29, 30, 34, 36, 33, 38, 36, 38, 43, 40, 40, 46, 44, 48, 43, 46, 47, 48, 47, 51, 53, 55, 53, 57, 57, 55, 57, 59, 58, 63, 58, 62, 63, 64, 67, 73, 66, 69, 68, 67, 73, 73, 74, 79, 80, 78, 71, 77, 76, 77, 79, 84, 79, 87, 82, 87, 89, 86, 89

Offset: 0

Clark Kimberling

nonn

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A001222, A001451, A022559.

Cf. A023978, A023979, A023980, A023981, A023983, A023984, A023985, A023986, A023987, A023990, A023991, A023992, A023993.

a[n_] := PrimeOmega[Multinomial[3*n, n, n]]; Array[a, 100, 0] (* Amiram Eldar, Jun 11 2025 *)

a(n) = bigomega((5*n)! / ((3*n)!*n!*n!)); \\ Amiram Eldar, Jun 11 2025

From Amiram Eldar, Jun 11 2025: (Start)
a(n) = A001222(A001451(n)).
a(n) = A022559(5*n) - A022559(3*n) - 2*A022559(n). (End)

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jun 11 2025

A001450 a(n) = binomial(5n,2n).

Original entry on oeis.org

1, 10, 210, 5005, 125970, 3268760, 86493225, 2319959400, 62852101650, 1715884494940, 47129212243960, 1300853625660225, 36052387482172425, 1002596421878664480, 27963143931814663880, 781879430625942976880, 21910242651571684460050, 615167304833936727234180

Offset: 0

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n=0..100
Peter Bala, A note on A001450
M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, Discrete Mathematics, Volume 339, Issue 3, 6 March 2016, Pages 1116-1139.

Cf. A001451, A001459, A001460, A000108, A060941, A066802, A259550.

[Binomial(5*n, 2*n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Maple
```
f := n->(5*n)!/((3*n)!*(2*n)!);
```

Table[Hypergeometric2F1[-3n,-2n,1,1],{n,0,60}] (* John M. Campbell, Jul 15 2011 *)
Table[Binomial[5n,2n],{n,0,20}] (* Harvey P. Dale, Nov 09 2011 *)

a(n) = binomial(5*n,2*n) \\ Altug Alkan, Oct 06 2015

a(n) = (5*n)!/((3*n)!*(2*n)!).
a(n) = 2F1[-3n,-2n,1,1] (see Mathematica code below). - John M. Campbell, Jul 15 2011
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/3, 1/2, 2/3], (3125/108)*x). - Robert Israel, Aug 07 2014
From Peter Bala, Oct 05 2015: (Start)
a(n) = [x^n] ( (1 + x)*C(x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
a(n) = 5*A259550(n) for n >= 1.
exp( (1/5) * Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 23*x^2 + 377*x^3 + ... is the o.g.f. for the sequence of Duchon numbers A060941. (End)
a(n) = [x^(2*n)] 1/(1 - x)^(3*n+1). - Ilya Gutkovskiy, Oct 10 2017
D-finite with recurrence 6*n*(3*n-1)*(2*n-1)*(3*n-2)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
a(n) = Sum_{k = 0..2*n} binomial(3*n+k-1, k). Cf. A066802. - Peter Bala, Jun 04 2024
Right-hand side of the identity Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(4*n-k, 2*n-k) = binomial(5*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(4*n-k, 2*n-k) = binomial(3*n, n). - Peter Bala, Jun 05 2024
From Karol A. Penson, May 07 2025: (Start)
G.f. denoted by h(x) satisfies the following algebraic equation of order 10:
8 - 3125*x + 20*(-13 + 3125*x)*h(x) - 45*(-74 + 9375*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^3 + 25*(1809 + 53125*x)*h(x)^4 + (34375*x - 738)*(3125*x - 108)*h(x)^5 + 15*(3125*x + 297)*(3125*x - 108)*h(x)^6 + 5*(3125*x - 108)^2*h(x)^7 + 135*(3125*x - 108)^2*h(x)^8 + (3125*x - 108)^3*h(x)^10=0.
a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, n>=0, where W(x) = W1(x)+W2(x)+W3(x)+W4(x) can be expressed with four generalized hypergeometric functions of type 4F3:
W1(x) = sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([1/5, 8/15, 7/10, 13/15], [2/5, 3/5, 4/5], (108*x)/3125)/(10*Pi*x^(4/5)),
W2(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([2/5, 11/15, 9/10, 16/15], [3/5, 4/5, 6/5], (108*x)/3125)/(50*Pi*x^(3/5)),
W3(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([3/5, 14/15, 11/10, 19/15], [4/5, 6/5, 7/5], (108*x)/3125)/(125*Pi*x^(2/5)), and
W4(x) = (7*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([4/5, 17/15, 13/10, 22/15], [6/5, 7/5, 8/5], (108*x)/3125))/(1250*Pi*x^(1/5)).
Using the formula for a(n) only, W(x) can be shown to be a positive function. It is singular at x=0 and at x=3125/108. This integral representation is unique since W(x) is the solution of the Hausdorff moment problem. (End)
From Peter Bala, Jun 21 2025: (Start)
a(n) = [x^(3*n)] 1/(1 - x)^(2*n+1).
a(n) = Sum_{k = 0..3*n} binomial(2*n+k-1, k). (End)

A184423 a(n) = (2n)!(3*n)!/n!^5.

Original entry on oeis.org

1, 12, 540, 33600, 2425500, 190702512, 15849497664, 1369618398720, 121821136479900, 11079206239530000, 1025579963180813040, 96310511463483233280, 9152842704012278107200, 878622906816654279840000

Offset: 0

Paul D. Hanna, Jan 13 2011

Denoted by h_3[n] by T. Piezas III. He also gives formulas for 1/Pi such as 1/Pi = 2 * Sum_{n>=0} a(n) * (-1)^n * (51*n + 7) / (12^3)^(n + 1/2). - Michael Somos, May 31 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w*y+w*z+x+y+z)). - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jun 28 2023: (Start)
The supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for primes p >= 5 and positive integers n and r. This follows from Meštrović equation 39, since a(n) = binomial(3*n,n) * binomial(2*n,n)^2.
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r. Cf. A362730 and A362732. (End)

			G.f.: A(x) = 1 + 12*x + 540*x^2 + 33600*x^3 + 2425500*x^4 +...
G.f. of A184424 equals A(x)^(1/2):
A(x)^(1/2) = 1 + 6*x + 252*x^2 + 15288*x^3 + 1089270*x^4 + 84963060*x^5 +...+ [(3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5)]*x^n +...

Gheorghe Coserea, Table of n, a(n) for n = 0..200
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Titus Piezas III, 0013: Article 3 (Pi Formulas and the Monster Group).

Cf. A184424, A275047.

Related to diagonal of rational functions: A268545-A268555.

Table[((2n)!(3n)!)/(n!)^5,{n,0,20}] (* Harvey P. Dale, Dec 18 2018 *)

PARI
```
{a(n)=(3*n)!*(2*n)!/n!^5}
```

{a(n)=polcoeff(sum(m=0,n,3^m*prod(k=1,m,(6*k-4)*(6*k-5))/m!^2*x^m+x*O(x^n))^2,n)}

Self-convolution of A184424:
a(n) = Sum_{k=0..n} A184424(k)*A184424(n-k) where A184424(n) = (3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5).
a(n) = 6 * (2*n - 1) * (3*n - 1) * (3*n - 2) / n^3 * a(n-1) if n>0. - Michael Somos, May 31 2012
0 = (x^2-108*x^3)*y''' + (3*x-486*x^2)*y''+ (1-348*x)*y' - 12*y, where y is g.f. - Gheorghe Coserea, Jul 15 2016
a(n) ~ 3^(1/2)/(2*Pi^(3/2)) * n^(-3/2) * 108^n. - Ilya Gutkovskiy, Jul 15 2016
a(n) = C(2*n,n)^2 * C(3*n,n) = ( [x^n](1 + x)^(2*n) )^2 * ( [x^n](1 + x)^(3*n) ) = [x^n]( F(x)^(12*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 11*x^2 + 350*x^3  + 15293*x^4  + 794433*x^5  + 45958617*x^6 + ... appears to have integral coefficients. Cf. A000897 and A001451. - Peter Bala, Dec 30 2019

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

cp's OEIS Frontend

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

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

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A023982 Sum of exponents in prime-power factorization of multinomial coefficient M(5n;3n,n,n).

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A001450 a(n) = binomial(5*n,2*n).

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A184423 a(n) = (2*n)!*(3*n)!/n!^5.

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

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

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

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

A001450 a(n) = binomial(5n,2n).

A184423 a(n) = (2n)!(3*n)!/n!^5.

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

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