A275653 a(n) = binomial(4*n,2*n)*binomial(3*n,2*n).

1, 18, 1050, 77616, 6370650, 554822268, 50199951984, 4664758248000, 442077195513690, 42533571002422500, 4141601026094832300, 407220411993767798400, 40363606408574136870000, 4028061310168832261158176, 404311537318239680601595200, 40785601782042745410592271616

Offset

0,2

Comments

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)* binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).

We also have Sum_{k = 0..4*n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).

Compare with the identities

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).

Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)* binomial(2*n,n) = A275655(n)

Sum_{k = 0..n} (-1)^(n+k)*binomial(8*n + k,8*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(8*n,4*n)* binomial(5*n,2*n)*binomial(2*n,n)/binomial(6*n,3*n).

See also A275652, A275654 and A275655.

Links

Table of n, a(n) for n=0..15.

Formula

a(n) = (4*n)!*(3*n)!/(n!*(2*n)!^3).

a(n) = A001448(n) * A005809(n).

Recurrence: a(n) = 3*(3*n - 1)*(3*n - 2)*(4*n - 1)*(4*n - 3)/(n^2*(2*n - 1)^2) * a(n-1).

a(n) = [x^n] ((1 + x)^2/(1 - x))^(2*n) * [x^n] (1 + x)^(3*n) = [x^n] G(x)^(6*n) where G(x) = 1 + 3*x + 38*x^2 + 1150*x^3 + 47099^x^4 + 2264968*x^5 + 120311611*x^6 + ... appears to have integer coefficients.

exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^6, where F(x) = 1 + 3*x + 92*x^2 + 4579*x^3 + 282605*x^4 + 19698991*x^5 + 1484923315*x^6 + ... appears to have integer coefficients.

a(n) ~ sqrt(3/2)*108^n/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016

From Peter Bala, Mar 23 2022: (Start)

a(n) = Sum_{k = 0..n} binomial(3*n-k-1,n-k)*binomial(4*n,k)^2.

a(n) = [x^n] (1 - x)^(2*n) * P(4*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.

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)

Maple

seq((4*n)!*(3*n)!/(n!*(2*n)!^3), n = 0..20);

Mathematica

Table[Binomial[4 n, 2 n] Binomial[3 n, 2 n], {n, 0, 15}] (* Michael De Vlieger, Aug 07 2016 *)

Crossrefs

Cf. A001448, A002894, A005809, A275652, A275654, A275655.

Sequence in context: A073960 A095786 A214181 * A214232 A146197 A301651

Adjacent sequences: A275650 A275651 A275652 * A275654 A275655 A275656

Keyword

nonn,easy

Author

Peter Bala, Aug 04 2016

Status

approved