cp's OEIS Frontend

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

%I A275653 #12 Mar 24 2022 10:33:43
%S A275653 1,18,1050,77616,6370650,554822268,50199951984,4664758248000,
%T A275653 442077195513690,42533571002422500,4141601026094832300,
%U A275653 407220411993767798400,40363606408574136870000,4028061310168832261158176,404311537318239680601595200,40785601782042745410592271616
%N A275653 a(n) = binomial(4*n,2*n)*binomial(3*n,2*n).
%C A275653 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).
%C A275653 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).
%C A275653 Compare with the identities
%C A275653 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).
%C A275653 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)
%C A275653 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).
%C A275653 See also A275652, A275654 and A275655.
%F A275653 a(n) = (4*n)!*(3*n)!/(n!*(2*n)!^3).
%F A275653 a(n) = A001448(n) * A005809(n).
%F A275653 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).
%F A275653 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.
%F A275653 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.
%F A275653 a(n) ~ sqrt(3/2)*108^n/(2*Pi*n). - _Ilya Gutkovskiy_, Aug 07 2016
%F A275653 From _Peter Bala_, Mar 23 2022: (Start)
%F A275653 a(n) = Sum_{k = 0..n} binomial(3*n-k-1,n-k)*binomial(4*n,k)^2.
%F A275653 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.
%F A275653 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)
%p A275653 seq((4*n)!*(3*n)!/(n!*(2*n)!^3), n = 0..20);
%t A275653 Table[Binomial[4 n, 2 n] Binomial[3 n, 2 n], {n, 0, 15}] (* _Michael De Vlieger_, Aug 07 2016 *)
%Y A275653 Cf. A001448, A002894, A005809, A275652, A275654, A275655.
%K A275653 nonn,easy
%O A275653 0,2
%A A275653 _Peter Bala_, Aug 04 2016