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a(n) = (7*n)!*(13*n/2)!*(n/2)! / ((3*n)!^2 * (7*n/2)!^2 * n!).
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%I #17 Oct 05 2023 08:37:13

%S 1,1716,20612592,328206021000,5876083665270000,112210544802995673216,

%T 2232092469681027490937400,45670179632369542491712236480,

%U 953926390279492216468973361270000,20241460048032081192591594667805420400,434878619369192244460121948456800558766592

%N a(n) = (7*n)!*(13*n/2)!*(n/2)! / ((3*n)!^2 * (7*n/2)!^2 * n!).

%C Fractional factorials are defined in terms of the gamma function; for example, (13*n/2)! = Gamma(1 + 13*n/2).

%C Row 3 of A365025.

%H Paolo Xausa, <a href="/A365027/b365027.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = Sum_{j = 0..3*n} binomial(7*n, 3*n-j)^2 * binomial(n+j-1, j).

%F a(n) ~ c^n * sqrt(91)/(42*Pi*n), where c = sqrt(13)*(13/3)^6.

%F Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all integers n and r.

%F a(n) = [x^n] G(x)^(78*n), where the power series G(x) = 1 + 22*x + 94622*x^2 + 821626080*x^3 + 9321370449728*x^4 + 122052794640882192*x^5 + 1748115226331150054950*x^6 + ... appears to have integer coefficients.

%F exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^78, where the power series F(x) = 1 + 22*x + 132374*x^2 + 1405498512*x^3 + 18873219886000*x^4 + 288319543590164888*x^5 + 4779239354183722040470*x^6 + ... appears to have integer coefficients.

%p seq( simplify((7*n)!*(13*n/2)!*(n/2)! / ((3*n)!^2 * (7*n/2)!^2 * n!)), n = 0..15);

%t A365027[n_]:=(7n)!(13n/2)!(n/2)!/((3n)!^2(7n/2)!^2n!);Array[A365027,10,0] (* _Paolo Xausa_, Oct 05 2023 *)

%o (Python)

%o from math import factorial

%o from sympy import factorial2

%o def A365027(n): return int(factorial(7*n)*factorial2(13*n)*factorial2(n)//((factorial2(7*n)*factorial(3*n))**2*factorial(n))) # _Chai Wah Wu_, Aug 24 2023

%Y Cf. A275652, A365025, A365026.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Aug 18 2023