A365026 - OEIS

%I #19 Oct 05 2023 08:37:18

%S 1,126,79380,65523780,60634147860,59774707082376,61346313465418800,

%T 64736852770959042240,69724035322703253191700,

%U 76277370761329867481375100,84482032811073922526904281880,94508142285721995026811874069200,106599928449546340546215262030974000

%N a(n) = (5*n)!*(9*n/2)!*(n/2)! / ((2*n)!^2 * (5*n/2)!^2 * n!).

%C Fractional factorials are defined in terms of the gamma function; for example, (9*n/2)! = Gamma(1 + 9*n/2).

%C Row 2 of A365025.

%H Paolo Xausa, <a href="/A365026/b365026.txt">Table of n, a(n) for n = 0..300</a>

%F a(n) = Sum_{j = 0..2*n} binomial(5*n, 2*n-j)^2 * binomial(n+j-1, j).

%F P-recursive: (5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(2*n)^2*(2*n-1)^2*(2*n-2)^2*(2*n-3)^2*a(n)= 9*(9*n-2)*(9*n-4)*(9*n-6)*(9*n-8)*(9*n-10)*(9*n-12)*(9*n-14)*(9*n-16)*(5*n-1)*(5*n-3)*(5*n-7)*(5*n-9)*a(n-2) with a(0) = 1 and a(1) = 126.

%F a(n) ~ c^n * 3*sqrt(5)/(20*Pi*n), where c = (3^9)/(2^4).

%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)^(9*n), where the power series G(x) = 1 + 14*x + 2744*x^2 + 1130724*x^3 + 615596785*x^4 + 388901411712*x^5 + 269588153179744*x^6 + ... appears to have integer coefficients.

%F exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^9, where the power series F(x) = 1 + 14*x + 4508*x^2 + 2489004*x^3 + 1728415009*x^4 + 1362984972918*x^5 + 1165343050808188*x^6 + ... appears to have integer coefficients.

%p seq( simplify((5*n)!*(9*n/2)!*(n/2)! / ((2*n)!^2 * (5*n/2)!^2 * n!)), n = 0..15);

%t A365026[n_]:=(5n)!(9n/2)!(n/2)!/((2n)!^2(5n/2)!^2n!);Array[A365026,15,0] (* _Paolo Xausa_, Oct 05 2023 *)

%o (Python)

%o from math import factorial

%o from sympy import factorial2

%o def A365026(n): return int(factorial(5*n)*factorial2(9*n)*factorial2(n)//((factorial2(5*n)*factorial(n<<1))**2*factorial(n))) # _Chai Wah Wu_, Aug 24 2023

%Y Cf. A275652, A365025, A365027.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Aug 17 2023