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a(n) = (9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!).
10

%I #10 Jul 16 2023 05:51:09

%S 1,128,43758,17039360,7012604550,2976412336128,1288415796384780,

%T 565399665327996928,250622090889055155270,111950839825145979207680,

%U 50312973039218473430585508,22723567527558510746926055424,10304958075870392958137083227804

%N a(n) = (9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!).

%C A295440, defined by A295440(n) = (18*n)!*(4*n)!*(3*n)! / ((9*n)!*(8*n)!*(6*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 10). Here we are essentially considering the sequence {A295440(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).

%C This sequence is only conjecturally an integer sequence.

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

%H J. W. Bober, <a href="http://arxiv.org/abs/0709.1977">Factorial ratios, hypergeometric series, and a family of step functions</a>, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.

%F a(n) ~ c^n * 1/sqrt(4*Pi*n), where c = (3^7)/(2^3) * sqrt(3) = 473.4993895191418....

%F a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(n*(n - 1)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 128.

%p seq( simplify((9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!)) , n = 0..15);

%Y Cf. A276100, A276101, A276102, A295431, A295440, A347854, A347855, A347856, A347857, A347858, A364172 - A364185.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 13 2023