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a(n) = (9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!).
0

%I #10 Jul 16 2023 05:51:37

%S 1,48,4862,549120,65132550,7945986048,987291797996,124259864002560,

%T 15789207515217990,2021092963752345600,260227401685879140612,

%U 33665720694993527504896,4372592850984736084611996,569819472537519480058675200,74468439316740019538310543000

%N a(n) = (9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!).

%C A295442, defined by A295442(n) = (18*n)!*(5*n)!*(3*n)!/((10*n)!*(9*n)!*(6*n)!*n!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 12). Here we are essentially considering the sequence {A295442(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(2*Pi*n), where c = 2*(3^7)/(5^3) * sqrt(15) = 135.5234332504899....

%F a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(5*n*(n - 1)*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 48.

%p seq( simplify((9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!)), n = 0..15)

%Y Cf. A276100, A276101, A276102, A295431, A295442, A347854, A347855, A347856, A347857, A347858, A364172 - A364185.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 13 2023