login
a(n) = (6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!).
0

%I #10 Jul 16 2023 05:52:03

%S 1,36,3564,408408,49697388,6249195036,802241960520,104466877291260,

%T 13746018177013356,1823169705017624880,243331037661693468564,

%U 32641262295291161362656,4396944340992842923469640,594371374049863341847620936,80586283761263090599592845140

%N a(n) = (6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!).

%C A295445, defined by A295445(n) = (18*n)!*(2*n)! / ((9*n)!*(6*n)!*(5*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 15). Here we are essentially considering the sequence {A295445(n/3) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (2*n/3)! := Gamma(1 + 2*n/3).

%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(5*Pi*n) where c = (1296/25)*20^(1/3) = 140.7154092442799....

%F a(n) = 93312*(2*n - 3)*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(5*n*(n - 1)*(n - 2)*(5*n - 3)*(5*n - 6)*(5*n - 9)*(5*n - 12))*a(n-3) with a(0) = 1, a(1) = 36 and a(2) = 3564.

%p seq( simplify((6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!)), n = 0..15);

%Y Cf. A276100, A276101, A276102, A295431, A295445, A347854, A347855, A347856, A347857, A347858, A364172 - A364185.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 13 2023