login
a(n) = (15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!).
0

%I #15 Jul 16 2023 05:46:23

%S 1,35840,5545451340,991901222174720,188242272043069768860,

%T 36901030731039027064995840,7383354803839076831124554790900,

%U 1498315221854950975184507333477662720,307213802011837003346320048243705086348060

%N a(n) = (15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!).

%C A295458, defined by A295458(n) = (30*n)!*(5*n)!*(4*n)! / ((15*n)!*(10*n)!*(8*n)!*(6*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 28). Here we are essentially considering the sequence {A295458(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (5*n/2)! := Gamma(1 + 5*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(12*Pi*n), where c = (3^4)*(5^5) * sqrt(3)/2.

%F a(n) = 43200*(15*n - 1)*(15*n - 7)*(15*n - 11)*(15*n - 13)*(15*n - 17)*(15*n - 19)*(15*n - 23)*(15*n - 29)/(n*(n - 1)*(3*n - 2)*(3*n - 4)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) with a(0) = 1 and a(1) = 35840.

%p seq( simplify((15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!)), n = 0..15);

%Y Cf. A276100, A276101, A276102, A295431, A295458, A347854, A347855, A347856, A347857, A347858, A364173 - A364185.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 13 2023