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a(n) = (6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!).
3

%I #17 Jul 16 2023 05:50:50

%S 1,45,6237,1021020,178719453,32427545670,6016814703900,

%T 1133540594837892,215925912619400925,41477110789150966020,

%U 8019784929635201045862,1558875476359831844951100,304331361887290342345862940,59629409730107012112361325820

%N a(n) = (6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!).

%C Fractional factorials are defined in terms of the gamma function; for example, (n/3)! := Gamma(n/3 + 1).

%C Given two sequences of numbers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. For a list of the 52 sporadic integral factorial ratio sequences see A295431.

%C It is usually assumed that the c's and d's are integers but here we allow for some of the c's and d's to be rational numbers.

%C A295437, defined by A295437(n) = (18*n)!*n! / ((9*n)!*(6*n)!*(4*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 7). Here we are essentially considering the sequence {A295437(n/3) : n >= 0}. 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 Peter Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>

%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) ~ 2^(4*n/3 - 3/2) * 3^(4*n) / sqrt(Pi*n). - _Vaclav Kotesovec_, Jul 13 2023

%F a(n) = 5832*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(n*(n - 1)*(n - 2)*(2*n - 3)*(4*n - 3)*(4*n - 9))*a(n-3) for n >= 3 with a(0) = 1, a(1) = 45 and a(2) = 6237.

%p seq( simplify((6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!)), n = 0..15);

%t Table[Product[36*(6*k - 5)*(6*k - 1)/(k*(3*k + n)), {k, 1, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Jul 13 2023 *)

%Y Cf. A276100, A276101, A276102, A295431, A295437, A347854, A347855, A347856, A347857, A347858, A364173 - A364185.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 12 2023