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a(n) = (6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!).
19

%I #19 Nov 27 2017 08:34:11

%S 1,50,8250,1636250,349456250,77636318760,17672894531250,

%T 4089765214843750,957711284472656250,226280605806640625000,

%U 53837289804317953893960,12880759628253295898437500

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

%C Fractional factorials are defined in terms of the gamma function, for example, (n/5)! := gamma(n/5 + 1).

%C This is only conjecturally an integer sequence. The similarly defined sequence (6*n)!*floor(n/5)!/((3*n)!*(2*n)!*floor(6*n/5)!) = A211418(6*n) is integral.

%C Let u(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!) = A211417(n). The three sequences u(1/2*n), u(1/3*n) and u(1/5*n) appear to be integral (checked up to n = 200). This is the sequence u(1/5*n). See A276100 ( u(1/2*n) ) and A276101 ( u(1/3*n) ).

%C The generating function for u(n) is Hypergeom([29/30, 23/30, 19/30, 17/30, 13/30, 11/30, 7/30, 1/30], [4/5, 3/5, 2/5, 1/5, 2/3, 1/3, 1/2], (2^14*3^9*5^5)*x) and is algebraic - see Rodriguez-Villegas. Are the generating functions for u(1/2*n), u(1/3*n) and u(1/5*n) also algebraic?

%C The o.g.f. for this sequence can be expressed as a sum of 5 generalized hypergeometric functions of type 8F7.

%H P. Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>

%H F. Rodriguez-Villegas, <a href="http://arxiv.org/abs/math/0701362">Integral ratios of factorials and algebraic hypergeometric functions</a>, arXiv:math/0701362 [math.NT], 2007.

%F a(n) ~ (2^14*3^9*5^5)^(n/5)/sqrt(12*Pi*n).

%p A211417 := proc(n)

%p (30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);

%p end proc:

%p seq(simplify(A211417(1/5*n)), n = 0..10);

%t Table[(6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!) // FullSimplify, {n, 0, 11}] (* _Jean-François Alcover_, Nov 27 2017 *)

%Y Cf. A211417, A276100, A276101.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Aug 22 2016