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A276102 a(n) = (6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!).

%I A276102 #19 Nov 27 2017 08:34:11
%S A276102 1,50,8250,1636250,349456250,77636318760,17672894531250,
%T A276102 4089765214843750,957711284472656250,226280605806640625000,
%U A276102 53837289804317953893960,12880759628253295898437500
%N A276102 a(n) = (6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!).
%C A276102 Fractional factorials are defined in terms of the gamma function, for example, (n/5)! := gamma(n/5 + 1).
%C A276102 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 A276102 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 A276102 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 A276102 The o.g.f. for this sequence can be expressed as a sum of 5 generalized hypergeometric functions of type 8F7.
%H A276102 P. Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>
%H A276102 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 A276102 a(n) ~ (2^14*3^9*5^5)^(n/5)/sqrt(12*Pi*n).
%p A276102 A211417 := proc(n)
%p A276102 (30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
%p A276102 end proc:
%p A276102 seq(simplify(A211417(1/5*n)), n = 0..10);
%t A276102 Table[(6*n)!*(n/5)!/((3*n)!*(2*n)!*(6*n/5)!) // FullSimplify, {n, 0, 11}] (* _Jean-François Alcover_, Nov 27 2017 *)
%Y A276102 Cf. A211417, A276100, A276101.
%K A276102 nonn,easy
%O A276102 0,2
%A A276102 _Peter Bala_, Aug 22 2016