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A276101 a(n) = (10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!).

%I A276101 #17 Nov 27 2017 08:34:03
%S A276101 1,1458,9723402,77636318760,665145965903562,5915482311008318958,
%T A276101 53837289804317953893960,497704257299202369371725086,
%U A276101 4653371135224869009103021872330,43880754270176401422739454033276880
%N A276101 a(n) = (10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!).
%C A276101 Fractional factorials are defined in terms of the gamma function, for example, (n/3)! := gamma(n/3 + 1).
%C A276101 This is only conjecturally an integer sequence. The similarly defined sequence (10*n)!*floor(n/3)!/((5*n)!*floor(10*n/3)!*(2*n)!) = A211418(10*n) is integral.
%C A276101 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(1/3*n). See A276100 ( u(1/2*n) ) and A276102 ( u(1/5*n) ).
%C A276101 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 known to be algebraic. Are the generating functions for u(1/2*n), u(1/3*n) and u(1/5*n) also algebraic?
%H A276101 P. Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>
%H A276101 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 A276101 O.g.f. A(x) = 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^3) + 1458*x*Hypergeom([29/30, 23/30, 17/30, 11/30, 13/10, 11/10, 9/10, 7/10], [17/15, 14/15, 11/15, 8/15, 5/6, 4/3, 2/3], (2^14*3^9*5^5)*x^3) + 9723402*x^2*Hypergeom([ 49/30, 43/30, 37/30, 31/30, 13/10, 11/10, 9/10, 7/10], [22/15, 19/15, 16/15, 13/15, 7/6, 5/3, 4/3],(2^14*3^9*5^5)*x^3).
%F A276101 a(n) ~ (2^14*3^9*5^5)^(n/3)/(sqrt(20*Pi*n)).
%p A276101 A211417 := proc(n)
%p A276101 (30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
%p A276101 end proc:
%p A276101 seq(simplify(A211417(1/3*n)), n = 0..10);
%t A276101 Table[(10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!) // FullSimplify, {n, 0, 9}] (* _Jean-François Alcover_, Nov 27 2017 *)
%Y A276101 Cf. A211417, A276100, A276102.
%K A276101 nonn,easy
%O A276101 0,2
%A A276101 _Peter Bala_, Aug 22 2016