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a(n) = (15*n)!*(n/2)!/((15*n/2)!*(5*n)!*(3*n)!).
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%I #27 Oct 04 2016 13:37:54

%S 1,114688,77636318760,62505037015810048,53837289804317953893960,

%T 48066503353826060675410034688,43880754270176401422739454033276880,

%U 40671547154451909281150562260837340282880,38113558705192522309151157825210540422513019720

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

%C Fractional factorials are defined in terms of the gamma function, for example, (n/2)! := gamma(n/2 + 1).

%C This is only conjecturally an integer sequence. The similarly defined sequence (15*n)!*floor(n/2)!/(floor(15*n/2)!*(5*n)!*(3*n)!) = A211418(15*n) is integral.

%C Let u(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!) = A211417(n). This sequence of ratios of factorials is integral and was used by Chebyshev in his estimate of the number of primes less than or equal to a fixed integer n. The three sequences u(1/2*n), u(1/3*n) and u(1/5*n) also appear to be integral (checked up to n = 200). This is the sequence u(1/2*n). See A276101( u(1/3*n) ) and A276102( u(1/5*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 known to be algebraic - see Rodriguez-Villegas. Are the generating functions for u(1/2*n), u(1/3*n) and u(1/5*n) also algebraic?

%H Carauleanu Marc, <a href="/A276100/b276100.txt">Table of n, a(n) for n = 0..111</a>

%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 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^2) + 114688*x*Hypergeom([22/15, 19/15, 17/15, 16/15, 14/15, 13/15, 11/15, 8/15], [13/10, 11/10, 9/10, 7/10, 7/6, 5/6, 3/2], (2^14*3^9*5^5)*x^2).

%F a(n) ~ (2^14*3^9*5^5)^(n/2)/sqrt(30*Pi*n).

%p A211417 := proc(n)

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

%p end proc:

%p seq(simplify(A211417(1/2*n)), n = 0..10);

%t Table[(15 n)!*(n/2)!/((15 n/2)!*(5 n)!*(3 n)!), {n, 0, 8}] (* _Michael De Vlieger_, Aug 28 2016 *)

%Y Cf. A211417, A211418, A276101, A276102.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Aug 22 2016