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

1, 50, 8250, 1636250, 349456250, 77636318760, 17672894531250, 4089765214843750, 957711284472656250, 226280605806640625000, 53837289804317953893960, 12880759628253295898437500

Offset: 0

Author

Peter Bala, Aug 22 2016

Keywords

• nonn
• easy

Comments

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

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.

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) ).

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?

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

Links

• P. Bala, Some integer ratios of factorials
• F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Crossrefs

Cf. A211417, A276100, A276101.

Programs

Maple

A211417 := proc(n)
(30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
end proc:
seq(simplify(A211417(1/5*n)), n = 0..10);

Mathematica

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

Formula

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