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A276101
a(n) = (10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!).
19
1, 1458, 9723402, 77636318760, 665145965903562, 5915482311008318958, 53837289804317953893960, 497704257299202369371725086, 4653371135224869009103021872330, 43880754270176401422739454033276880
OFFSET
0,2
COMMENTS
Fractional factorials are defined in terms of the gamma function, for example, (n/3)! := gamma(n/3 + 1).
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.
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) ).
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?
FORMULA
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).
a(n) ~ (2^14*3^9*5^5)^(n/3)/(sqrt(20*Pi*n)).
MAPLE
A211417 := proc(n)
(30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
end proc:
seq(simplify(A211417(1/3*n)), n = 0..10);
MATHEMATICA
Table[(10*n)!*(n/3)!/((5*n)!*(10*n/3)!*(2*n)!) // FullSimplify, {n, 0, 9}] (* Jean-François Alcover, Nov 27 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Aug 22 2016
STATUS
approved