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A276100
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a(n) = (15*n)!*(n/2)!/((15*n/2)!*(5*n)!*(3*n)!).
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21
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1, 114688, 77636318760, 62505037015810048, 53837289804317953893960, 48066503353826060675410034688, 43880754270176401422739454033276880, 40671547154451909281150562260837340282880, 38113558705192522309151157825210540422513019720
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OFFSET
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0,2
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COMMENTS
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Fractional factorials are defined in terms of the gamma function, for example, (n/2)! := gamma(n/2 + 1).
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.
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) ).
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?
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LINKS
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FORMULA
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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).
a(n) ~ (2^14*3^9*5^5)^(n/2)/sqrt(30*Pi*n).
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MAPLE
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(30*n)!*(n)!/((15*n)!(10*n)!(6*n)!);
end proc:
seq(simplify(A211417(1/2*n)), n = 0..10);
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MATHEMATICA
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Table[(15 n)!*(n/2)!/((15 n/2)!*(5 n)!*(3 n)!), {n, 0, 8}] (* Michael De Vlieger, Aug 28 2016 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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