A166384 a(n) = (5*n)!/(2*n)!.

1, 60, 151200, 1816214400, 60339831552000, 4274473667143680000, 553761949463615692800000, 118528911546113729396736000000, 38996486014317601426443730944000000, 18684042257398689569086457362513920000000, 12501158328406120266757143916231576780800000000

Offset: 0

Author

Karol A. Penson, Oct 13 2009

Keywords

• nonn

Comments

Integral representation as n-th moment of a positive function on a positive halfaxis (solution of the Stieltjes moment problem), in Maple notation:

a(n) = Integral_{x=0..oo} ( x^n*((1/3125)*sqrt(5)*((625/4)*Pi*csc((2/5)*Pi)*csc((1/5)*Pi)*sin((3/10)*Pi)*Gamma(7/10)*2^(2/5)*hypergeom([7/10], [2/5, 3/5, 4/5], -(4/3125)*x)/(Gamma(4/5)*x^(4/5)) - (625/4)*Pi*sec((3/10)*Pi) *csc((2/5)*Pi)*sin((1/10)*Pi)*Gamma(9/10)*2^(4/5)*hypergeom([9/10], [3/5, 4/5, 6/5], -(4/3125)*x)/(Gamma(3/5)*x^(3/5)) - (125/8)*Pi*sec((3/10)*Pi)*Gamma(3/5)*2^(1/5)*hypergeom([11/10], [4/5, 6/5, 7/5], -(4/3125)*x)/(Gamma(9/10)*x^(2/5)) + (125/8)*Pi*csc((2/5)*Pi)*Gamma(4/5)*2^(3/5)*hypergeom([13/10], [6/5, 7/5,8/5], -(4/3125)*x)/(Gamma(7/10)*x^(1/5)))/Pi^(3/2)) ).

This solution is not unique.

From G. C. Greubel, May 12 2016: (Start)

2^n * 10^(floor(n/2)) | a(n), for n>=0.

2^(2*n) * 10^(floor(n/2)) | a(n), for n>=2. (End)

Links

• G. C. Greubel, Table of n, a(n) for n = 0..100

Programs

Mathematica

Table[(5*n)!/(2*n)!, {n, 0, 25}] (* G. C. Greubel, May 12 2016 *)

Formula

G.f.: Sum_{n>=0} a(n)*x^n/((n!)^4) = hypergeom([1/5, 2/5, 3/5, 4/5],[1/2, 1, 1, 1], (3125/4)*x).

Asymptotics: a(n) = sqrt(10)*(1/2-1/(80*n)+1/(6400*n^2)+(619/3840000)*sqrt(2)/n^3+O(1/n^4))*(5^n)^5/(((1/n)^n)^3*(exp(n))^3*(2^n)^2).

From Seiichi Manyama, May 25 2025: (Start)

a(n) = RisingFactorial(2*n+1,3*n).

a(n) = (3*n)! * [x^(3*n)] 1/(1 - x)^(2*n+1). (End)