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(5*n+2)!/(2*(n!)^5), n >= 0.
0

%I #15 Jul 27 2022 08:36:01

%S 1,2520,7484400,22870848000,70579794285000,218799620836917120,

%T 679953587124305894400,2116187746296592370688000,

%U 6592431144164903462359935000,20550499897066845200729434200000,64091912654977017603465324370118400,199956261330234671205699024876891648000

%N (5*n+2)!/(2*(n!)^5), n >= 0.

%C Although limit( a(n)^(1/n), n=infinity ) = 5^5, apparently this sequence is not a Hausdorff moment sequence of any positive function on (0,5^5).

%F In Maple notation:

%F O.g.f. : hypergeom([3/5, 4/5, 6/5, 7/5], [1, 1, 1], 5^5*z);

%F E.g.f. : hypergeom([3/5, 4/5, 6/5, 7/5], [1, 1, 1, 1], 5^5*z);

%F Asymptotics: a(n) -> (25*n^2+5*n-2)*(5^(5*n+1/2))* n^(-2)/(8*Pi^2), for n -> infinity.

%F D-finite with recurrence (n^4)*a(n) -5*(5*n+1)*(5*n+2)*(5*n-2)*(5*n-1)*a(n-1)=0. - _R. J. Mathar_, Jul 27 2022

%p seq((5*n+2)!/(2*(n!)^5), n=0..11).

%t Table[(5n+2)!/(2(n!)^5),{n,0,15}] (* _Harvey P. Dale_, Aug 04 2019 *)

%Y Cf. A002544.

%K nonn

%O 0,2

%A _Karol A. Penson_, Aug 09 2013