%I #11 Jul 03 2014 16:26:47
%S 1,19,961,101101,19116721,5895818671,2767218413569,1876349280125881,
%T 1765053702368618401,2229877686380646862891,3684146198939103437432641,
%U 7785613784940909310055130469,20653334793956901864746843979601,67675129289987844087403319678358151
%N a(n) = (n+2)!^2*(n+1)!/4*hypergeom([-n],[2,3,3],-1).
%C Special values of hypergeometric function of type 1F3.
%C For all n it seems that the last digit of a(n) is either 1 or 9.
%p Digits:= 1100:
%p a:= n-> round(evalf((n+2)!^2*(n+1)!/4*hypergeom([-n], [2, 3, 3], -1))):
%p seq(a(n), n=0..20); # _Alois P. Heinz_, Mar 04 2014
%t Table[(n+2)!^2*(n+1)!/4*HypergeometricPFQ[{-n},{2,3,3},-1],{n,0,10}] (* _Vaclav Kotesovec_, Mar 04 2014 *)
%K nonn
%O 0,2
%A _Karol A. Penson_, Mar 04 2014