%I #16 Feb 16 2024 06:48:14
%S 1,120,63900,63148000,85136103750,137629764435024,250331826090382280,
%T 494436455370401985600,1037731227148399567352625,
%U 2281874234819846601146115000,5205960892339635531670022801628,12237148815599682784939438806708960,29483782935554473122496294160376815950
%N G.f.: exp(Sum_{k>=1} (5*k)!/k!^5 * x^k/k).
%C In general, if r>=2, m>0 and g.f. = exp(m * Sum_{k>=1} (r*k)!/k!^r * x^k/k), then a(n) ~ c(r,m) * m * r^(r*n + 1/2) / ((2*Pi)^((r-1)/2) * n^((r+1)/2)) , where c(r,m) = exp((m * r! / r^r) * HypergeometricPFQ[{1, 1, (r+1)/r, (r+2)/r, ... , (2*r-1)/r}, {2, 2, ...r-times... 2, 2}, 1]). - _Vaclav Kotesovec_, Feb 16 2024
%F a(n) ~ c * 5^(5*n)/n^3, where c = sqrt(5) * exp(24*HypergeometricPFQ[{1, 1, 6/5, 7/5, 8/5, 9/5}, {2, 2, 2, 2, 2}, 1] / 625) / (4*Pi^2) = 0.05943406... - _Vaclav Kotesovec_, Mar 06 2020, updated Feb 16 2024
%F a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A008978(k) * a(n-k). - _Seiichi Manyama_, Feb 09 2024
%t CoefficientList[Series[Exp[Sum[(5*k)!/k!^5*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
%t CoefficientList[Series[Exp[120*x*HypergeometricPFQ[{1, 1, 6/5, 7/5, 8/5, 9/5}, {2, 2, 2, 2, 2}, 3125*x]], {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 09 2024 *)
%Y Cf. A000108, A008978, A229451, A333042, A370295.
%K nonn
%O 0,2
%A _Vaclav Kotesovec_, Mar 06 2020