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G.f.: exp(Sum_{k>=1} (5*k)!/(5!*k!^5) * x^k/k).
5

%I #11 Feb 15 2024 17:23:21

%S 1,1,473,467606,637121154,1039792179805,1905441263652576,

%T 3785382599457953517,7981116324798212651066,

%U 17613760342120835610374245,40303877398793645855018120732,94970269248783993542201925505548,229287077006842005926064077532676555,565001770629439341048001870559581136157

%N G.f.: exp(Sum_{k>=1} (5*k)!/(5!*k!^5) * x^k/k).

%C In general, for m>=2, if g.f. = exp(Sum_{k>=1} (m*k)!/(m!*k!^m) * x^k/k), then a(n,m) ~ c(m) * m^(m*n) / n^((m+1)/2), where c(m) = exp(HypergeometricPFQ[{1, 1, (m+1)/m, (m+2)/m, ... , (2*m-1)/m}, {2, 2, ...m-times... 2, 2}, 1] / m^m) / (m! * (2*Pi)^((m-1)/2) / sqrt(m)).

%C Limit_{m->oo} c(m) / (exp(m)/(m^m*(2*Pi)^(m/2))) = 1.

%F G.f. A(x) = G(x)^(1/120), where G(x) is the g.f. for A333043.

%F a(n) ~ c * 5^(5*n)/n^3, where c = exp(HypergeometricPFQ[{1, 1, 6/5, 7/5, 8/5, 9/5}, {2, 2, 2, 2, 2}, 1] / 3125) / (96*sqrt(5)*Pi^2) = 0.00047219161473962545263459216995582653262467228952818554361164671183728...

%t CoefficientList[Series[Exp[Sum[(5*k)!/(5!*k!^5)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]

%t CoefficientList[Series[Exp[x*HypergeometricPFQ[{1, 1, 6/5, 7/5, 8/5, 9/5}, {2, 2, 2, 2, 2}, 3125*x]], {x, 0, 20}], x]

%Y Cf. A008978, A322252, A229452, A370294, A333043.

%K nonn

%O 0,3

%A _Vaclav Kotesovec_, Feb 14 2024