A347023 E.g.f.: 1 / (1 - 6 * log(1 + x))^(1/6).

1, 1, 6, 72, 1254, 28794, 819888, 27869316, 1101032100, 49570797780, 2505156062472, 140417898936336, 8644973807845368, 579908437058338920, 42098286646367326368, 3288252917244250703664, 274974019392668843164176, 24510436934573885695407504, 2319947117871178825560902112

Offset: 0

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

• nonn

Comments

In general, for k > 1, if e.g.f. = 1 / (1 - k*log(1 + x))^(1/k), then a(n) ~ n! * exp(1/k^2) / (Gamma(1/k) * k^(1/k) * n^(1 - 1/k) * (exp(1/k) - 1)^(n + 1/k)). - Vaclav Kotesovec, Aug 14 2021

Crossrefs

Cf. A006252, A008542, A320343, A346985, A347019, A347020, A347021, A347022.

Programs

Mathematica

nmax = 18; CoefficientList[Series[1/(1 - 6 Log[1 + x])^(1/6), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[StirlingS1[n, k] 6^k Pochhammer[1/6, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * A008542(k).

a(n) ~ n! * exp(1/36) / (Gamma(1/6) * 6^(1/6) * n^(5/6) * (exp(1/6) - 1)^(n + 1/6)). - Vaclav Kotesovec, Aug 14 2021