A305550 Expansion of e.g.f. Product_{k>=1} (1 + (exp(x) - 1)^k).

1, 1, 3, 19, 135, 1171, 12543, 156619, 2185095, 33787171, 579341583, 10927420219, 223956672855, 4940901389971, 116678668726623, 2938719256363819, 78709685812037415, 2234633592020685571, 67005923560416063663, 2114549937496479803419, 70024572874029038582775, 2427790107567416812409971

Offset: 0

Author

Ilya Gutkovskiy, Jun 15 2018

Keywords

• nonn

Comments

Stirling transform of A088311.

From Peter Bala, Jul 08 2022: (Start)

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 16 we obtain the sequence [1, 1, 3, 3, 7, 3, 15, 11, 7, 3, 15, 11, 7, 3, 15, 11, ...], with an apparent period of 4 beginning at a(4). Cf. A167137.

More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series. (End)

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..400
• N. J. A. Sloane, Transforms
• Eric Weisstein's World of Mathematics, Stirling Transform

Crossrefs

Cf. A000009, A088311, A167137, A306045, A320350.

Programs

Maple

b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*add(
     `if`(d::odd, d, 0), d=numtheory[divisors](j)), j=1..n)/n)
    end:
a:= n-> add(Stirling2(n, k)*k!*b(k), k=0..n):
seq(a(n), n=0..25);  # Alois P. Heinz, Jun 15 2018

Mathematica

nmax = 21; CoefficientList[Series[Product[(1 + (Exp[x] - 1)^k), {k, 1, nmax}], {x, 0, nmax}], x] Range[0, nmax]!
nmax = 21; CoefficientList[Series[Exp[Sum[(-1)^k (Exp[x] - 1)^k/(k ((Exp[x] - 1)^k - 1)), {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[StirlingS2[n, k] PartitionsQ[k] k!, {k, 0, n}], {n, 0, 21}]

Formula

E.g.f.: exp(Sum_{k>=1} (-1)^k*(exp(x) - 1)^k/(k*((exp(x) - 1)^k - 1))).

a(n) = Sum_{k=0..n} Stirling2(n,k)*A088311(k).

From Vaclav Kotesovec, Jun 17 2018: (Start)

a(n) ~ n! * exp(Pi*sqrt(n/(6*log(2))) + (1/log(2) - 1) * Pi^2/48) / (2^(9/4) * 3^(1/4) * n^(3/4) * (log(2))^(n + 1/4)).

a(n) ~ sqrt(Pi) * exp(Pi*sqrt(n/(6*log(2))) + (1/log(2) - 1) * Pi^2/48 - n) * n^(n + 1/2) / (2^(7/4) * 3^(1/4) * n^(3/4) * (log(2))^(n + 1/4)).

(End)