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%I #73 Jul 11 2022 08:36:08
%S 1,1,3,19,135,1171,12543,156619,2185095,33787171,579341583,
%T 10927420219,223956672855,4940901389971,116678668726623,
%U 2938719256363819,78709685812037415,2234633592020685571,67005923560416063663,2114549937496479803419,70024572874029038582775,2427790107567416812409971
%N Expansion of e.g.f. Product_{k>=1} (1 + (exp(x) - 1)^k).
%C Stirling transform of A088311.
%C From _Peter Bala_, Jul 08 2022: (Start)
%C 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.
%C 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)
%H Vaclav Kotesovec, <a href="/A305550/b305550.txt">Table of n, a(n) for n = 0..400</a>
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StirlingTransform.html">Stirling Transform</a>
%F E.g.f.: exp(Sum_{k>=1} (-1)^k*(exp(x) - 1)^k/(k*((exp(x) - 1)^k - 1))).
%F a(n) = Sum_{k=0..n} Stirling2(n,k)*A088311(k).
%F From _Vaclav Kotesovec_, Jun 17 2018: (Start)
%F 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)).
%F 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)).
%F (End)
%p b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*add(
%p `if`(d::odd, d, 0), d=numtheory[divisors](j)), j=1..n)/n)
%p end:
%p a:= n-> add(Stirling2(n, k)*k!*b(k), k=0..n):
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Jun 15 2018
%t nmax = 21; CoefficientList[Series[Product[(1 + (Exp[x] - 1)^k), {k, 1, nmax}], {x, 0, nmax}], x] Range[0, nmax]!
%t 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]!
%t Table[Sum[StirlingS2[n, k] PartitionsQ[k] k!, {k, 0, n}], {n, 0, 21}]
%Y Cf. A000009, A088311, A167137, A306045, A320350.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Jun 15 2018
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