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A346983 Expansion of e.g.f. 1 / (5 - 4 * exp(x))^(1/4).

%I A346983 #23 Nov 16 2023 11:50:42
%S A346983 1,1,6,61,891,16996,400251,11217781,364638336,13486045291,
%T A346983 559192836771,25691965808026,1295521405067181,71131584836353861,
%U A346983 4224255395774155566,269791923787785076921,18439806740525320993551,1342957106015632474616956,103824389511747541791086511
%N A346983 Expansion of e.g.f. 1 / (5 - 4 * exp(x))^(1/4).
%C A346983 Stirling transform of A007696.
%H A346983 Seiichi Manyama, <a href="/A346983/b346983.txt">Table of n, a(n) for n = 0..360</a>
%F A346983 a(n) = Sum_{k=0..n} Stirling2(n,k) * A007696(k).
%F A346983 a(n) ~ n! / (Gamma(1/4) * 5^(1/4) * n^(3/4) * log(5/4)^(n + 1/4)). - _Vaclav Kotesovec_, Aug 14 2021
%F A346983 O.g.f. (conjectural): 1/(1 - x/(1 - 5*x/(1 - 5*x/(1 - 10*x/(1 - 9*x/(1 - 15*x/(1 - ... - (4*n-3)*x/(1 - 5*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type. - _Peter Bala_, Aug 22 2023
%F A346983 a(0) = 1; a(n) = Sum_{k=1..n} (4 - 3*k/n) * binomial(n,k) * a(n-k). - _Seiichi Manyama_, Sep 09 2023
%F A346983 a(0) = 1; a(n) = a(n-1) - 5*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - _Seiichi Manyama_, Nov 16 2023
%p A346983 g:= proc(n) option remember; `if`(n<2, 1, (4*n-3)*g(n-1)) end:
%p A346983 b:= proc(n, m) option remember;
%p A346983      `if`(n=0, g(m), m*b(n-1, m)+b(n-1, m+1))
%p A346983     end:
%p A346983 a:= n-> b(n, 0):
%p A346983 seq(a(n), n=0..18);  # _Alois P. Heinz_, Aug 09 2021
%t A346983 nmax = 18; CoefficientList[Series[1/(5 - 4 Exp[x])^(1/4), {x, 0, nmax}], x] Range[0, nmax]!
%t A346983 Table[Sum[StirlingS2[n, k] 4^k Pochhammer[1/4, k], {k, 0, n}], {n, 0, 18}]
%Y A346983 Cf. A000670, A007696, A305404, A346982, A346984, A346985, A352117, A352118, A352119.
%Y A346983 Cf. A094417, A354242, A365567.
%K A346983 nonn,easy
%O A346983 0,3
%A A346983 _Ilya Gutkovskiy_, Aug 09 2021