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%I #25 Nov 16 2023 11:50:18
%S 1,1,5,41,477,7201,133685,2945881,75145677,2177900241,70687244965,
%T 2539879312521,100086803174077,4291845333310081,198954892070938645,
%U 9914294755149067961,528504758009562261677,30010032597449931644721,1808359960001658961070725
%N Expansion of e.g.f. 1 / (4 - 3 * exp(x))^(1/3).
%C Stirling transform of A007559.
%H Seiichi Manyama, <a href="/A346982/b346982.txt">Table of n, a(n) for n = 0..373</a>
%F a(n) = Sum_{k=0..n} Stirling2(n,k) * A007559(k).
%F a(n) ~ n! / (Gamma(1/3) * 2^(2/3) * n^(2/3) * log(4/3)^(n + 1/3)). - _Vaclav Kotesovec_, Aug 14 2021
%F From _Peter Bala_, Aug 22 2023: (Start)
%F O.g.f. (conjectural): 1/(1 - x/(1 - 4*x/(1 - 4*x/(1 - 8*x/(1 - 7*x/(1 - 12*x/(1 - ... - (3*n-2)*x/(1 - 4*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type (S-fraction).
%F More generally, it appears that the o.g.f. of the sequence whose e.g.f. is equal to 1/(r+1 - r*exp(s*x))^(m/s) corresponds to the S-fraction 1/(1 - r*m*x/(1 - s*(r+1)*x/(1 - r*(m+s)*x/(1 - 2*s(r+1)*x/(1 - r*(m+2*s)*x/(1 - 3*s(r+1)*x/( 1 - ... ))))))). This is the case r = 3, s = 1, m = 1/3. (End)
%F a(0) = 1; a(n) = Sum_{k=1..n} (3 - 2*k/n) * binomial(n,k) * a(n-k). - _Seiichi Manyama_, Sep 09 2023
%F a(0) = 1; a(n) = a(n-1) - 4*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - _Seiichi Manyama_, Nov 16 2023
%p g:= proc(n) option remember; `if`(n<2, 1, (3*n-2)*g(n-1)) end:
%p b:= proc(n, m) option remember;
%p `if`(n=0, g(m), m*b(n-1, m)+b(n-1, m+1))
%p end:
%p a:= n-> b(n, 0):
%p seq(a(n), n=0..18); # _Alois P. Heinz_, Aug 09 2021
%t nmax = 18; CoefficientList[Series[1/(4 - 3 Exp[x])^(1/3), {x, 0, nmax}], x] Range[0, nmax]!
%t Table[Sum[StirlingS2[n, k] 3^k Pochhammer[1/3, k], {k, 0, n}], {n, 0, 18}]
%Y Cf. A000670, A007559, A305404, A346983, A346984, A346985, A352117, A352118, A352119.
%Y Cf. A032033, A365558.
%K nonn,easy
%O 0,3
%A _Ilya Gutkovskiy_, Aug 09 2021
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