%I #37 Nov 16 2023 11:50:46

%S 1,2,14,158,2486,50222,1239254,36126638,1214933846,46299580142,

%T 1971815255894,92809525295918,4784166929982806,268050260650705262,

%U 16219498558371118934,1054102762745609325998,73229184033780135425366,5415407651703010175897582

%N Expansion of e.g.f. 1/sqrt(5 - 4 * exp(x)).

%C From _Peter Bala_, Jul 07 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, 2, 14, 14, 6, 14, 6, 14, 6, ...], with an apparent period of 2 beginning at a(3). Cf. A354253.

%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 Seiichi Manyama, <a href="/A354242/b354242.txt">Table of n, a(n) for n = 0..360</a>

%F E.g.f.: Sum_{k>=0} binomial(2*k,k) * (exp(x) - 1)^k.

%F a(n) = Sum_{k=0..n} (2*k)! * Stirling2(n,k)/k!.

%F a(n) ~ sqrt(2/5) * n^n / (exp(n) * log(5/4)^(n + 1/2)). - _Vaclav Kotesovec_, Jun 04 2022

%F Conjectural o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 2*x/(1 - 5*x/(1 - 6*x/(1 - 10*x/(1 - 10*x/(1 - 15*x/(1 - ... - (4*n-2)*x/(1 - 5*n*x/(1 - ...))))))))). - _Peter Bala_, Jul 07 2022

%F a(0) = 1; a(n) = Sum_{k=1..n} (4 - 2*k/n) * binomial(n,k) * a(n-k). - _Seiichi Manyama_, Sep 09 2023

%F a(0) = 1; a(n) = 2*a(n-1) - 5*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - _Seiichi Manyama_, Nov 16 2023

%o (PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(5-4*exp(x))))

%o (PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(2*k, k)*(exp(x)-1)^k)))

%o (PARI) a(n) = sum(k=0, n, (2*k)!*stirling(n, k, 2)/k!);

%Y Cf. A305404, A354252, A354253.

%Y Cf. A000670, A001813, A094417, A316747, A354240, A354241.

%K nonn

%O 0,2

%A _Seiichi Manyama_, May 20 2022