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A354253 Expansion of e.g.f. 1/sqrt(9 - 8 * exp(x)).

%I A354253 #30 Nov 17 2023 11:20:13
%S A354253 1,4,52,1108,32980,1261204,58928212,3253363348,207225008980,
%T A354253 14958174725524,1206698072485972,107589343503498388,
%U A354253 10505997552329149780,1115087729794287434644,127819745001180490920532,15736779719362919373550228,2071062794354825889656471380
%N A354253 Expansion of e.g.f. 1/sqrt(9 - 8 * exp(x)).
%C A354253 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 7 we obtain the sequence [1, 4, 3, 2, 3, 0, 0, 4, 3, 2, 3, 0, 0, 4, 3, 2, 3, 0, 0, ...], with a preperiod of length 1 and an apparent period thereafter of 6 = phi(7). Cf. A354242. - _Peter Bala_, Jul 08 2022
%H A354253 Seiichi Manyama, <a href="/A354253/b354253.txt">Table of n, a(n) for n = 0..331</a>
%F A354253 E.g.f.: Sum_{k>=0} binomial(2*k,k) * (2 * (exp(x) - 1))^k.
%F A354253 a(n) = Sum_{k=0..n} 2^k * (2*k)! * Stirling2(n,k)/k!.
%F A354253 a(n) ~ sqrt(2) * n^n / (3 * exp(n) * log(9/8)^(n + 1/2)). - _Vaclav Kotesovec_, Jun 04 2022
%F A354253 Conjectural o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 4*x/(1 - 9*x/(1 - 12*x/(1 - 18*x/(1 - 20*x/(1 - 27*x/(1 - ... - (8*n-4)*x/(1 - 9*n*x/(1 - ...))))))))). - _Peter Bala_, Jul 08 2022
%F A354253 a(0) = 1; a(n) = Sum_{k=1..n} (8 - 4*k/n) * binomial(n,k) * a(n-k). - _Seiichi Manyama_, Sep 09 2023
%F A354253 a(0) = 1; a(n) = 4*a(n-1) - 9*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - _Seiichi Manyama_, Nov 17 2023
%o A354253 (PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(9-8*exp(x))))
%o A354253 (PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(2*k, k)*(2*(exp(x)-1))^k)))
%o A354253 (PARI) a(n) = sum(k=0, n, 2^k*(2*k)!*stirling(n, k, 2)/k!);
%Y A354253 Cf. A305404, A354242, A354252.
%Y A354253 Cf. A144828, A238465.
%K A354253 nonn
%O A354253 0,2
%A A354253 _Seiichi Manyama_, May 21 2022