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A354253
Expansion of e.g.f. 1/sqrt(9 - 8 * exp(x)).
5
1, 4, 52, 1108, 32980, 1261204, 58928212, 3253363348, 207225008980, 14958174725524, 1206698072485972, 107589343503498388, 10505997552329149780, 1115087729794287434644, 127819745001180490920532, 15736779719362919373550228, 2071062794354825889656471380
OFFSET
0,2
COMMENTS
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
LINKS
FORMULA
E.g.f.: Sum_{k>=0} binomial(2*k,k) * (2 * (exp(x) - 1))^k.
a(n) = Sum_{k=0..n} 2^k * (2*k)! * Stirling2(n,k)/k!.
a(n) ~ sqrt(2) * n^n / (3 * exp(n) * log(9/8)^(n + 1/2)). - Vaclav Kotesovec, Jun 04 2022
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
a(0) = 1; a(n) = Sum_{k=1..n} (8 - 4*k/n) * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023
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
PROG
(PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(9-8*exp(x))))
(PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(2*k, k)*(2*(exp(x)-1))^k)))
(PARI) a(n) = sum(k=0, n, 2^k*(2*k)!*stirling(n, k, 2)/k!);
CROSSREFS
KEYWORD
nonn
AUTHOR
Seiichi Manyama, May 21 2022
STATUS
approved