A355409 Expansion of e.g.f. 1/(1 + exp(2*x) - exp(3*x)).

1, 1, 7, 55, 571, 7471, 117307, 2148175, 44958571, 1058555791, 27693129307, 796934764495, 25018548004171, 850870651904911, 31163746960955707, 1222922731101304015, 51189052318085027371, 2276586205163067346831, 107204914362429152404507

Offset

0,3

Comments

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with period dividing phi(k) = A000010(k). For example, modulo 9 we obtain the sequence [1, 1, 7, 1, 4, 1, 1, 1, 7, 1, 4, 1, 1, 1, 7, 1, 4, 1, 1, ...] with an apparent period of 6 = phi(9) beginning at a(1). Cf. A354242. - Peter Bala, Apr 16 2024

Links

Table of n, a(n) for n=0..18.

Formula

a(0) = 1; a(n) = Sum_{k=1..n} (3^k - 2^k) * binomial(n,k) * a(n-k).

a(n) ~ n! / ((3 + r^2) * log(r)^(n+1)), where r = (1 + 2*cosh(log((29 + 3*sqrt(93))/2)/3))/3. - Vaclav Kotesovec, Jul 01 2022

Prog

(PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(1/(1+exp(2*x)-exp(3*x))))
(PARI) a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, (3^j-2^j)*binomial(i, j)*v[i-j+1])); v;

Crossrefs

Cf. A371460 (binomial transform).

Cf. A000010, A354242, A355381, A355408, A370092.

Sequence in context: A178922 A306046 A362080 * A112243 A083836 A326885

Adjacent sequences: A355406 A355407 A355408 * A355410 A355411 A355412

Keyword

nonn

Author

Seiichi Manyama, Jul 01 2022

Status

approved