A294213 E.g.f.: exp(1/((1-x)*(1-x^2)) - 1).

1, 1, 5, 25, 193, 1601, 16741, 190345, 2509025, 35825473, 569012581, 9716400761, 180303804385, 3569527588225, 75681964322693, 1700163418683241, 40499757023856961, 1016190431274596225, 26843084299482509125, 743180975111364212953

Offset

0,3

Links

Seiichi Manyama, Table of n, a(n) for n = 0..435

Formula

a(n) ~ exp(-61/96 + 3*n^(1/3)/4 + 3*n^(2/3)/2 - n) * n^(n - 1/6) / sqrt(3) * (1 + 25/(64*n^(1/3))). - Vaclav Kotesovec, Oct 26 2017

a(n) = n*a(n-1) + (n-1)*(2*n - 1)*a(n-2) - 2*(n-3)*(n-2)*(n-1)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4) + (n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5). - Vaclav Kotesovec, Dec 02 2021

From Peter Bala, Oct 17 2023: (Start)

a(n+k) == a(n) (mod k) for all n and k >= 1. Hence for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. Cf. A047974.

a(5*n + 2) == a(5*n + 3) == 0 (mod 5);

a(25*n + 3) == a(25*n + 8) == a(25*n + 13) == a(25*n + 17) == a(25*n + 18) == a(25*n + 23) == 0 (mod 5^2);

a(125*n + 18) == a(125*n + 67) == a(125*n + 93) == a(125*n + 118) == 0 (mod 5^3). (End)

Mathematica

nmax = 20; CoefficientList[Series[E^(1/((1-x)*(1-x^2)) - 1), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 26 2017 *)

Prog

(PARI) N=66; x='x+O('x^N); Vec(serlaplace(exp(1/((1-x)*(1-x^2))-1)))

Crossrefs

Column k=2 of A294212.

Sequence in context: A265184 A096245 A294363 * A330198 A346269 A337041

Adjacent sequences: A294210 A294211 A294212 * A294214 A294215 A294216

Keyword

nonn,easy

Author

Seiichi Manyama, Oct 25 2017

Status

approved