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
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
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Jul 01 2022
STATUS
approved