A337556 a(0) = 1; a(n) = (1/3) * Sum_{k=1..n} binomial(n,k) * (4^k-1) * a(n-k).
1, 1, 7, 57, 607, 8121, 130527, 2447257, 52435327, 1263925881, 33851510047, 997303255257, 32052722909247, 1116001351341241, 41845542004844767, 1681112968022124057, 72039936723424794367, 3280036569708658302201, 158127582939120607830687, 8046697501049910668173657
Offset: 0
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 0..387
Programs
-
Maple
E:= 3 / (3 + exp(x) - exp(4*x)): S:= series(E,x,41): seq(n!*coeff(S,x,n),n=0..40); # Robert Israel, Oct 13 2020
-
Mathematica
a[0] = 1; a[n_] := a[n] = (1/3) Sum[Binomial[n, k] (4^k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}] nmax = 19; CoefficientList[Series[3/(3 + Exp[x] - Exp[4 x]), {x, 0, nmax}], x] Range[0, nmax]! -
PARI
seq(n)={Vec(serlaplace(3 / (3 + exp(x + O(x*x^n)) - exp(4*x + O(x*x^n)))))} \\ Andrew Howroyd, Aug 31 2020
Formula
E.g.f.: 3 / (3 + exp(x) - exp(4*x)).
a(n) ~ n! * (r^3 - 1) * (4*r^3 - 16*r^2 + 64*r - 3) / (771 * log(r)^(n+1)), where r = 1.452626878833844... is the positive real root of the equation r*(r^3 - 1) = 3. - Vaclav Kotesovec, Aug 31 2020