OFFSET
0,5
FORMULA
a(n) = n! * Sum_{k=0..floor(n/4)} 1/(k!^2 * (2*k)! * (n-4*k)!) = Sum_{k=0..floor(n/4)} binomial(n,4*k) * A000897(k).
From Vaclav Kotesovec, Mar 20 2023: (Start)
Recurrence: (n-2)*n^2*a(n) = (4*n^3 - 12*n^2 + 10*n - 3)*a(n-1) - (n-1)*(6*n^2 - 18*n + 13)*a(n-2) + 4*(n-2)^2*(n-1)*a(n-3) + 63*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ (1 + 2*sqrt(2))^(n+1) / (4*Pi*n). (End)
MATHEMATICA
Table[n!*Sum[1/(k!^2*(2*k)!*(n - 4*k)!), {k, 0, n/4}], {n, 0, 30}] (* Vaclav Kotesovec, Mar 20 2023 *)
PROG
(PARI) a(n) = n!*sum(k=0, n\4, 1/(k!^2*(2*k)!*(n-4*k)!));
CROSSREFS
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Mar 19 2023
STATUS
approved