A365525 a(n) = Sum_{k=0..floor(n/4)} Stirling2(n,4*k).

1, 0, 0, 0, 1, 10, 65, 350, 1702, 7806, 34855, 157630, 770529, 4432220, 31307432, 259090260, 2316320073, 21172354778, 193091210857, 1744478148866, 15627203762926, 139526376391986, 1251976261264071, 11417796498945894, 107280845105151601

Offset

0,6

Links

Robert Israel, Table of n, a(n) for n = 0..574

Formula

Let A(0)=1, B(0)=0, C(0)=0 and D(0)=0. Let B(n+1) = Sum_{k=0..n} binomial(n,k)*A(k), C(n+1) = Sum_{k=0..n} binomial(n,k)*B(k), D(n+1) = Sum_{k=0..n} binomial(n,k)*C(k) and A(n+1) = Sum_{k=0..n} binomial(n,k)*D(k). a(n) = A(n), A365526(n) = B(n), A365527(n) = C(n) and A099948(n) = D(n).

G.f.: Sum_{k>=0} x^(4*k) / Product_{j=1..4*k} (1-j*x).

Maple

f:= proc(n) local k; add(Stirling2(n, 4*k), k=0..n/4) end proc:
map(f, [$0..30]); # Robert Israel, Sep 11 2024

Mathematica

a[n_] := Sum[StirlingS2[n, 4*k], {k, 0, Floor[n/4]}]; Array[a, 25, 0] (* Amiram Eldar, Sep 11 2023 *)

Prog

(PARI) a(n) = sum(k=0, n\4, stirling(n, 4*k, 2));
(Python)
from sympy.functions.combinatorial.numbers import stirling
def A365525(n): return sum(stirling(n, k<<2) for k in range((n>>2)+1)) # Chai Wah Wu, Sep 08 2023

Crossrefs

Cf. A099948, A365526, A365527.

Sequence in context: A056280 A000453 A365532 * A327505 A346954 A346895

Adjacent sequences: A365522 A365523 A365524 * A365526 A365527 A365528

Keyword

nonn

Author

Seiichi Manyama, Sep 08 2023

Status

approved