A225116 a(n) = 3^n*A_{n, 1/3}(-1) where A_{n, k}(x) are the generalized Eulerian polynomials.

1, 5, 24, 110, 480, 2000, 8064, 32240, 130560, 531200, 2095104, 8030720, 33546240, 156569600, 536838144, 243660800, 8589803520, 244224819200, 137438429184, -28539130347520, 2199021158400, 4960294141952000, 35184363700224, -1015283149035274240, 562949919866880

Offset: 0

Author

Peter Luschny, Apr 29 2013

Keywords

• sign

Links

• Peter Luschny, Generalized Eulerian polynomials.

Crossrefs

Cf. A155585(n) = 1^n*A_{n, 1/1}(-1), A119881(n) = 2^n*A_{n, 1/2}(-1).

Programs

Maple

EulerianPolynomial := proc(n,k,x) local j; if x = 1 then k^n*n! else (1-x)^(1+n)*(1+add(binomial(n,j)* polylog(-j,x)*k^j, j = 0..n)) fi end:
A225116 := n -> 3^n*EulerianPolynomial(n, 1/3, -1);
seq(round(evalf(A225116(i), 24)), i = 0..24);

Mathematica

Table[2^(t+1)*(Zeta[-t]*(1-2^(t+1))+(2^t-1)), {t,0,24}] (* Peter Luschny, Jul 20 2013 *)
Table[EulerE[n, 3] 2^n , {n, 0, 20}] (* Vladimir Reshetnikov, Oct 21 2015 *)

Sage

from mpmath import mp, polylog
mp.dps = 32; mp.pretty = True
def A225116(n): return 2^(1+n)*(3^n+add(binomial(n,j)*polylog(-j,-1) *3^(n-j) for j in (0..n)))
[int(A225116(n)) for n in (0..24)]

Formula

a(n) = 2^(1+n)*(3^n+sum_{j=0..n}(binomial(n,j)*Li_{-j}(-1)*3^(n-j))).

a(n) = 2^(t+1)*(zeta(-t)*(1-2^(t+1))+(2^t-1)). - Peter Luschny, Jul 20 2013