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%I #19 Mar 19 2020 16:38:13
%S 1,5,24,110,480,2000,8064,32240,130560,531200,2095104,8030720,
%T 33546240,156569600,536838144,243660800,8589803520,244224819200,
%U 137438429184,-28539130347520,2199021158400,4960294141952000,35184363700224,-1015283149035274240,562949919866880
%N a(n) = 3^n*A_{n, 1/3}(-1) where A_{n, k}(x) are the generalized Eulerian polynomials.
%H Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/EulerianPolynomialsGeneralized">Generalized Eulerian polynomials</a>.
%F a(n) = 2^(1+n)*(3^n+sum_{j=0..n}(binomial(n,j)*Li_{-j}(-1)*3^(n-j))).
%F a(n) = 2^(t+1)*(zeta(-t)*(1-2^(t+1))+(2^t-1)). - _Peter Luschny_, Jul 20 2013
%p 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:
%p A225116 := n -> 3^n*EulerianPolynomial(n, 1/3, -1);
%p seq(round(evalf(A225116(i), 24)), i = 0..24);
%t Table[2^(t+1)*(Zeta[-t]*(1-2^(t+1))+(2^t-1)), {t,0,24}] (* _Peter Luschny_, Jul 20 2013 *)
%t Table[EulerE[n, 3] 2^n , {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 21 2015 *)
%o (Sage)
%o from mpmath import mp, polylog
%o mp.dps = 32; mp.pretty = True
%o def A225116(n): return 2^(1+n)*(3^n+add(binomial(n,j)*polylog(-j,-1) *3^(n-j) for j in (0..n)))
%o [int(A225116(n)) for n in (0..24)]
%Y Cf. A155585(n) = 1^n*A_{n, 1/1}(-1), A119881(n) = 2^n*A_{n, 1/2}(-1).
%K sign
%O 0,2
%A _Peter Luschny_, Apr 29 2013
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