A225147 a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.

-1, 2, 5, -46, -205, 3362, 22265, -515086, -4544185, 135274562, 1491632525, -54276473326, -718181418565, 30884386347362, 476768795646785, -23657073914466766, -417370516232719345, 23471059057478981762, 465849831125196593045, -29279357851856595135406

Offset: 0

Author

Peter Luschny, Apr 30 2013

Keywords

• sign
• easy

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..300
• Peter Luschny, Generalized Eulerian polynomials.

Crossrefs

Cf. A000810 (real part (up to sign)), A212435 (k=2), A122045 (k=1), A002439.

Bisections are A002438 and A000191.

Programs

Maple

B := proc(n, u, k) option remember;
if n = 1 then if (u < 0) or (u >= 1) then 0 else 1 fi
else k*u*B(n-1, u, k) + k*(n-u)*B(n-1, u-1, k) fi end:
EulerianPolynomial := proc(n, k, x) local m; if x = 0 then RETURN(1) fi;
add(B(n+1, m+1/k, k)*u^m, m = 0..n); subs(u=x, %) end:
seq(Im((1-I)^(1-n)*EulerianPolynomial(n, 3, I)), n=0..19);

Mathematica

CoefficientList[Series[-E^(-2*x)*Sech[3*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Sep 29 2014 after Sergei N. Gladkovskii *)
Table[-6^n EulerE[n,1/6], {n,0,19}] (* Peter Luschny, Nov 16 2016 after Peter Bala *)

Sage

from mpmath import mp, polylog, im
mp.dps = 32; mp.pretty = True
def A225147(n): return im(-2*I*(1+add(binomial(n,j)*polylog(-j,I)*3^j for j in (0..n))))
[int(A225147(n)) for n in (0..19)]

Formula

a(n) = Im(-2*i*(1+Sum_{j=0..n}(binomial(n,j)*Li{-j}(i)*3^j))).

For a recurrence see the Maple program.

G.f.: conjecture -T(0)/(1+2*x), where T(k) = 1 - 9*x^2*(k+1)^2/(9*x^2*(k+1)^2 + (1+2*x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013

a(n) = -(-3)^n*skp(n, 2/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014

G.f.: A225147 = -1/T(0), where T(k) = 1 + 2*x + (k+1)^2*(3*x)^2/ T(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 29 2014

E.g.f.: -exp(-2*x)*sech(3*x). - Sergei N. Gladkovskii, Sep 29 2014

a(n) ~ n! * (sqrt(3)*sin(Pi*n/2) - cos(Pi*n/2)) * 2^(n+1) * 3^n / Pi^(n+1). - Vaclav Kotesovec, Sep 29 2014

From Peter Bala, Nov 13 2016: (Start)

a(n) = - 6^n*E(n,1/6), where E(n,x) denotes the Euler polynomial of order n.

a(2*n) = (-1)^(n+1)*A002438(n); a(2*n+1) = (1/2)*(-1)^n*A002439(n). (End)