A210246 Polylogarithm li(-n,-1/3) multiplied by (4^(n+1))/3.

1, -1, -2, 2, 40, 104, -1232, -13168, 16000, 1483904, 9695488, -151161088, -3287997440, 146760704, 866038110208, 10263094740992, -169941494497280, -6324725967978496, -15215847186563072, 2895126258819203072, 54295929047166484480

Offset

0,3

Comments

Given n, consider the series s(n) = li(-n,-1/3) = SUM((-1)^k (k^n)/3^k) for k=0,1,2,... . Then a(n)=s(n)*(4^(n+1))/3. For more details, see A212846.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..436 (terms 0..100 from Stanislav Sykora)

OEIS-Wiki, Eulerian polynomials

S. Sykora, Finite and Infinite Sums of the Power Series (k^p)(x^k), Stan's Library Vol. I, April 2006, updated March 2012. See Eq.(29).

Eric Weisstein's World of Mathematics, Polylogarithm

Formula

Recurrence: s(n+1)=(-1/4)*SUM(C(n+1,i)*s(i)), where i=0,1,2,...,n, C(n,m) are binomial coefficients, and the starting value is s(0)=SUM((-1/3)^k)=3/4.

From Peter Bala, Mar 12 2013: (Start)

E.g.f.: A(x) = 4/(3 + exp(4*x)) = 1 - x - 2*x^2/2! + 2*x^3/3! + 40*x^4/4! + ....

The compositional inverse (A(-x) - 1)^(-1) = x + 2*x^2/2 + 7*x^3/3 + 20*x^4/4 + 61*x^5/5 + ... is the logarithmic generating function for A015518.

Recurrence equation: a(n+1) = -4*a(n) + 3*sum {k = 0..n} binomial(n,k)*a(k)*a(n-k), with a(0) = 1.

(End)

G.f.: 1 + x/Q(0), where Q(k) = 2*x*(k+1) - 1 + 3*x^2*(k+1)*(k+2)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Sep 22 2013

G.f.: 1/Q(0), where Q(k) = 1 + x*(k+1)/( 1 - 3*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 17 2013

E.g.f.: 2 - W(0), where W(k) = 1 + x/( 4*k+1 - x/( 1 + 4*x/( 4*k+3 - 4*x/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2014

a(n) = Sum_{k=0..n} k! * (-1)^k * 4^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022

a(n) ~ n! * cos((n+1)*arctan(Pi/log(3))) * 2^(2*n + 3) / (3 * (Pi^2 + log(3)^2)^((n+1)/2)). - Vaclav Kotesovec, May 17 2022

Example

a(5) = polylog(-5,-1/3)*4^6/3 = 104.

Maple

seq(add((-1)^(n-k)*combinat[eulerian1](n, k)*3^k, k=0..n), n=0..20); # Peter Luschny, Apr 21 2013

Mathematica

Table[PolyLog[-n, -1/3] (4^(n+1))/3, {n, 30}] (* T. D. Noe, Mar 23 2012 *)
a[ n_] := If[ n < 1, Boole[n == 0], PolyLog[ -n, -1/3] 4^(n + 1) / 3]; (* Michael Somos, Nov 01 2014 *)

Prog

(PARI) /* See in A212846, run limnpq(nmax, 1, 3) */
(PARI) x='x+O('x^66); Vec(serlaplace( 4/(3+exp(4*x)) )) \\ Joerg Arndt, Apr 21 2013
(PARI) a(n) = sum(k=0, n, k!*(-1)^k*4^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

Crossrefs

Similar to A210244. Cf. A210247 (sign changes).

Cf. A212846 (li(-n,-1/2)), A212847 (li(-n,-2/3)).

CF. A213127 through A213157.

Sequence in context: A001121 A202711 A222668 * A116567 A215049 A087541

Adjacent sequences: A210243 A210244 A210245 * A210247 A210248 A210249

Keyword

sign

Author

Stanislav Sykora, Mar 19 2012

Status

approved