OFFSET
1,4
COMMENTS
Given an integer n>0, consider the infinite series s(n) = li(-n,-1/2) = Sum_{k>=1} (-1)^k*k^n/2^k. Then s(n)=2*a(n)/A131137(n+1).
LINKS
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 W. Weisstein, MathWorld: Polylogarithm
FORMULA
Recurrence: s(n+1)=(-1/3)*Sum_{i=0..n} binomial(n+1,i)*s(i), with the starting value of s(0)=2/3.
EXAMPLE
s(1)=-2/9, s(2)=-2/27, s(3)=+2/27, s(4)=+10/81.
MATHEMATICA
PROG
(PARI) a(n)=numerator(polylog(-n, -1/2)/2) \\ Charles R Greathouse IV, Jul 15 2014
CROSSREFS
KEYWORD
sign,frac
AUTHOR
Stanislav Sykora, Mar 19 2012
STATUS
approved