OFFSET
1,2
COMMENTS
PolyLog[n,z] = Sum[ z^k/k^n, {k,1,Infinity} ]. PolyLog[ -n,1/n] = Sum[ k^n/n^k, {k,1,Infinity}] for n>1. n divides a(n). p^k divides a(p^k) for all prime p and integer k>0. p^k divides a(p^k-1) for prime p>2 and integer k>0.
LINKS
Eric Weisstein's World of Mathematics, Polylogarithm.
FORMULA
a(n) = Numerator[ PolyLog[ -n, 1/n ] ]. For n>1 a(n) = Numerator[ (-1)^n * PolyLog[ -n, n ] ].
For n>1, a(n) is the numerator of n*A122778(n)/(n-1)^(n+1) = Sum[k=0..n] A(n,k)*n^(k+1)/(n-1)^(n+1). For n>1, a(n) = n * A122778(n)/gcd(A122778(n),(n-1)^(n+1)). - Max Alekseyev, Sep 11 2006
EXAMPLE
PolyLog[ -n,1/n] begins -1/12, 6, 33/8, 380/81, 3535/512, 189714/15625, ...
MATHEMATICA
Numerator[Table[PolyLog[ -n, 1/n], {n, 1, 40}]]
PROG
(PARI) a(n)=numerator(polylog(-n, 1/n)) \\ Charles R Greathouse IV, Jul 14 2014
CROSSREFS
KEYWORD
frac,sign
AUTHOR
Alexander Adamchuk, Sep 06 2006
STATUS
approved