1,1

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. Numerators of PolyLog[ -n, 1/n ] are listed in A121376[n] = {-1,6,33,380,3535,189714,285929,...}. a(p+1) = p^(p+1) for prime p. a(p^k+1) = p^( k*p^k + 2*k - (p^k - 1)/(p - 1) ) for prime p and integer k>0. Prime divisors of a(n) are the same as prime divisors of (n-1).

It appears that for the most square-free (n-1) if q is the largest prime divisor of (n-1) then q^(n - (n-1)/q + 1) divides a(n).

PolyLog[ -n, 1/n ] = A121376[n] / A121985[n] = Sum[ Eulerian[n,k] * n^(n-k+1), {k,0,n} ] / (n-1)^(n+1) = n*A122778[n] = Sum[ Eulerian[n,k] * n^k, {k,0,n} ] / (n-1)^(n+1) = A122020[n] for n>1.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics. Polylogarithm.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics. Eulerian number

a(n) = Denominator[ PolyLog[ -n, 1/n ] ]. a(n) = Denominator[ (-1)^(n+1) * PolyLog[ -n, n ] ].

PolyLog[ -n, 1/n ] begins -1/12, 6, 33/8, 380/81, 3535/512, 189714/15625, ...

a(3) = 2^3, a(4) = 3^4, a(200) = 199^200,

a(257) = 2^1809, a(290) = 17^564,

a(319) = 2^7 * 3^164 * 53^314, where 2*3*53 = 318 = 319 - 1 and 314 = 319 - 319/53 + 1,

a(709) = 2^716 * 3^360 * 59^698, a(710) = 709^710.

Table[Denominator[PolyLog[ -n, 1/n]], {n, 1, 30}]

Cf. A121376 = numerators.

Cf. A120020, A122778.

Sequence in context: A140378 A085094 A010214 this_sequence A068329 A010215 A059857

Adjacent sequences: A121982 A121983 A121984 this_sequence A121986 A121987 A121988

frac,nonn

Alexander Adamchuk (alex(AT)kolmogorov.com), Sep 10 2006, Sep 14 2006