OFFSET
0,5
COMMENTS
In general, for m >= 1, [x^n] polylog(2,x)^m ~ m*zeta(2)^(m-1)/n^2 = m * Pi^(2*m-2) / (6^(m-1) * n^2).
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..250
Vaclav Kotesovec, Recurrence (of order 10)
Eric Weisstein's World of Mathematics, Polylogarithm.
Wikipedia, Polylogarithm.
FORMULA
a(n)/(n!)^2 ~ Pi^6 / (54*n^2).
MATHEMATICA
CoefficientList[Series[PolyLog[2, x]^4, {x, 0, 20}], x] * Range[0, 20]!^2
Table[n!^2 * Sum[Sum[1/(k*(j-k))^2, {k, 1, j-1}] * Sum[1/(k*(n-j-k))^2, {k, 1, n-j-1}], {j, 1, n-1}], {n, 0, 20}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Feb 12 2024
STATUS
approved