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A210247
a(n) = sign of the polylogarithm li(-n,-1/3) for n > 0, with a(0) = 1.
4
1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, -1
OFFSET
0
COMMENTS
a(n) is the sign of s(n) = li(-n,-1/3) = Sum_{k>=1} ((-1)^k)*k^n/3^k.
Should s(n) be 0, the sign would be set to 0 as well.
However, it is not known whether this ever happens.
LINKS
Robert Israel, Table of n, a(n) for n = 0..2000 (first 200 terms from Stanislav Sykora)
Stanislav 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
a(n) = sign(A210246(n)).
Conjecture: a(n) = -a(n-28). - Mikhail Kurkov, Aug 27 2018
That is not quite true: the first counterexample is n=578, where a(578)=a(578+28)=-1. - Robert Israel, Sep 05 2018
a(n) is the same as the sign of cos((n+1)*arctan(Pi/log(3))). - Mikhail Kurkov, Apr 13 2021 [verification needed]
EXAMPLE
a(5) = sign(A210246(5)) = sign(104) = +1.
MAPLE
S:= series(4/(3 + exp(4*x)), x, 201):
seq(signum(coeff(S, x, j)), j=0..200); # Robert Israel, Sep 05 2018
MATHEMATICA
Join[{1}, Table[Sign[PolyLog[-n, -1/3]], {n, 1, 100}]] (* G. C. Greubel, Oct 27 2018 *)
PROG
(PARI) for(n=0, 100, print1(if(n==0, 1, sign(polylog(-n, -1/3))), ", ")) \\ G. C. Greubel, Oct 27 2018
CROSSREFS
Sequence in context: A319117 A063747 A210245 * A256175 A319116 A337004
KEYWORD
sign
AUTHOR
Stanislav Sykora, Mar 19 2012
STATUS
approved