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%I #82 Feb 13 2023 07:56:44
%S 1,-1,-2,2,40,104,-1232,-13168,16000,1483904,9695488,-151161088,
%T -3287997440,146760704,866038110208,10263094740992,-169941494497280,
%U -6324725967978496,-15215847186563072,2895126258819203072,54295929047166484480
%N Polylogarithm li(-n,-1/3) multiplied by (4^(n+1))/3.
%C Given n, consider the series s(n) = li(-n,-1/3) = SUM((-1)^k (k^n)/3^k) for k=0,1,2,... . Then a(n)=s(n)*(4^(n+1))/3. For more details, see A212846.
%H Seiichi Manyama, <a href="/A210246/b210246.txt">Table of n, a(n) for n = 0..436</a> (terms 0..100 from Stanislav Sykora)
%H OEIS-Wiki, <a href="http://oeis.org/wiki/Eulerian_polynomials">Eulerian polynomials</a>
%H S. Sykora, <a href="http://dx.doi.org/10.3247/SL1Math06.002">Finite and Infinite Sums of the Power Series (k^p)(x^k)</a>, Stan's Library Vol. I, April 2006, updated March 2012. See Eq.(29).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Polylogarithm.html">Polylogarithm</a>
%F Recurrence: s(n+1)=(-1/4)*SUM(C(n+1,i)*s(i)), where i=0,1,2,...,n, C(n,m) are binomial coefficients, and the starting value is s(0)=SUM((-1/3)^k)=3/4.
%F From _Peter Bala_, Mar 12 2013: (Start)
%F E.g.f.: A(x) = 4/(3 + exp(4*x)) = 1 - x - 2*x^2/2! + 2*x^3/3! + 40*x^4/4! + ....
%F The compositional inverse (A(-x) - 1)^(-1) = x + 2*x^2/2 + 7*x^3/3 + 20*x^4/4 + 61*x^5/5 + ... is the logarithmic generating function for A015518.
%F Recurrence equation: a(n+1) = -4*a(n) + 3*sum {k = 0..n} binomial(n,k)*a(k)*a(n-k), with a(0) = 1.
%F (End)
%F G.f.: 1 + x/Q(0), where Q(k) = 2*x*(k+1) - 1 + 3*x^2*(k+1)*(k+2)/Q(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Sep 22 2013
%F G.f.: 1/Q(0), where Q(k) = 1 + x*(k+1)/( 1 - 3*x*(k+1)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Dec 17 2013
%F E.g.f.: 2 - W(0), where W(k) = 1 + x/( 4*k+1 - x/( 1 + 4*x/( 4*k+3 - 4*x/W(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Oct 22 2014
%F a(n) = Sum_{k=0..n} k! * (-1)^k * 4^(n-k) * Stirling2(n,k). - _Seiichi Manyama_, Mar 13 2022
%F a(n) ~ n! * cos((n+1)*arctan(Pi/log(3))) * 2^(2*n + 3) / (3 * (Pi^2 + log(3)^2)^((n+1)/2)). - _Vaclav Kotesovec_, May 17 2022
%e a(5) = polylog(-5,-1/3)*4^6/3 = 104.
%p seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*3^k,k=0..n),n=0..20); # _Peter Luschny_, Apr 21 2013
%t Table[PolyLog[-n, -1/3] (4^(n+1))/3, {n, 30}] (* _T. D. Noe_, Mar 23 2012 *)
%t a[ n_] := If[ n < 1, Boole[n == 0], PolyLog[ -n, -1/3] 4^(n + 1) / 3]; (* _Michael Somos_, Nov 01 2014 *)
%o (PARI) /* See in A212846, run limnpq(nmax,1,3) */
%o (PARI) x='x+O('x^66); Vec(serlaplace( 4/(3+exp(4*x)) )) \\ _Joerg Arndt_, Apr 21 2013
%o (PARI) a(n) = sum(k=0, n, k!*(-1)^k*4^(n-k)*stirling(n, k, 2)); \\ _Seiichi Manyama_, Mar 13 2022
%Y Similar to A210244. Cf. A210247 (sign changes).
%Y Cf. A212846 (li(-n,-1/2)), A212847 (li(-n,-2/3)).
%Y CF. A213127 through A213157.
%K sign
%O 0,3
%A _Stanislav Sykora_, Mar 19 2012
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