OFFSET
0,4
COMMENTS
Define the Eulerian numbers A(n,k) as the number of permutations of {1,2,..,n} with k ascents and the Eulerian polynomials A_{0}(x) = 1; A_{n}(x) = sum_{k=0..n-1} A(n,k) x^k for n > 0. Then a(n) = A_{n}(-n) are the values of the Eulerian polynomials for n = 0,-1,-2,-3,...
LINKS
OEIS Wiki, Eulerian polynomials.
FORMULA
a(n) = -(n+1)^(n+1)*Li_{-n}(n)/n, where Li_{n}(z) denotes the polylogarithm. For n = 0, interpret it as a limit for continuous n -> 0, that gives a(0) = 1. - Vladimir Reshetnikov, Oct 15 2016
a(n) = n! * [x^n] (n + 1) / (n + exp(-(n + 1)*x)). - Ilya Gutkovskiy, Jun 28 2020
MAPLE
c := proc(n, m) local k; add((-1)^k*binomial(n+1, k)*(m+1-k)^n, k=0..m) end:
a := proc(n) local k; `if`(n=0, 1, add(c(n, k)*(-n)^k, k=0..n-1)) end:
MATHEMATICA
a[0] = 1; a[n_] := -(n + 1)^(n + 1) PolyLog[-n, -n]/n; Table[a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
PROG
(PARI) a(n) = if (n==0, 1, -(n + 1)^(n + 1)*polylog(-n, -n)/n); \\ Michel Marcus, May 30 2018
CROSSREFS
KEYWORD
sign
AUTHOR
Peter Luschny, Aug 12 2010
EXTENSIONS
More terms from Michel Marcus, May 30 2018
STATUS
approved