OFFSET
0,3
COMMENTS
Let M(p, n) denote the n-th central moment of the geometric distribution p(1-p)^x. The sums of the polynomial coefficients of M(p, n)*p^n, ( {}, {1, -1}, {2, -3, 1}, {9, -18, 10, -1}, {44, -110, 90, -25}, ... ), are zero and the sum of their absolute values is 2*a(n). - Federico Provvedi, Sep 09 2020
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
FORMULA
a(n) = Sum_{k=0..n}(C(n, k)*(-1)^(n-k)*Sum_{i=1..k}(i!*i*Stirling2(k, i))).
E.g.f.: ((exp(x)-1)/(2-exp(x))^2)*exp(-x).
a(n) = (A000670(n+1)+(-1)^(n+1))/4. - Vladeta Jovovic, Jan 17 2005
G.f.: x/(1+x)/Q(0), where Q(k) = 1 - x*(3*k+4) - 2*x^2*(k+1)*(k+3)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 03 2013
a(n) ~ n! * n / (8 * (log(2))^(n+2)). - Vaclav Kotesovec, Nov 27 2017
a(n) = (1/2) * (-1)^n * Phi(2, -n-1, 2), where Phi(z, s, a) is the Lerch transcendantal function. - Federico Provvedi, Sep 04 2020
a(n ) = (-1)^n * (PolyLog(-1 - n, 2) - 2) / 8. - Peter Luschny, Nov 09 2020
MATHEMATICA
Table[Sum[Binomial[n, k](-1)^(n-k)Sum[i!i StirlingS2[k, i], {i, 1, k}], {k, 0, n}], {n, 0, 20}]
Table[(-1)^n LerchPhi[2, -n-1, 2]/2, {n, 0, 20}] (* Federico Provvedi, Sep 04 2020 *)
a[n_] := (-1)^n (PolyLog[-1 - n, 2] - 2) / 8;
Table[a[n], {n, 0, 20}] (* Peter Luschny, Nov 09 2020 *)
a[n_] := (-1)^n HurwitzLerchPhi[2, -n-1, 2] / 2;
Table[a[n], {n, 0, 20}] (* Federico Provvedi, Nov 11 2020 *)
PROG
(PARI) a(n) = sum(k=0, n, binomial(n, k)*(-1)^(n-k)*sum(i=1, k, i!*i*stirling(k, i, 2))); \\ Michel Marcus, Jun 25 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Mario Catalani (mario.catalani(AT)unito.it), Jan 02 2004
STATUS
approved