A091346 Binomial convolution of A069321(n), where A069321(0)=0, with the sequence of all 1's alternating in sign.

0, 1, 3, 19, 135, 1171, 11823, 136459, 1771815, 25561891, 405658143, 7022891899, 131714587095, 2660335742611, 57570797744463, 1328913670495339, 32592691757283975, 846383665814211331, 23200396829832102783, 669421949061096050779, 20281206249626018470455

Offset: 0

Author

Mario Catalani (mario.catalani(AT)unito.it), Jan 02 2004

Keywords

• easy
• nonn

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

Crossrefs

Cf. A083410.

Programs

Maple

b:= proc(n, k) option remember;
     `if`(n=0, k!, k*b(n-1, k)+b(n-1, k+1))
    end:
a:= n-> (b(n+1, 0)-(-1)^n)/4:
seq(a(n), n=0..23);  # Alois P. Heinz, Feb 14 2025

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 *)

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

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