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%I #48 Nov 10 2020 09:20:52
%S 0,1,3,19,135,1171,11823,136459,1771815,25561891,405658143,7022891899,
%T 131714587095,2660335742611,57570797744463,1328913670495339,
%U 32592691757283975,846383665814211331,23200396829832102783
%N Binomial convolution of A069321(n), where A069321(0)=0, with the sequence of all 1's alternating in sign.
%C 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
%H Vincenzo Librandi, <a href="/A091346/b091346.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = Sum_{k=0..n}(C(n, k)*(-1)^(n-k)*Sum_{i=1..k}(i!*i*Stirling2(k, i))).
%F E.g.f.: ((exp(x)-1)/(2-exp(x))^2)*exp(-x).
%F a(n) = (A000670(n+1)+(-1)^(n+1))/4. - _Vladeta Jovovic_, Jan 17 2005
%F 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
%F a(n) ~ n! * n / (8 * (log(2))^(n+2)). - _Vaclav Kotesovec_, Nov 27 2017
%F 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
%F a(n ) = (-1)^n * (PolyLog(-1 - n, 2) - 2) / 8. - _Peter Luschny_, Nov 09 2020
%t Table[Sum[Binomial[n, k](-1)^(n-k)Sum[i!i StirlingS2[k, i], {i, 1, k}], {k, 0, n}], {n, 0, 20}]
%t Table[(-1)^n LerchPhi[2, -n-1, 2]/2, {n, 0, 20}] (* _Federico Provvedi_, Sep 04 2020 *)
%t a[n_] := (-1)^n (PolyLog[-1 - n, 2] - 2) / 8;
%t Table[a[n], {n, 0, 20}] (* _Peter Luschny_, Nov 09 2020 *)
%t a[n_] := (-1)^n HurwitzLerchPhi[2, -n-1, 2] / 2;
%t Table[a[n], {n, 0, 20}] (* _Federico Provvedi_, Nov 11 2020 *)
%o (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
%Y Cf. A083410.
%K easy,nonn
%O 0,3
%A Mario Catalani (mario.catalani(AT)unito.it), Jan 02 2004
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