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A091346
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Binomial convolution of A069321(n), where A069321(0)=0, with the sequence of all 1's alternating in sign.
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10
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0, 1, 3, 19, 135, 1171, 11823, 136459, 1771815, 25561891, 405658143, 7022891899, 131714587095, 2660335742611, 57570797744463, 1328913670495339, 32592691757283975, 846383665814211331, 23200396829832102783
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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).
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) = (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
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MATHEMATICA
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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;
a[n_] := (-1)^n HurwitzLerchPhi[2, -n-1, 2] / 2;
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Mario Catalani (mario.catalani(AT)unito.it), Jan 02 2004
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STATUS
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approved
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