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%I #31 Sep 08 2022 08:45:58
%S 1,1,4,39,592,12965,378276,14062363,643946920,35426253465,
%T 2295988778440,172565368741931,14847924324645996,1446814927797156541,
%U 158201328106874927980,19258822568210913998955,2592339296719295037808336,383513887126740027040942577
%N E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n / n!.
%C More generally, the following sums are equal:
%C (1) Sum_{n>=0} (q^n + p)^n * r^n/n!,
%C (2) Sum_{n>=0} q^(n^2) * exp(p*q^n*r) * r^n/n!;
%C here, q = (1+x) and p = -1, r = 1. - _Paul D. Hanna_, Jun 21 2019
%H Vaclav Kotesovec, <a href="/A192935/b192935.txt">Table of n, a(n) for n = 0..280</a>
%F E.g.f.: Sum_{n>=0} (1+x)^(n^2) * exp(-(1+x)^n) / n!. - _Paul D. Hanna_, Jun 21 2019
%F a(n) = (1/n!)* Sum_{k=0..n} Stirling1(n,k)*A108459(k), where the e.g.f. of A108459 = Sum_{n>=0} (exp(n*x)-1)^n/n! (see Vladeta Jovovic's formula in A122400).
%e E.g.f: A(x) = 1 + x + 4*x^2/2! + 39*x^3/3! + 592*x^4/4! + 12965*x^5/5! + 378276*x^6/6! + 14062363*x^7/7! + 643946920*x^8/8! + 35426253465*x^9/9! + 2295988778440*x^10/10! +...
%e such that
%e A(x) = 1 + ((1+x) - 1) + ((1+x)^2 - 1)^2/2! + ((1+x)^3 - 1)^3/3! + ((1+x)^4 - 1)^4/4! + ((1+x)^5 - 1)^5/5! + ((1+x)^6 - 1)^6/6! + ((1+x)^7 - 1)^7/7! + ...
%e also
%e A(x) = 1 + (1+x)*exp(-(1+x)) + (1+x)^4*exp(-(1+x)^2)/2! + (1+x)^9*exp(-(1+x)^3)/3! + (1+x)^16*exp(-(1+x)^4)/4! + (1+x)^25*exp(-(1+x)^5)/5! + (1+x)^36*exp(-(1+x)^6)/6! + (1+x)^49*exp(-(1+x)^7)/7! + ...
%e RELATED SERIES.
%e Expansion of ((1+x)^n-1)^n suggests that the e.g.f. is related to LambertW(x):
%e ((1+x)^2-1)^2 = 4*x^2 + 4*x^3 + x^4;
%e ((1+x)^3-1)^3 = 27*x^3 + 81*x^4 + 108*x^5 + 81*x^6 + 36*x^7 + 9*x^8 + x^9;
%e ((1+x)^4-1)^4 = 256*x^4 + 1536*x^5 + 4480*x^6 + 8320*x^7 + 10896*x^8 +...
%t With[{m = 20}, CoefficientList[Series[Sum[If[n==0, 1, ((1+x)^n -1)^n/n!], {n,0,m+2}], {x,0,m}], x]*Range[0, m]!] (* _G. C. Greubel_, Feb 06 2019 *)
%o (PARI) {a(n)=n!*polcoeff(sum(m=0,n,((1+x+x*O(x^n))^m-1)^m/m!),n)}
%o for(n=0, 30, print1(a(n)*n!, ", "))
%o (PARI) {Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
%o {a(n)=sum(k=0, n, Stirling1(n, k)*k!*polcoeff(sum(m=0,k,(exp(m*x+x*O(x^n))-1)^m/m!),k))}
%o for(n=0, 30, print1(a(n)*n!, ", "))
%o (Magma) m:=20; R<x>:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (&+[((1+x)^n -1)^n/Factorial(n): n in [0..m+2]]) )); [Factorial(n-1)*b[n]: n in [1..m]]; // _G. C. Greubel_, Feb 06 2019
%o (Sage) m = 20; T = taylor(sum(((1+x)^k-1)^k/factorial(k) for k in range(m+2)), x, 0, m); [factorial(n)*T.coefficient(x, n) for n in (0..m)] # _G. C. Greubel_, Feb 06 2019
%Y Cf. A192985, A108459, A122400.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 13 2011
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