A192935 E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n / n!.
1, 1, 4, 39, 592, 12965, 378276, 14062363, 643946920, 35426253465, 2295988778440, 172565368741931, 14847924324645996, 1446814927797156541, 158201328106874927980, 19258822568210913998955, 2592339296719295037808336, 383513887126740027040942577
Offset: 0
Keywords
Examples
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! +... such that 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! + ... also 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! + ... RELATED SERIES. Expansion of ((1+x)^n-1)^n suggests that the e.g.f. is related to LambertW(x): ((1+x)^2-1)^2 = 4*x^2 + 4*x^3 + x^4; ((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; ((1+x)^4-1)^4 = 256*x^4 + 1536*x^5 + 4480*x^6 + 8320*x^7 + 10896*x^8 +...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..280
Programs
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Magma
m:=20; R
:=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 -
Mathematica
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 *) -
PARI
{a(n)=n!*polcoeff(sum(m=0,n,((1+x+x*O(x^n))^m-1)^m/m!),n)} for(n=0, 30, print1(a(n)*n!, ", ")) -
PARI
{Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)} {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))} for(n=0, 30, print1(a(n)*n!, ", ")) -
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
Formula
E.g.f.: Sum_{n>=0} (1+x)^(n^2) * exp(-(1+x)^n) / n!. - Paul D. Hanna, Jun 21 2019
Comments