A352410 Expansion of e.g.f. LambertW( -x/(1-x) ) / (-x).
1, 2, 9, 67, 717, 10141, 179353, 3816989, 95076537, 2714895433, 87457961421, 3138260371225, 124147801973605, 5368353187693757, 251928853285058433, 12752446755011776741, 692625349011401620209, 40178978855796929378065, 2479383850197948228950293
Offset: 0
Examples
E.g.f.: A(x) = 1 + 2*x + 9*x^2/2! + 67*x^3/3! + 717*x^4/4! + 10141*x^5/5! + 179353*x^6/6! + 3816989*x^7/7! + ... such that A(x) = exp(x*A(x)) / (1-x), where exp(x*A(x)) = 1 + x + 5*x^2/2! + 40*x^3/3! + 449*x^4/4! + 6556*x^5/5! + 118507*x^6/6! + ... + A052868(n)*x^n/n! + ... which equals LambertW(-x/(1-x)) * (1-x)/(-x). Related table. Another defining property of the e.g.f. A(x) is illustrated here. The table of coefficients of x^k/k! in 1/A(x)^n begins: n=1: [1, -2, -1, -7, -71, -961, -16409, -339571, ...]; n=2: [1, -4, 6, -2, -24, -362, -6644, -144538, ...]; n=3: [1, -6, 21, -33, -3, -63, -1395, -34275, ...]; n=4: [1, -8, 44, -148, 232, -4, -152, -4876, ...]; n=5: [1, -10, 75, -395, 1305, -2045, -5, -355, ...]; n=6: [1, -12, 114, -822, 4224, -13806, 21636, -6, ...]; n=7: [1, -14, 161, -1477, 10381, -52507, 170401, -267043, -7, ...]; ... from which we can illustrate that the partial sum of coefficients of x^k, k=0..n, in 1/A(x)^n equals zero, for n > 1, as follows: n=1:-1 = 1 + -2; n=2: 0 = 1 + -4 + 6/2!; n=3: 0 = 1 + -6 + 21/2! + -33/3!; n=4: 0 = 1 + -8 + 44/2! + -148/3! + 232/4!; n=5: 0 = 1 + -10 + 75/2! + -395/3! + 1305/4! + -2045/5!; n=6: 0 = 1 + -12 + 114/2! + -822/3! + 4224/4! + -13806/5! + 21636/6!; n=7: 0 = 1 + -14 + 161/2! + -1477/3! + 10381/4! + -52507/5! + 170401/6! + -267043/7!; ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..370
Programs
-
Mathematica
terms = 19; A[] = 0; Do[A[x] = Exp[x*A[x]]/(1-x) + O[x]^terms // Normal, terms]; CoefficientList[A[x], x]Range[0,terms-1]! (* Stefano Spezia, Mar 24 2025 *) With[{nn=20},CoefficientList[Series[LambertW[-x/(1-x)]/-x,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 24 2025 *)
-
PARI
{a(n) = n!*polcoeff( (1/x)*serreverse( x/(exp(x +x^2*O(x^n)) + x) ),n)} for(n=0,30,print1(a(n),", ")) -
PARI
my(x='x+O('x^30)); Vec(serlaplace(lambertw(-x/(1-x))/(-x))) \\ Michel Marcus, Mar 17 2022 -
PARI
a(n) = n!*sum(k=0, n, (k+1)^(k-1)*binomial(n, k)/k!); \\ Seiichi Manyama, Sep 24 2022
Formula
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = LambertW( -x/(1-x) ) / (-x).
(2) A(x) = exp( x*A(x) ) / (1-x).
(3) A(x) = log( (1-x) * A(x) ) / x.
(4) A( x/(exp(x) + x) ) = exp(x) + x.
(5) A(x) = (1/x) * Series_Reversion( x/(exp(x) + x) ).
(6) Sum_{k=0..n} [x^k] 1/A(x)^n = 0, for n > 1.
(7) [x^(n+1)/(n+1)!] 1/A(x)^n = -n for n >= (-1).
a(n) ~ (1 + exp(1))^(n + 3/2) * n^(n-1) / exp(n + 1/2). - Vaclav Kotesovec, Mar 15 2022
a(n) = n! * Sum_{k=0..n} (k+1)^(k-1) * binomial(n,k)/k!. - Seiichi Manyama, Sep 24 2022
Comments