A304866 E.g.f. A(x) satisfies: Sum_{n>=0} (n*x - A(x))^n / n! = 1.

1, 1, 5, 40, 434, 5921, 97152, 1861224, 40757712, 1003975497, 27471881920, 826643389408, 27127000901376, 964082532067097, 36887864803320832, 1511803871474266800, 66072899130124806144, 3067434610357079350833, 150748671875250474491904, 7818483194884075165619704, 426760505829170289041408000, 24454432374649526694779591985, 1467784259166552629365984329728

Offset: 1

Author

Paul D. Hanna, Jun 16 2018

Keywords

• nonn

Examples

			E.g.f.: A(x) = x + x^2/2! + 5*x^3/3! + 40*x^4/4! + 434*x^5/5! + 5921*x^6/6! + 97152*x^7/7! + 1861224*x^8/8! + 40757712*x^9/9! + 1003975497*x^10/10! + ...
such that
1  =  1 + (x - A(x)) + (2*x - A(x))^2/2! + (3*x - A(x))^3/3! + (4*x - A(x))^4/4! + (5*x - A(x))^5/5! + (6*x - A(x))^6/6! + (7*x - A(x))^7/7! + ...
Also,
-LambertW(-x)  =  A(x) + A(x)*(2*x - A(x))/2! + A(x)*(3*x - A(x))^2/3! + A(x)*(4*x - A(x))^3/4! + A(x)*(5*x - A(x))^4/5! + A(x)*(6*x - A(x))^5/6! + ...  =  x + 2*x^2/2! + 3^2*x^3/3! + 4^3*x^4/4! + 5^4*x^5/5! + 6^5*x^6/6! + ...
Further,
LambertW(-x)/(-x)  =  1 + (2*x - A(x)) + (3*x - A(x))^2/2! + (4*x - A(x))^3/3! + (5*x - A(x))^4/4! + (6*x - A(x))^5/5! + (7*x - A(x))^6/6! + ...  =  1 + x + 3*x^2/2! + 4^2*x^3/3! + 5^3*x^4/4! + 6^4*x^5/5! + 7^5*x^6/6! + ...
Note the following series relations involving e.g.f. A(x):
Sum_{n>=0} -A(x) * (n*x - A(x))^(n-1) / n!  =  1 - Sum_{n>=1} n^(n-1) * x^n / n!  =  1 + LambertW(-x),
Sum_{n>=0} ((n + m)*x - A(x))^n / n!  =  Sum_{n>=0} m*(n + m)^(n-1) * x^n / n!  =  ( LambertW(-x)/(-x) )^m.
These relations imply
( LambertW(-x) / (-x) )^(-A(x)/x)  =  1 + LambertW(-x)
giving A(x) = log(1 + LambertW(-x)) * x / LambertW(-x).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A300014 (exp(A(x))), A306090.

Programs

Mathematica

nmax = 20; Rest[CoefficientList[Series[Log[1 + LambertW[-x]] * x / LambertW[-x], {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Sep 01 2020 *)

PARI

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] = polcoeff(sum(m=1, #A, (m*x - x*Ser(A))^m/m!), #A)); n!*A[n]}
for(n=1, 30, print1(a(n), ", "))

Formula

E.g.f. A(x) = Sum_{n>=1} a(n) * x^n / n! satisfies:

(1) Sum_{n>=0} (n*x - A(x))^n / n! = 1.

(2) Sum_{n>=0} (n*x - 2*A(x))^n / n! = 1 + LambertW(-x).

(3) Sum_{n>=0} (n*x - m*A(x))^n / n! = (1 + LambertW(-x))^(m-1).

(4) Sum_{n>=0} -A(x) * (n*x - A(x))^(n-1) / n! = 1 + LambertW(-x).

(5) Sum_{n>=0} ((n + 1)*x - A(x))^n / n! = LambertW(-x)/(-x).

(6) Sum_{n>=0} ((n + m)*x - A(x))^n / n! = ( LambertW(-x)/(-x) )^m.

(7) Sum_{n>=0} ((n + m)*x + p*A(x))^n / n! = ( LambertW(-x)/(-x) )^m / (1 + LambertW(-x))^(p+1).

(8) ( LambertW(-x) / (-x) )^(-A(x)/x) = 1 + LambertW(-x).

(9) A(x) = log(1 + LambertW(-x)) * x / LambertW(-x).

a(n) ~ sqrt(Pi/2) * exp(-1) * n^(n - 1/2) * (1 - log(n)/sqrt(2*Pi*n)). - Vaclav Kotesovec, Sep 01 2020