A352448 Expansion of e.g.f. LambertW( -2*x/(1-x) ) / (-2*x).

1, 3, 22, 278, 5128, 125592, 3850000, 142013328, 6129705088, 303238991744, 16920975718144, 1051612647426816, 72045481821580288, 5394849460316820480, 438392509692455286784, 38424395486908104071168, 3613476161122656804438016

Offset: 0

Author

Paul D. Hanna, Mar 16 2022

Keywords

• nonn

Comments

An interesting property of this e.g.f. A(x) is that the sum of coefficients of x^k, k=0..n, in 1/A(x)^n equals zero, for n > 1.

Examples

			E.g.f.: A(x) = 1 + 3*x + 22*x^2/2! + 278*x^3/3! + 5128*x^4/4! + 125592*x^5/5! + 3850000*x^6/6! + 142013328*x^7/7! + ...
such that A(x) = exp( 2*x*A(x) ) / (1-x), where
exp( 2*x*A(x) ) = 1 + 2*x + 16*x^2/2! + 212*x^3/3! + 4016*x^4/4! + 99952*x^5/5! + 3096448*x^6/6! + 115063328*x^7/7! + ...
Related table.
Another interesting 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,  -3,  -4,   -44,  -736,  -16832, -491168, ...];
n=2: [1,  -6,  10,   -16,  -320,   -8064, -249344, ...];
n=3: [1,  -9,  42,   -78,   -48,   -1776,  -66528, ...];
n=4: [1, -12,  92,  -392,   728,    -128,   -8960, ...];
n=5: [1, -15, 160, -1120,  4600,   -8520,    -320, ...];
n=6: [1, -18, 246, -2424, 16104,  -64752,  119952, ...];
...
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:-2 = 1 +  -3;
n=2: 0 = 1 +  -6 +  10/2!;
n=3: 0 = 1 +  -9 +  42/2! +   -78/3!;
n=4: 0 = 1 + -12 +  92/2! +  -392/3! +   728/4!;
n=5: 0 = 1 + -15 + 160/2! + -1120/3! +  4600/4! +   -8520/5!;
n=6: 0 = 1 + -18 + 246/2! + -2424/3! + 16104/4! +  -64752/5! +  119952/6!;
...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..344

Crossrefs

Cf. A352410, A352411, A352412.

Cf. A361068.

Programs

Mathematica

terms = 17; A[] = 0; Do[A[x] = Exp[2x*A[x]]/(1-x) + O[x]^terms // Normal, terms]; CoefficientList[A[x], x]Range[0,terms-1]! (* Stefano Spezia, Mar 24 2025 *)

PARI

{a(n) = n!*polcoeff( (1/x)*serreverse( x/(exp(2*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(-2*x/(1-x))/(-2*x))) \\ Michel Marcus, Mar 17 2022

PARI

a(n) = n!*sum(k=0, n, 2^k*(k+1)^(k-1)*binomial(n, k)/k!); \\ Seiichi Manyama, Mar 03 2023

Formula

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

(1) A(x) = LambertW( -2*x/(1-x) ) / (-2*x).

(2) A(x) = exp( 2*x*A(x) ) / (1-x).

(3) A(x) = log( (1-x) * A(x) ) / (2*x).

(4) A( x/(exp(2*x) + x) ) = exp(2*x) + x.

(5) A(x) = (1/x) * Series_Reversion( x/(exp(2*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 = -2^(n+1) * n for n >= (-1).

a(n) ~ (1 + 2*exp(1))^(n + 3/2) * n^(n-1) / (2^(3/2) * exp(n + 1/2)). - Vaclav Kotesovec, Mar 18 2022

a(n) = n! * Sum_{k=0..n} 2^k * (k+1)^(k-1) * binomial(n,k)/k!. - Seiichi Manyama, Mar 03 2023