A352411 E.g.f.: x / LambertW( x/(1-x) ).

1, 0, -1, 1, -7, 31, -281, 2381, -28015, 346879, -5149009, 82769149, -1499707991, 29444151023, -632715633577, 14631547277101, -364321853163871, 9686058045625471, -274387229080161569, 8241211775883617405, -261766195805536280839, 8763341168691985628719

Offset: 0

Author

Paul D. Hanna, Mar 15 2022

Keywords

• sign

Comments

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

Examples

			E.g.f.: A(x) = 1 + 0*x - x^2/2! + x^3/3! - 7*x^4/4! + 31*x^5/5! - 281*x^6/6! + 2381*x^7/7! - 28015*x^8/8! + ...
such that A(x) = (1-x) * exp(x/A(x)), where
exp(x/A(x)) = 1 + x + x^2/2! + 4*x^3/3! + 9*x^4/4! + 76*x^5/5! + 175*x^6/6! + 3606*x^7/7! + 833*x^8/8! + ...
Related series.
The e.g.f. A(x) satisfies A( x/(exp(-x) + x) ) = 1/(exp(-x) + x), where
1/(exp(-x) + x) = 1 - x^2/2! + x^3/3! + 5*x^4/4! - 19*x^5/5! - 41*x^6/6! + 519*x^7/7! - 183*x^8/8! + ...
Related table.
Another defining property of the e.g.f. A(x) is illustrated here.
The table of coefficients of x^k/k! in A(x)^n begins:
n=1: [1, 0, -1, 1,  -7,   31, -281, 2381, -28015, ...];
n=2: [1, 0, -2, 2,  -8,   42, -332, 2970, -33392, ...];
n=3: [1, 0, -3, 3,  -3,   33, -243, 2397, -26631, ...];
n=4: [1, 0, -4, 4,   8,    4, -104, 1292, -15712, ...];
n=5: [1, 0, -5, 5,  25,  -45,   -5,  285,  -6095, ...];
n=6: [1, 0, -6, 6,  48, -114,  -36,    6,   -720, ...];
n=7: [1, 0, -7, 7,  77, -203, -287, 1085,     -7, ...];
n=8: [1, 0, -8, 8, 112, -312, -848, 4152,  -1856, 8, ...];
...
from which we can illustrate that the partial sum of coefficients of x^k, k=0..n, in A(x)^n equals zero, for n > 1, as follows:
n=1: 1 = 1 + 0;
n=2: 0 = 1 + 0 + -2/2!;
n=3: 0 = 1 + 0 + -3/2! + 3/3!;
n=4: 0 = 1 + 0 + -4/2! + 4/3! +   8/4!;
n=5: 0 = 1 + 0 + -5/2! + 5/3! +  25/4! +  -45/5!;
n=6: 0 = 1 + 0 + -6/2! + 6/3! +  48/4! + -114/5! +  -36/6!;
n=7: 0 = 1 + 0 + -7/2! + 7/3! +  77/4! + -203/5! + -287/6! + 1085/7!;
n=8: 0 = 1 + 0 + -8/2! + 8/3! + 112/4! + -312/5! + -848/6! + 4152/7! + -1856/8!;
...
		

Crossrefs

Cf. A352410, A352412, A352448.

Programs

PARI

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

Formula

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

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

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

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

(4) A( x/(exp(-x) + x) ) = 1/(exp(-x) + x).

(5) A(x) = x / Series_Reversion( x/(exp(-x) + x) ).

(6) Sum_{k=0..n} [x^k] A(x)^n = 0, for n > 1.

(7) [x^(n+1)/(n+1)!] A(x)^n = (-1)^n * n for n >= (-1).

a(n) ~ (-1)^(n+1) * exp(-1) * (1 - exp(-1))^(n - 1/2) * n^(n-1). - Vaclav Kotesovec, Mar 15 2022