%I #30 Mar 04 2024 10:03:39

%S 1,2,9,67,717,10141,179353,3816989,95076537,2714895433,87457961421,

%T 3138260371225,124147801973605,5368353187693757,251928853285058433,

%U 12752446755011776741,692625349011401620209,40178978855796929378065,2479383850197948228950293

%N Expansion of e.g.f. LambertW( -x/(1-x) ) / (-x).

%C 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.

%H Seiichi Manyama, <a href="/A352410/b352410.txt">Table of n, a(n) for n = 0..370</a>

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

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

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

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

%F (4) A( x/(exp(x) + x) ) = exp(x) + x.

%F (5) A(x) = (1/x) * Series_Reversion( x/(exp(x) + x) ).

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

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

%F a(n) ~ (1 + exp(1))^(n + 3/2) * n^(n-1) / exp(n + 1/2). - _Vaclav Kotesovec_, Mar 15 2022

%F a(n) = n! * Sum_{k=0..n} (k+1)^(k-1) * binomial(n,k)/k!. - _Seiichi Manyama_, Sep 24 2022

%e 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! + ...

%e such that A(x) = exp(x*A(x)) / (1-x), where

%e 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! + ...

%e which equals LambertW(-x/(1-x)) * (1-x)/(-x).

%e Related table.

%e Another defining property of the e.g.f. A(x) is illustrated here.

%e The table of coefficients of x^k/k! in 1/A(x)^n begins:

%e n=1: [1, -2, -1, -7, -71, -961, -16409, -339571, ...];

%e n=2: [1, -4, 6, -2, -24, -362, -6644, -144538, ...];

%e n=3: [1, -6, 21, -33, -3, -63, -1395, -34275, ...];

%e n=4: [1, -8, 44, -148, 232, -4, -152, -4876, ...];

%e n=5: [1, -10, 75, -395, 1305, -2045, -5, -355, ...];

%e n=6: [1, -12, 114, -822, 4224, -13806, 21636, -6, ...];

%e n=7: [1, -14, 161, -1477, 10381, -52507, 170401, -267043, -7, ...];

%e ...

%e 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:

%e n=1:-1 = 1 + -2;

%e n=2: 0 = 1 + -4 + 6/2!;

%e n=3: 0 = 1 + -6 + 21/2! + -33/3!;

%e n=4: 0 = 1 + -8 + 44/2! + -148/3! + 232/4!;

%e n=5: 0 = 1 + -10 + 75/2! + -395/3! + 1305/4! + -2045/5!;

%e n=6: 0 = 1 + -12 + 114/2! + -822/3! + 4224/4! + -13806/5! + 21636/6!;

%e n=7: 0 = 1 + -14 + 161/2! + -1477/3! + 10381/4! + -52507/5! + 170401/6! + -267043/7!;

%e ...

%o (PARI) {a(n) = n!*polcoeff( (1/x)*serreverse( x/(exp(x +x^2*O(x^n)) + x) ),n)}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) my(x='x+O('x^30)); Vec(serlaplace(lambertw(-x/(1-x))/(-x))) \\ _Michel Marcus_, Mar 17 2022

%o (PARI) a(n) = n!*sum(k=0, n, (k+1)^(k-1)*binomial(n, k)/k!); \\ _Seiichi Manyama_, Sep 24 2022

%Y Cf. A352411, A352412, A352448, A052868.

%Y Cf. A102743, A108919, A331726.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Mar 15 2022