1, 2, 9, 67, 717, 10141, 179353, 3816989, 95076537, 2714895433, 87457961421, 3138260371225, 124147801973605, 5368353187693757, 251928853285058433, 12752446755011776741, 692625349011401620209, 40178978855796929378065, 2479383850197948228950293
Offset: 0
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!;
...
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