It appears that the INVERT transform of factorial numbers
A000142 gives 1, 2, 5, 15, 54, 235, 1237, ... -
Antti Karttunen, May 30 2003
This is true: translating the defining recurrence to a generating function identity yields A(x) = 1/(1 - (0!*x + 1!*x^2 + 2!*x^3 + ...)) which is the INVERT formula.
In other words: let F(x) = Sum_{n>=0} n!*x^n then the g.f. is 1/(1-x*F(x)), cf.
A052186 (g.f. F(x)/(1+x*F(x))). -
Joerg Arndt, Apr 25 2011
G.f. A(x) satisfies: A(x) = (1-x)*A(x)^2 - x^2*A'(x). -
Paul D. Hanna, Aug 02 2008
G.f.: A(x) = 1/(1-x/(1-1*x/(1-1*x/(1-2*x/(1-2*x/(1-3*x/(1-3*x...))))))) (continued fraction). -
Paul Barry, Sep 25 2008
a(n) = upper left term in M^n, M = an infinite square production matrix in which a column of 1's is prepended to Pascal's triangle, as follows:
1, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 2, 1, 0, ...
1, 1, 3, 3, 1, ...
...
Also, a(n+1) = sum of top row terms of M^n. (End)
G.f.: 1+x/(U(0)-x) where U(k) = 1 + x*k - x*(k+1)/U(k+1); (continued fraction, 1-step). -
Sergei N. Gladkovskii, Oct 10 2012
G.f.: 1/(U(0) - x) where U(k) = 1 - x*(k+1)/(1 - x*(k+1)/U(k+1)); (continued fraction, 2-step). -
Sergei N. Gladkovskii, Nov 12 2012
a(n) ~ (n-1)! * (1 + 2/n + 7/n^2 + 31/n^3 + 165/n^4 + 1025/n^5 + 7310/n^6 + 59284/n^7 + 543702/n^8 + 5618267/n^9 + 65200918/n^10), for coefficients see
A260532. -
Vaclav Kotesovec, Jul 28 2015