OFFSET
1,2
COMMENTS
The coefficient of x^n in the n-th iteration of x/(1-x) = n^(n-1) = A000169(n); does this variant have a simple formula for a(n)?
FORMULA
a(n) = [x^n] F_n(x) where F_n(x) = F_{n-1}(F(x)) with F(x) = x/(1-x)^2.
EXAMPLE
The successive iterations of F(x) = x/(1-x)^2 begin:
F(x) = (1)x + 2x^2 + 3x^3 + 4x^4 + 5x^5 + 6x^6 + 7x^7 + 8x^8 +...
F(F(x)) = x + (4)x^2 + 14x^3 + 46x^4 + 145x^5 + 444x^6 + 1331x^7 +...
F(F(F(x))) = x + 6x^2 + (33)x^3 + 174x^4 + 892x^5 + 4480x^6 +...
F(F(F(F(x)))) = x + 8x^2 + 60x^3 + (436)x^4 + 3102x^5 + 21728x^6 +...
F(F(F(F(F(x))))) = x + 10x^2 + 95x^3 + 880x^4 + (8015)x^5 +72090x^6+..
F(F(F(F(F(F(x)))))) = x + 12x^2+138x^3+1554x^4+17255x^5+(189596)x^6+..
PROG
(PARI) {a(n)=local(F=x/(1-x)^2, G=x+x*O(x^n)); if(n<1, 0, for(i=1, n, G=subst(F, x, G)); return(polcoeff(G, n, x)))}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 01 2006
STATUS
approved