OFFSET
0,3
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..150
FORMULA
Given g.f. A(x), let G(x) = A(x*G(x)), then A(x) = G(x/A(x)) = 1 + x*(G(x) + x*G'(x)) / (G(x) - x*G(x)^2).
a(n)/a(n-1) ~ n/LambertW(1). - Vaclav Kotesovec, Sep 14 2024
EXAMPLE
G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 50*x^4 + 424*x^5 + 4472*x^6 + 55760*x^7 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0 = [1],0, 0, 0, 0, 0, 0, 0, 0, ...;
A^1 = [1, 1], 2, 8, 50, 424, 4472, 55760, 797022, ...;
A^2 = [1, 2, 5], 20, 120, 980, 10056, 122960, 1732736, ...;
A^3 = [1, 3, 9, 37], 216, 1704, 17006, 203760, 2829030, ...;
A^4 = [1, 4, 14, 60, 345], 2640, 25632, 300744, 4111472, ...;
A^5 = [1, 5, 20, 90, 515, 3841], 36310, 417000, 5609960, ...;
A^6 = [1, 6, 27, 128, 735, 5370, 49493], 556212, 7359480, ...;
A^7 = [1, 7, 35, 175, 1015, 7301, 65723, 722765], 9400986, ...;
A^8 = [1, 8, 44, 232, 1366, 9720, 85644, 921864, 11782417], ...; ...
then a(n) equals the sum of the coefficients of x^k, k=0..n-1, in A(x)^(n-1) (shown above in brackets) for n>=1:
a(1) = 1 = 1;
a(2) = 1 + 1 = 2;
a(3) = 1 + 2 + 5 = 8;
a(4) = 1 + 3 + 9 + 37 = 50;
a(5) = 1 + 4 + 14 + 60 + 345 = 424;
a(6) = 1 + 5 + 20 + 90 + 515 + 3841 = 4472;
a(7) = 1 + 6 + 27 + 128 + 735 + 5370 + 49493 = 55760;
a(8) = 1 + 7 + 35 + 175 + 1015 + 7301 + 65723 + 722765 = 797022; ...
Also, from a diagonal in the above table we can obtain the coefficients:
[1/1, 2/2, 9/3, 60/4, 515/5, 5370/6, 65723/7, 921864/8, ...]
to form the power series
G(x) = 1 + x + 3*x^2 + 15*x^3 + 103*x^4 + 895*x^5 + 9389*x^6 + 115233*x^7 +...
that satisfies: A(x) = G(x/A(x)) = 1 + x*(G(x) + x*G'(x))/(G(x) - x*G(x)^2).
PROG
(PARI) {a(n)=local(A=1+x); if(n==0, 1, for(i=1, n,
A=1+sum(k=1, n-1, sum(j=0, k-1, polcoeff(A^(k-1)+x*O(x^j), j))*x^k)+x*O(x^n));
sum(j=0, n-1, polcoeff(A^(n-1)+x*O(x^j), j)))}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 09 2013
STATUS
approved