OFFSET
0,3
COMMENTS
Compare to the trivial identity:
G(x) = -1 + x + G(x) + 1/G(x) is satisfied when G(x) = 1/(1-x).
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..200
Vaclav Kotesovec, Recurrence
FORMULA
G.f. satisfies: 0 = 1 - (1-x)*A(x)^3 - A(x)^4 + A(x)^6.
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 34.05333752261152244... is the root of the equation 729 - 4374*d + 2079*d^2 + 20844*d^3 + 1431*d^4 - 56214*d^5 + 1649*d^6 = 0 and c = 0.03096414033189124248457768352179346154431144356... - Vaclav Kotesovec, Sep 19 2013
EXAMPLE
G.f.: A(x) = 1 + x + 9*x^2 + 153*x^3 + 3255*x^4 + 77577*x^5 + 1981126*x^6 +...
where
A(x)^3 = 1 + 3*x + 30*x^2 + 514*x^3 + 10953*x^4 + 261225*x^5 + 6673593*x^6 +...
1/A(x)^3 = 1 - 3*x - 21*x^2 - 361*x^3 - 7698*x^4 - 183648*x^5 - 4692467*x^6 -...
so that A(x) = -1 + x + A(x)^3 + 1/A(x)^3.
MAPLE
a:= n-> coeff(series(RootOf(A= -1+x+A^3+1/A^3, A), x, n+1), x, n):
seq(a(n), n=0..20); # Alois P. Heinz, Sep 19 2013
MATHEMATICA
nmax=20; aa=ConstantArray[0, nmax]; aa[[1]]=1; Do[AGF=1+Sum[aa[[n]]*x^n, {n, 1, j-1}]+koef*x^j; sol=Solve[Coefficient[1 - (1-x)*AGF^3 - AGF^4 + AGF^6, x, j]==0, koef][[1]]; aa[[j]]=koef/.sol[[1]], {j, 2, nmax}]; Flatten[{1, aa}] (* Vaclav Kotesovec, Sep 19 2013 *)
PROG
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=-1+x + A^3 + 1/(A^3 +x*O(x^n))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 06 2013
STATUS
approved