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A213091
G.f. satisfies: A(x) = 1 + x/A(-x*A(x)^2).
23
1, 1, 1, 2, 4, 11, 31, 98, 317, 1070, 3685, 12928, 45924, 164552, 593398, 2148288, 7796846, 28328601, 102948125, 373955584, 1357252616, 4921292287, 17828236695, 64546901169, 233660589210, 846258569786, 3068523234989, 11147449003438, 40600425590874, 148330067463010
OFFSET
0,4
COMMENTS
Compare definition of g.f. to:
(1) B(x) = 1 + x/B(-x*B(x)) when B(x) = 1/(1-x).
(2) C(x) = 1 + x/C(-x*C(x)^3)^2 when C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
(3) D(x) = 1 + x/D(-x*D(x)^5)^3 when D(x) = 1 + x*D(x)^3 is the g.f. of the ternary tree numbers (A001764).
The first negative term is a(80). - Georg Fischer, Feb 16 2019
LINKS
EXAMPLE
G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 11*x^5 + 31*x^6 + 98*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 3*x^2 + 6*x^3 + 13*x^4 + 34*x^5 + 96*x^6 + 296*x^7 +...
A(-x*A(x)^2) = 1 - x - x^2 - x^3 - 4*x^4 - 10*x^5 - 34*x^6 - 107*x^7 -...
MATHEMATICA
nmax = 29; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x/A[(-x) A[x]^2]) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)
PROG
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1+x/subst(A, x, -x*subst(A^2, x, x+x*O(x^n))) ); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
KEYWORD
sign
AUTHOR
Paul D. Hanna, Jun 05 2012
STATUS
approved