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%I #12 Dec 30 2021 23:20:32
%S 1,1,3,8,28,95,351,1309,5056,19787,78847,317705,1294673,5321598,
%T 22047985,91957296,385832452,1627351862,6896087775,29345806842,
%U 125353612440,537303633158,2310270577888,9962069922553,43070357677938,186663591654655,810799482641934,3529141491880136,15390864728209348
%N G.f. A(x) satisfies: A(x*B(x)) = x^2 - x^3, where A(B(x)) = x.
%H Paul D. Hanna, <a href="/A273162/b273162.txt">Table of n, a(n) for n = 1..515</a>
%F G.f. A(x) satisfies: A(x*A(x)) = A(x)^2 - A(x)^3.
%F G.f.: A(x) = Product_{n>=0} F(n), where F(0) = x, F(1) = 1-x, and F(n+1) = 1 - (1 - F(n))^2 * F(n) for n > 0. - _Paul D. Hanna_, Dec 30 2021
%e G.f.: A(x) = x + x^2 + 3*x^3 + 8*x^4 + 28*x^5 + 95*x^6 + 351*x^7 + 1309*x^8 + 5056*x^9 + 19787*x^10 + 78847*x^11 + 317705*x^12 +...
%e such that A(x*B(x)) = x^2 - x^3, where A(B(x)) = x.
%e RELATED SERIES.
%e A(x)^2 = x^2 + 2*x^3 + 7*x^4 + 22*x^5 + 81*x^6 + 294*x^7 + 1124*x^8 + 4338*x^9 + 17140*x^10 + 68476*x^11 + 277229*x^12 + 1132716*x^13 + 4669477*x^14 +...
%e A(x)^3 = x^3 + 3*x^4 + 12*x^5 + 43*x^6 + 168*x^7 + 648*x^8 + 2574*x^9 + 10284*x^10 + 41691*x^11 + 170333*x^12 + 702273*x^13 + 2915100*x^14 +...
%e A(x*A(x)) = A(x)^2 - A(x)^3 where
%e A(x)^2 - A(x)^3 = x^2 + x^3 + 4*x^4 + 10*x^5 + 38*x^6 + 126*x^7 + 476*x^8 + 1764*x^9 + 6856*x^10 + 26785*x^11 + 106896*x^12 + 430443*x^13 + 1754377*x^14 +...
%e Let B(x) be the series reversion of g.f. A(x) so that A(B(x)) = x, then
%e B(x) = x - x^2 - x^3 + 2*x^4 - 2*x^5 + 3*x^6 - x^7 - 7*x^8 + 10*x^9 + 2*x^10 - 15*x^11 + 2*x^12 + 34*x^13 - 51*x^14 + 17*x^15 + 73*x^16 - 218*x^17 + 323*x^18 - 135*x^19 - 467*x^20 + 1139*x^21 - 1279*x^22 + 430*x^23 + 1587*x^24 +...
%e such that A(x*B(x)) = x^2 - x^3,
%e also, B(x) = B(x^2 - x^3)/x.
%e From _Paul D. Hanna_, Dec 30 2021: (Start)
%e GENERATING METHOD.
%e Define F(n), a polynomial in x of order 3^(n-1), by the following recurrence:
%e F(0) = x,
%e F(1) = (1 - x),
%e F(2) = (1 - x^2 * (1-x)),
%e F(3) = (1 - x^4 * (1-x)^2 * F(2)),
%e F(4) = (1 - x^8 * (1-x)^4 * F(2)^2 * F(3)),
%e F(5) = (1 - x^16 * (1-x)^8 * F(2)^4 * F(3)^2 * F(4)),
%e ...
%e F(n+1) = 1 - (F(n) - 1)^2 * F(n)
%e ...
%e Then the series reversion B(x) equals the infinite product:
%e B(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...
%e that is,
%e B(x) = x * (1-x) * (1 - x^2*(1-x)) * (1 - x^4*(1-x)^2*(1 - x^2*(1-x))) * (1 - x^8*(1-x)^4*(1 - x^2*(1-x))^2*(1 - x^4*(1-x)^2*(1 - x^2*(1-x)))) * ...
%e (End)
%o (PARI) {a(n) = my(A=[1, 1], F, B); for(i=1, n, A=concat(A, 0); F=x*Ser(A); B=serreverse(F); A[#A] = Vec(subst(F, x, x*B))[#A]); A[n]}
%o for(n=1, 50, print1(a(n), ", "))
%o (PARI) /* Using Infinite Product Formula for Series Reversion */
%o {F(n) = my(G=x); if(n==0,G=x, if(n==1,G=1-x, G = 1 - (F(n-1) - 1)^2 * F(n-1) ));G}
%o {a(n) = my(A, B = prod(k=0,#binary(n), F(k) +x*O(x^n)));
%o A = serreverse(B); polcoeff(A,n)}
%o for(n=1,50,print1(a(n),", ")) \\ _Paul D. Hanna_, Dec 30 2021
%Y Cf. A273203, A273095.
%K nonn
%O 1,3
%A _Paul D. Hanna_, May 16 2016
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