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G.f. A(x) = Sum_{n>=1} a(n)*x^(2*n-1) satisfies: A(x^2*R(x)) = x^3 - x^5, where A(R(x)) = x.
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%I #7 Sep 01 2022 12:00:19

%S 1,1,3,13,62,318,1721,9660,55710,328159,1965990,11941986,73377955,

%T 455278070,2848419729,17949945086,113831136362,725895938586,

%U 4651935111855,29944166532941,193516292714718,1255118085879061,8167158264606613,53303153257919970,348836650405783954

%N G.f. A(x) = Sum_{n>=1} a(n)*x^(2*n-1) satisfies: A(x^2*R(x)) = x^3 - x^5, where A(R(x)) = x.

%F G.f.: A(x) = Series_Reversion( Product_{n>=0} F(n) ), where F(0) = x, F(1) = 1-x^2, and F(n+1) = 1 - (1 - F(n))^3 * F(n)^2 for n > 0.

%e G.f.: A(x) = x + x^3 + 3*x^5 + 13*x^7 + 62*x^9 + 318*x^11 + 1721*x^13 + 9660*x^15 + 55710*x^17 + ...

%e The series reversion is here denoted R(x) so that R(A(x)) = x where

%e R(x) = x - x^3 - x^7 + 3*x^9 - 3*x^11 + x^13 - x^19 + 7*x^21 - 21*x^23 + 38*x^25 + ... + A350479(n)*x^(2*n-1) + ...

%e and which by definition also satisfies A(x^2*R(x)) = x^3 - x^5.

%e GENERATING METHOD.

%e One may obtain the g.f. A(x) from the following method used to generate the series reversion R(x).

%e Define F(n), a polynomial in x of order 2*5^(n-1), by the following recurrence:

%e F(0) = x,

%e F(1) = (1 - x^2),

%e F(2) = (1 - x^6 * (1-x^2)^2),

%e F(3) = (1 - x^18 * (1-x^2)^6 * F(2)^2),

%e F(4) = (1 - x^54 * (1-x^2)^18 * F(2)^6 * F(3)^2),

%e F(5) = (1 - x^162 * (1-x^2)^54 * F(2)^18 * F(3)^6 * F(4)^2),

%e ...

%e F(n+1) = 1 - (1 - F(n))^3 * F(n)^2

%e ...

%e Then the series reversion R(x) equals the infinite product:

%e R(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...

%e that is,

%e R(x) = x * (1-x^2) * (1 - x^6*(1-x^2)^2) * (1 - x^18*(1-x^2)^6*(1 - x^6*(1-x^2)^2)^2) * (1 - x^54*(1-x^2)^18*(1 - x^6*(1-x^2)^2)^6*(1 - x^18*(1-x^2)^6*(1 - x^6*(1-x^2)^2)^2)^2) * ...

%e The g.f. of this sequence is then obtained as the series reversion of this infinite product.

%o (PARI) {a(n) = my(A=[1,0]); for(i=1,n, A=concat(A,0);

%o A[#A] = -polcoeff( x^3*(1 - x^2) - subst(x*Ser(A),x, x^2 * serreverse(x*Ser(A))), #A+2) );A[n]}

%o for(n=1,30,print1(a(2*n-1),", "))

%o (PARI) /* Using Infinite Product Formula for Series Reversion */

%o N = 300; \\ set limit on order of polynomials to be 2 times desired number of terms

%o {F(n) = my(G=x); if(n==0,G=x, if(n==1, G = (1-x^2), G = 1 - (1 - F(n-1))^3 * F(n-1)^2 +x*O(x^N) ));G}

%o {a(n) = my(A = prod(k=0,#binary(n), F(k) +x*O(x^n))); polcoeff(serreverse(A),n)}

%o for(n=1,30,print1(a(2*n-1),", "))

%Y Cf. A350479 (inverse), A273162, A273203, A350434, A350474, A350476, A350480, A350482.

%K nonn

%O 1,3

%A _Paul D. Hanna_, Jan 01 2022