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and which by definition also satifies satisfies R(x^3*A(x)) = x^4 - x^6.

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G.f. A(x) satisfies: A(x) = A(x^4 - x^6)/x^3.

1, -1, 0, 0, -1, 3, -3, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 9, -36, 84, -123, 93, 81, -459, 978, -1346, 1152, -132, -1649, 3681, -5010, 4690, -2496, -858, 4147, -6201, 6396, -5002, 3003, -1365, 455, -105, 15, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 33

1,6

G.f. A(x) = Sum_{n>=1} a(n)*x^(2*n-1) satisfies:

(1) A(x) = A(x^4 - x^6)/x^3.

(2) R(x^3*A(x)) = x^4 - x^6, where R(A(x)) = x.

(3) A(x) = Product_{n>=0} F(n), where F(0) = x, F(1) = 1-x^2, and F(n+1) = 1 - (1 - F(n))^4 * F(n)^2 for n > 0.

G.f.: A(x) = x - x^3 - x^9 + 3*x^11 - 3*x^13 + x^15 - x^33 + 9*x^35 - 36*x^37 + 84*x^39 - 123*x^41 + 93*x^43 + 81*x^45 + ...

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

R(x) = x + x^3 + 3*x^5 + 12*x^7 + 56*x^9 + 282*x^11 + 1494*x^13 + 8207*x^15 + 46332*x^17 + ... + A350482(n)*x^(2*n-1) + ...

and which by definition also satifies R(x^3*A(x)) = x^4 - x^6.

GENERATING METHOD.

One may generate the g.f. A(x) using the following method.

Define F(n), a polynomial in x of order 2*6^(n-1), by the following recurrence:

F(0) = x,

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

F(2) = (1 - x^8 * (1-x^2)^2),

F(3) = (1 - x^32 * (1-x^2)^8 * F(2)^2),

F(4) = (1 - x^128 * (1-x^2)^32 * F(2)^8 * F(3)^2),

F(5) = (1 - x^512 * (1-x^2)^128 * F(2)^32 * F(3)^8 * F(4)^2),

...

F(n+1) = 1 - (1 - F(n))^4 * F(n)^2

...

Then the g.f. A(x) equals the infinite product:

A(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...

that is,

A(x) = x * (1-x^2) * (1 - x^8*(1-x^2)^2) * (1 - x^32*(1-x^2)^8*(1 - x^8*(1-x^2)^2)^2) * (1 - x^128*(1-x^2)^32*(1 - x^8*(1-x^2)^2)^8*(1 - x^32*(1-x^2)^8*(1 - x^8*(1-x^2)^2)^2)^2) * ...

SPECIFIC VALUES.

The infinite product formula allows us to evaluate the function A(x) at certain x rather quickly.

A(1/2) = (1/2) * (3/2^2) * (4087/2^12) * (4722366482760053097487/2^72) * ... = 0.37417602538194148451978837081...

A(2/3) = (2/3) * (5/3^2) * (525041/3^12) * ... = 0.36591009281837971406458290316...

A(1/3) = (1/3) * (8/3^2) * (531377/3^12) * ... = 0.29626061413597559076118753086...

The first relative maximum value of A(x) is given by

A(0.5712201306311149010325669...) = 0.3828554098922613628968808...

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

R[#R] = -polcoeff( x^4*(1 - x^2) - subst(x*Ser(R), x, x^3 * serreverse(x*Ser(R))), #R+3) );

A=Vec(serreverse(x*Ser(R))); H=A; A[n]}

for(n=1, 70, print1(a(2*n-1), ", "))

(PARI) /* Using Infinite Product Formula */

N = 400; \\ set limit on order of polynomials to be 2 times desired number of terms

{F(n) = my(G=x); if(n==0, G=x, if(n==1, G = (1-x^2), G = 1 - (1 - F(n-1))^4 * F(n-1)^2 +x^2*O(x^N) )); G}

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

for(n=1, 70, print1(a(2*n-1), ", "))

Cf. A350482 (inverse), A273218, A350431, A350433, A350475, A350477, A350479, A350481.

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Paul D. Hanna, Jan 01 2022

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