%I #7 Sep 01 2022 11:59:20

%S 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,

%T -1346,1152,-132,-1649,3681,-5010,4690,-2496,-858,4147,-6201,6396,

%U -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

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

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

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

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

%F (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.

%e 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 + ...

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

%e 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) + ...

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

%e GENERATING METHOD.

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

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

%e F(0) = x,

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

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

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

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

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

%e ...

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

%e ...

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

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

%e that is,

%e 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) * ...

%e SPECIFIC VALUES.

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

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

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

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

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

%e A(0.5712201306311149010325669...) = 0.3828554098922613628968808...

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

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

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

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

%o (PARI) /* Using Infinite Product Formula */

%o N = 400; \\ 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))^4 * F(n-1)^2 +x^2*O(x^N) ));G}

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

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

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

%K sign

%O 1,6

%A _Paul D. Hanna_, Jan 01 2022