%I #7 Sep 01 2022 11:55:57

%S 1,-1,0,-1,3,-3,1,0,0,-1,7,-21,38,-62,129,-262,412,-543,748,-1111,

%T 1491,-1675,1705,-1782,1881,-1749,1290,-716,305,-249,989,-4024,13098,

%U -36463,92766,-223338,509455,-1093243,2218041,-4315247,8126815,-14817936,26133655

%N G.f. A(x) satisfies: A(x) = A(x^3 - x^5)/x^2.

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

%F (1) A(x) = A(x^3 - x^5)/x^2.

%F (2) R(x^2*A(x)) = x^3 - x^5, 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))^3 * F(n)^2 for n > 0.

%e G.f.: A(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 - 62*x^27 + 129*x^29 - 262*x^31 + ...

%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 + 13*x^7 + 62*x^9 + 318*x^11 + 1721*x^13 + 9660*x^15 + 55710*x^17 + ... + A350478(n)*x^(2*n-1) + ...

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

%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*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 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^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 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) * (1015/2^10) * (1125899155808599/2^50) * ... = 0.37170385361645629840998279262...

%e A(2/3) = (2/3) * (5/3^2) * (57449/3^10) * (717897986779793260667849/3^50) * ... = 0.36032797749163984225405820293...

%e A(1/3) = (1/3) * (8/3^2) * (58985/3^10) * (717897986779793260667849/3^50) * ... = 0.29597515647738646568618474726...

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

%e A(0.5648072584544076680095600...) = 0.3788377227391210352270204...

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

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

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

%o for(n=1,60,print1(a(2*n-1),", "))

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

%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^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=1,60,print1(a(2*n-1),", "))

%Y Cf. A350478 (inverse), A273218, A350431, A350433, A350475, A350477, A350481, A350483.

%K sign

%O 1,5

%A _Paul D. Hanna_, Jan 01 2022