A350479 G.f. A(x) satisfies: A(x) = A(x^3 - x^5)/x^2.

1, -1, 0, -1, 3, -3, 1, 0, 0, -1, 7, -21, 38, -62, 129, -262, 412, -543, 748, -1111, 1491, -1675, 1705, -1782, 1881, -1749, 1290, -716, 305, -249, 989, -4024, 13098, -36463, 92766, -223338, 509455, -1093243, 2218041, -4315247, 8126815, -14817936, 26133655

Offset

1,5

Links

Table of n, a(n) for n=1..43.

Formula

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

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

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

Example

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 + ...
The series reversion is here denoted R(x) so that R(A(x)) = x where
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) + ...
and which by definition also satisfies R(x^2*A(x)) = x^3 - x^5.
GENERATING METHOD.
One may generate the g.f. A(x) using the following method.
Define F(n), a polynomial in x of order 2*5^(n-1), by the following recurrence:
F(0) = x,
F(1) = (1 - x^2),
F(2) = (1 - x^6 * (1-x^2)^2),
F(3) = (1 - x^18 * (1-x^2)^6 * F(2)^2),
F(4) = (1 - x^54 * (1-x^2)^18 * F(2)^6 * F(3)^2),
F(5) = (1 - x^162 * (1-x^2)^54 * F(2)^18 * F(3)^6 * F(4)^2),
...
F(n+1) = 1 - (1 - F(n))^3 * 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^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) * ...
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) * (1015/2^10) * (1125899155808599/2^50) * ... = 0.37170385361645629840998279262...
A(2/3) = (2/3) * (5/3^2) * (57449/3^10) * (717897986779793260667849/3^50) * ... = 0.36032797749163984225405820293...
A(1/3) = (1/3) * (8/3^2) * (58985/3^10) * (717897986779793260667849/3^50) * ... = 0.29597515647738646568618474726...
The first relative maximum value of A(x) is given by
A(0.5648072584544076680095600...) = 0.3788377227391210352270204...

Prog

(PARI) {a(n) = my(A, R=[1, 0]); for(i=1, n, R=concat(R, 0);
R[#R] = -polcoeff( x^3*(1 - x^2) - subst(x*Ser(R), x, x^2 * serreverse(x*Ser(R))), #R+2) );
A=Vec(serreverse(x*Ser(R))); A[n]}
for(n=1, 60, print1(a(2*n-1), ", "))
(PARI) /* Using Infinite Product Formula */
N = 300; \\ 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))^3 * 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, 60, print1(a(2*n-1), ", "))

Crossrefs

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

Sequence in context: A261477 A205826 A131193 * A130632 A229158 A290454

Adjacent sequences: A350476 A350477 A350478 * A350480 A350481 A350482

Keyword

sign

Author

Paul D. Hanna, Jan 01 2022

Status

approved