OFFSET
1,4
FORMULA
G.f.: A(x) = Series_Reversion( Product_{n>=0} F(n) ), where F(0) = x, F(1) = 1+x, and F(n+1) = 1 + (F(n) - 1)^2 * F(n) for n > 0.
EXAMPLE
G.f.: A(x) = x - x^2 + x^3 - 2*x^4 + 6*x^5 - 17*x^6 + 45*x^7 - 123*x^8 + 360*x^9 - 1085*x^10 + 3271*x^11 - 9905*x^12 + ...
Let R(x) be the series reversion of A(x),
R(x) = x + x^2 + x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 5*x^7 + 9*x^8 + 14*x^9 + 18*x^10 + 25*x^11 + 38*x^12 + ... + A350432(n)*x^n + ...
then R(x) and g.f. A(x) satisfy:
(1) A(R(x)) = x,
(2) A(x*R(x)) = x^2 + x^3.
GENERATING METHOD.
Define F(n), a polynomial in x of order 3^(n-1), by the following recurrence:
F(0) = x,
F(1) = (1 + x),
F(2) = (1 + x^2 * (1+x)),
F(3) = (1 + x^4 * (1+x)^2 * F(2)),
F(4) = (1 + x^8 * (1+x)^4 * F(2)^2 * F(3)),
F(5) = (1 + x^16 * (1+x)^8 * F(2)^4 * F(3)^2 * F(4)),
...
F(n+1) = 1 + (F(n) - 1)^2 * F(n)
...
Then the series reversion R(x) equals the infinite product:
R(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...
that is,
R(x) = x * (1+x) * (1 + x^2*(1+x)) * (1 + x^4*(1+x)^2*(1 + x^2*(1+x))) * (1 + x^8*(1+x)^4*(1 + x^2*(1+x))^2*(1 + x^4*(1+x)^2*(1 + x^2*(1+x)))) * ...
PROG
(PARI) /* Using Functional Equation in Definition */
{a(n) = my(A=[1, -1], B); for(i=1, n, A = concat(A, 0);
R = serreverse(x*Ser(A));
A[#A] = -polcoeff( x^2 + x^3 - subst(x*Ser(A), x, x*R), #A+1) ); H=A; A[n]}
for(n=1, 30, print1(a(n), ", "))
(PARI) /* Using Infinite Product Formula for Series Reversion */
{F(n) = my(G=x); if(n==0, G=x, if(n==1, G=1+x, G = 1 + (F(n-1) - 1)^2 * F(n-1) )); G}
{a(n) = my(A, R = prod(k=0, #binary(n), F(k) +x*O(x^n)));
A = serreverse(R); polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Dec 30 2021
STATUS
approved