OFFSET
1,4
COMMENTS
The radius of convergence r of g.f. A(x) solves r*(1+r)^3 = 1 where r = 0.3802775690976141156733...
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..500
FORMULA
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
(1) A(x) = A( x^2*(1+x)^3 ) / (x*(1+x)^2).
(2) R(x*(1+x)^2*A(x)) = x^2*(1+x)^3, where R(A(x)) = x.
(3) A(x) = x * Product_{n>=1} F(n), where F(1) = 1+x, and F(n+1) = 1 + (F(n) - 1)^2 * F(n)^3 for n >= 1.
(4) A(x)^3 = x^2*B(x) where B(x) is the g.f. of A369553.
EXAMPLE
G.f.: A(x) = x + x^2 + x^3 + 4*x^4 + 7*x^5 + 11*x^6 + 26*x^7 + 75*x^8 + 222*x^9 + 592*x^10 + 1404*x^11 + 3187*x^12 + ...
RELATED SERIES.
A(x)^3/x^2 = x + 3*x^2 + 6*x^3 + 19*x^4 + 51*x^5 + 114*x^6 + 280*x^7 + 750*x^8 + 2064*x^9 + 5701*x^10 + ... + A369553(n)*x^n + ...
Let R(x) be the series reversion of A(x),
R(x) = x - x^2 + x^3 - 4*x^4 + 13*x^5 - 32*x^6 + 92*x^7 - 327*x^8 + 1066*x^9 - 3287*x^10 + 11143*x^11 - 39080*x^12 + ...
then R(x) and g.f. A(x) satisfy:
(1) R(A(x)) = x,
(2) R(x*(1+x)^2*A(x)) = x^2*(1 + x)^3.
GENERATING METHOD.
Define F(n), a polynomial in x of order 5^(n-1), by the following recurrence:
F(1) = (1 + x),
F(2) = (1 + x^2 * (1+x)^3),
F(3) = (1 + x^4 * (1+x)^6 * F(2)^3),
F(4) = (1 + x^8 * (1+x)^12 * F(2)^6 * F(3)^3),
F(5) = (1 + x^16 * (1+x)^24 * F(2)^12 * F(3)^6 * F(4)^3),
...
F(n+1) = 1 + (F(n) - 1)^2 * F(n)^3
...
Then the g.f. A(x) equals the infinite product:
A(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...
PROG
(PARI) {a(n) = my(A=[1], F); for(i=1, n, A=concat(A, 0); F=x*Ser(A); A[#A] = polcoeff( subst(F, x, x^2*(1 + x)^3 ) - x*(1 + x)^2*F , #A+1) ); A[n]}
for(n=1, 35, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 25 2024
STATUS
approved