OFFSET
1,2
COMMENTS
a(n+1)/a(n) tends to 2.193... - Vaclav Kotesovec, Feb 27 2024
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..310
FORMULA
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
Let Q(x) = 1 - 2*Sum_{n>=1} x^(n^2), then
(1) Q(x) = Sum_{n>=0} Product_{k=1..n} (x^(2*k-1) - 3*A(x)).
(2) Q(x) = Sum_{n>=0} x^(n^2) / Product_{k=0..n} (1 + 3*x^(2*k)*A(x)).
(3) Q(x) = 1/(1 - F(1)), where F(n) = (x^(2*n-1) - 3*A(x))/(1 + x^(2*n-1) - 3*A(x) - F(n+1)), a continued fraction.
EXAMPLE
G.f. A(x) = x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 32*x^6 + 66*x^7 + 142*x^8 + 316*x^9 + 712*x^10 + 1596*x^11 + 3526*x^12 + ...
Let Q(x) = 1 - 2*Sum_{n>=1} x^(n^2)
then A = A(x) satisfies
(1) Q(x) = 1 + (x - 3*A) + (x - 3*A)*(x^3 - 3*A) + (x - 3*A)*(x^3 - 3*A)*(x^5 - 3*A) + (x - 3*A)*(x^3 - 3*A)*(x^5 - 3*A)*(x^7 - 3*A) + (x - 3*A)*(x^3 - 3*A)*(x^5 - 3*A)*(x^7 - 3*A)*(x^9 - 3*A) + ...
also
(2) Q(x) = 1/(1 + 3*A) + x/((1 + 3*A)*(1 + 3*x^2*A)) + x^4/((1 + 3*A)*(1 + 3*x^2*A)*(1 + 3*x^4*A)) + x^9/((1 + 3*A)*(1 + 3*x^2*A)*(1 + 3*x^4*A)*(1 + 3*x^6*A)) + x^16/((1 + 3*A)*(1 + 3*x^2*A)*(1 + 3*x^4*A)*(1 + 3*x^6*A)*(1 + 3*x^8*A)) + ...
Further, A = A(x) satisfies the continued fraction given by
(3) Q(x) = 1/(1 - (x - 3*A)/(1 + x - 3*A - (x^3 - 3*A)/(1 + x^3 - 3*A - (x^5 - 3*A)/(1 + x^5 - 3*A - (x^7 - 3*A)/(1 + x^7 - 3*A - (x^9 - 3*A)/(1 + x^9 - 3*A - (x^11 - 3*A)/(1 + x^11 - 3*A - (x^13 - 3*A)/(1 - ...)))))))).
where
Q(x) = 1 - 2*x - 2*x^4 - 2*x^9 - 2*x^16 - 2*x^25 - 2*x^36 + ... + -2*x^(n^2) + ...
PROG
(PARI) {a(n, y=3) = my(A=[0, 1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff( (sum(m=1, #A, prod(k=1, m, x^(2*k-1) - y*Ser(A) ) ) + (y-1)*sum(m=1, sqrtint(#A+1), x^(m^2) ) )/(y), #A-1) ); H=A; A[n+1]}
for(n=1, 40, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 20 2024
STATUS
approved