OFFSET
0,2
COMMENTS
A continued fraction identity from page 46 of Ramanujan's lost notebook gives:
if k = a*b, a = (sqrt(1 + 4*k) + 1)/2, and b = (sqrt(1 + 4*k) - 1)/2, then
1/(1 + (k + q)/(1 + (k + q^2)/(1 + (k + q^3)/(1 + ...)))) = 1/(a + q/(a+b*q + q^2/(a+b*q^2 + q^3/(a+b*q^3 + ...)))).
Here we set k = x and q = x, with a = 1 - C(-x) and b = -C(-x), where C(x) = x + C(x)^2 is a g.f. of the Catalan numbers (A000108).
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
FORMULA
G.f.: 1/(1 + (x + x)/(1 + (x + x^2)/(1 + (x + x^3)/(1 + ...))))), a continued fraction.
G.f.: 1/(a + x/(a+b*x + x^2/(a+b*x^2 + x^3/(a+b*x^3 + ...))))), a continued fraction where a = (sqrt(1 + 4*x) + 1)/2 and b = (sqrt(1 + 4*x) - 1)/2.
a(n) ~ (-1)^n * c * 4^n/n^(3/2), where c = 5.404096784701640214096763058949993... - Vaclav Kotesovec, Mar 03 2022
EXAMPLE
G.f.: A(x) = 1 - 2*x + 6*x^2 - 18*x^3 + 56*x^4 - 182*x^5 + 610*x^6 - 2090*x^7 + 7284*x^8 - 25732*x^9 + 91908*x^10 + ...
where the g.f. equals the continued fraction
A(x) = 1/(1 + (x + x)/(1 + (x + x^2)/(1 + (x + x^3)/(1 + (x + x^4)/(1 + (x + x^5)/(1 + (x + x^6)/(1 ...)))))),
and also equals the continued fraction given by
A(x) = 1/(a + x/(a+b*x + x^2/(a+b*x^2 + x^3/(a+b*x^3 + x^4/(a+b*x^4 + x^5/(a+b*x^5 + x^6/(a+b*x^6 + ...)))))),
where a = x/b = 1 + b, and b = (sqrt(1 + 4*x) - 1)/2, which begins
b = x - x^2 + 2*x^3 - 5*x^4 + 14*x^5 - 42*x^6 + 132*x^7 - 429*x^8 + 1430*x^9 + ... + (-1)^n*A000108(n)*x^(n+1) + ...
PROG
(PARI) {a(n) = my(R=1); for(k=0, n-1,
R = 1/(1 + (x + x^(n-k))*R +x*O(x^n))); polcoeff(R, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n) = my(R=1, C = (sqrt(1 + 4*x +x^2*O(x^n)) - 1)/2);
for(k=0, n-2, R = 1/(1+C + C*x^(n-k-1) + x^(n-k)*R +x^2*O(x^n))); R = 1/(1+C + x*R); polcoeff(R, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Mar 02 2022
STATUS
approved