Consider
$$f(1) = g(1) = 1$$ $$f(2) = A,g(2) = B$$ $$f(3) = 1 + A,g(3) = 1+B$$
And for $n>3$ :
$$f(n) = f(n-1) + f(n-2) + f(n/2)$$ $$g(n) = g(n-1) + g(n-2)$$
where we take the integer part of the fraction $n/2$ so $5/2 = 2,7/2=3,...$
Let us define this limit that always converges:
$$t(A,B) = \lim_{n \to \infty} \frac{f(n)}{g(n)}$$
Now it appears that
$$t(2,4)^2 = t(3,4) =2$$ $$t(2,1) = t(8,5) = 4$$ $$t(10,4)= t(3,1) = 6$$
Are any of those equations true ?
Can this be explained by fibonacci alone ?
EDIT
$$\lim_{n \to \infty} \frac{f(n+1) - t(A,B)g(n+1)}{f(n) - t(A,B)g(n)} = G$$
Where $G$ is the golden mean.
It converges a bit slower than for the fibonacci ratio
$$\lim_{n \to \infty} \frac{g(n+1)}{g(n)} = G$$
but eventually it does.
I have no hard evidence of it though. Clearly it cannot be bigger than $G$ since that would make $f$ grow way too fast. But $G - 0.00001$ or so is more difficult to reject at first sight.