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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.

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    $\begingroup$ $g(n)\sim\frac{\frac{3-B}{\sqrt{5}}+B-1}{2}\varphi^n$ isn't too hard to find out, where $\varphi$ is the golden ratio. Looking at the numbers, I think $f(n)=\frac{R_{n-1}-S_{n-6}}{2}+R_{n-1}A$, where $S_n$ is given by A001595 and $R_n$ is given by A298352. Both $R_n$ and $S_n$ are in $\Theta(\varphi^n)$ according to the OEIS links, although I couldn't find an exact constant for $R_n$. $\endgroup$
    – Varun Vejalla
    Commented Jun 5, 2023 at 7:00
  • $\begingroup$ @VarunVejalla thanks $\endgroup$
    – mick
    Commented Jun 5, 2023 at 11:19

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