Why does this alternate Fibonacci relation appear to be true? So the normal fibonacci relation is famously as follows:
$$\begin{aligned}F_0 &= 0 \\
F_1 &= 1 \\
F_n &= F_{n-1} + F_{n-2}\end{aligned}$$
My friend appears to have discovered this alternate relation that allows you to skip by 3s, but I can't find it anywhere else to confirm why it works.
$$\begin{aligned}F_0 &= 0 \\
F_3 &= 2 \\
F_n &= 4F_{n-3} + F_{n-6}\end{aligned}$$
I think it's just a case of "simplification", but I wanted to confirm that. Does this work look correct?
$$\begin{aligned}
F_n &= F_{n-1} + F_{n-2} \\
F_n &= F_{n-2} + F_{n-3} + F_{n-2} \\
F_n &= 2F_{n-2} + F_{n-3} \\
F_n &= 2(F_{n-3} + F_{n-4}) + F_{n-3} \\
F_n &= 3F_{n-3} + 2F_{n-4} \\
F_n &= 3F_{n-3} + F_{n-4} + F_{n-5} + F_{n-6} \\
F_n &= 4F_{n-3} + F_{n-6}
\end{aligned}$$
Edit: I get that it doesn't do the entire sequence this way. We only needed the even terms for what we were working on. That's why I even called out that it skipped by 3s. To "fix" it all you would need to do is define the first 6 anyways.

That alternative definition "forgets" the terms $F_n$ when $n$ is not a multiple of three. So $F_5$, say, could be a crazy elephant.
