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.
Your derivation that $F_n=4F_{n-3} + F_{n-6}$ is correct ... so this is indeed true for the Fibonacci numbers.
However, your alternative formulation does not define $F_1$ and $F_2$, so you can't call it an alternative formulation. And even if you define $F_1$ and $F_2$, what are $F_4$ and $F_5$?
Indeed, given that the recursive part of your formulation makes reference to $F_{n-6}$, you need to define all 'base' values $F_0$ through $F_5$ separately in order for this to work as an alternative formulation.