We have a sequence $$a(2n+1, p, q) = a(n-2^{f(n)}, p, q), a(2n, p, q) = pa(n, p, q)+qa(2n-2^{f(n)}, p, q), a(0)=a(1)=1$$ where $f(n)$ is A007814, exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.

Then $|a((4^n-1)/3, p, q)|$ is a cube of an integer for any $p,q \in \mathbb{Z}$, $n\in \left\lbrace0, \mathbb{N}\right\rbrace$.

Is there a way to prove it?

Let $p$ and $q$ be integers.

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.

Then we have an integer sequence given by \begin{align} a(0)=a(1)&=1\\ a(2n)& = pa(n)+qa(2n-2^{f(n)})\\ a(2n+1) &= a(n-2^{f(n)}) \end{align}

I conjecture that $a(\frac{4^n-1}{3})$ is always a cube of an integer for any $p,q \in \mathbb{Z}$, $n\in \left\lbrace0, \mathbb{N}\right\rbrace$.

Is there a way to prove it?