Let $f(n)$ is A059894, $$f(2n+1) = f(n) + 2^{\left\lfloor\log_2{n}\right\rfloor}, f(2n) = f(n) + 2^{\left\lfloor\log_2{n}\right\rfloor+1}, f(1)=1$$ and we have a sequence $$a(2n+1) = a(n), a(2n) = pa(n) + qa(n - 2^{g(n)}) + a(2n - 2^{g(n)}), a(0)=1$$ where $g(n)$ is A007814, $$g(2n+1) = 0, g(2n) = g(n) + 1, g(1) = 0$$ so if we have $$b(2n+1) = b(n), b(2n) = qb(n) + pb(n - 2^{g(n)}) + b(2n - 2^{g(n)}), b(0)=1$$ then it's true that $$a(2n) = b(2f(n)), b(2n) = a(2f(n))$$ Here $p,q\in\mathbb{R}$.
Is there a way to prove it?