Let $F(n)$ be A000045 i.e. Fibonacci numbers. Here $$ F(n) = F(n-1) + F(n-2), \\ F(0) = 0, F(1) = 1 $$
Let $a(n)$ be A066258 i.e. $$ a(n) = F(n)^2F(n+1) $$
Let $b(n)$ be A345253 i.e. maximal Fibonacci tree: Arrangement of the positive integers as labels of a complete binary tree.
Let $c(n)$ be an integer sequence such that $$ c(2n+1) = b(2^{n+1} - 1) + \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{\frac{4^i}{2}} b(2^{n + i + 1} - 4^i + j - 1), \\ c(2n) = b(2^{n+1} - 2) + \sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{4^i} b(2^{n + i + 1} - 2\cdot4^i + j - 1) $$
I conjecture that $$c(n)=a(n).$$
I also conjecture that each $b(k)$ is used in the summation of $c(m)$ no more than once.
Here is the PARI/GP program to check it numerically:
a(n) = fibonacci(n)^2 * fibonacci(n+1)
b(n) = my(x=0, y=0); for(i=0, logint(n, 2), [x, y]=[y+1, x+y]; if(bittest(n, i), [x, y]=[y, x+y])); y;
c1(n) = b(2^(n+1) - 1) + sum(i=1, n, sum(j=1, 4^i / 2, b(2^(n + i + 1) - 4^i + j - 1)))
c2(n) = b(2^(n+1) - 2) + sum(i=1, n-1, sum(j=1, 4^i, b(2^(n + i + 1) - 2*4^i + j - 1)))
test(n) = my(A = n\2); if(n%2, c1(A), c2(A)) == a(n)
Is there a way to prove it?
