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 $$ \ell(n) = \left\lfloor\log_2 n\right\rfloor $$
Let $$ T(n, k) = \left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2 $$ Here $T(n,k)$ is the $(k+1)$-th bit from the right side in the binary expansion of $n$.
Let $a(n)$ be an integer sequence such that we start with $x=0, y=0, z=0$ and then for $0 \leqslant i \leqslant \ell(n)$ consecutively apply $$ [x, y] = [y+1, x+y], \\ [z, x] = [x+1, z+x], \\ [y, z] = [z+1, y+z]. $$ After that, at each step we also consecutively apply $$ [x, y] = [y, x+y], \\ [z, x] = [x, z+x], \\ [y, z] = [z, y+z]. $$ if $T(n, i)=1$. Then $a(n)=z-x-1$ where we take values after the whole transformation.
Let $s(n)$ be the set of the numbers $k$ such that $a(k)=n$.
I conjecture that $$ \sum\limits_{i\in s(n)} 1 = F(n), \\ \sum\limits_{i\in s(n)} (\ell(i) + 1) = \sum\limits_{i=0}^{n+1}F(i)F(n-i+1), \\ \sum\limits_{i\in s(n)} \sum\limits_{j=0}^{\ell(i)} T(i, j) = nF(n) - (n-1)F(n-1). $$
Here is the PARI/GP program to check it numerically:
a(n) = my(x = 0, y = 0, z = 0); for(i = 0, logint(n, 2), [x, y] = [y+1, x+y]; [z, x] = [x+1, z+x]; [y, z] = [z+1, y+z]; if(bittest(n, i), [x, y] = [y, x+y]; [z, x] = [x, z+x]; [y, z] = [z, y+z])); z - x - 1
test1(n) = my(v1, s = 1); v1 = vector(fibonacci(n), i, 0); for(i=1, #v1, while(!(a(s) == n), s++); v1[i] = logint(s, 2) + 1; s++); vecsum(v1) == sum(i = 0, n+1, fibonacci(i) * fibonacci(n - i + 1))
test2(n) = my(v1, s = 1); v1 = vector(fibonacci(n), i, 0); for(i=1, #v1, while(!(a(s) == n), s++); v1[i] = hammingweight(s); s++); vecsum(v1) == n * fibonacci(n) - (n-1) * fibonacci(n-1)
Is there a way to prove it?
