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 A345253 i.e. maximal Fibonacci tree: arrangement of the positive integers as labels of a complete binary tree.
There is a PARI/GP program due to Kevin Ryde:
a(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;
We can slightly modify it by
bittest(n, i) -> !bittest(n, i)
- Let $b(n)$ be an integer sequence obtained from the program given above with mentioned modification.
The sequence begins with $$ 0, 1, 1, 4, 3, 3, 2, 12, 9, 8, 6, 8, 6, 5, 4, 33, 25, 22, 17, 21 $$
- Let $c(n)$ be an integer sequence such that $c(1)=1$ and if $x$ is a term, then so are $2x$, $4x+1$ and $4x+3$.
The sequence begins with $$ 1, 2, 4, 5, 7, 8, 9, 10, 11, 14, 16, 17, 18, 19, 20, 21, 22, 23 $$
- Let $d(n)$ be A000975 i.e. the $n$-th number without consecutive equal binary digits.
The sequence begins with $$ 0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845 $$
- Let $e(n)$ be A066258 i.e. $$ e(n) = F(n)^2F(n+1) $$
I conjecture that the sequence $b(c(n))+1$ is a permutation of natural numbers.
I also conjecture that $$ \sum\limits_{i=d(n-1)+1}^{d(n)}(b(c(i))+1)=e(n). $$
Here is the PARI/GP program to check it numerically:
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;
b1(n) = if(n < 2, n, (n%2 == 0 && b1(n/2)) || (n%4 == 1 && b1((n-1)/4)) || (n%4 == 3 && b1((n-3)/4)))
d(n) = (4*2^n - 3 - (-1)^n) / 6
b2(n) = my(s = 2^(n-1), A = 0); for(i = 1, d(n) - d(n-1), while(!b1(s), s++); A += (b(s)+1); s++); A
test(n) = b2(n) == fibonacci(n)^2*fibonacci(n+1)
Is there a way to prove it?
