Let $$\varphi=\frac{1+\sqrt{5}}{2}$$ Let $$a_1(n)=\left\lfloor n\varphi \right\rfloor, a_2(n)=n+a_1(n)$$ Let $\operatorname{tr}(n)$ be A007814, i.e., the number of trailing zeros in the binary representation of $n$. Here $$\operatorname{tr}(2n+1)=0, \operatorname{tr}(2n)=\operatorname{tr}(n)+1$$
Let $b_1(n)$ and $b_2(n)$ be a pair of infinite complementary sequences (that is, each natural number occurs exactly once in one of the sequences) such that
- $b_1(1) = 1$ and $b_1(n)-b_1(n-m)$ belongs to $\left\lbrace a_2(m), a_2(m)+1 \right\rbrace$ for any $n>m>0$.
- $b_2(1) = 2$ and $b_2(n)-b_2(n-m)$ belongs to $\left\lbrace a_1(m), a_1(m)+1 \right\rbrace$ for any $n>m>0$.
- if $T(n,k)$ is infinite array such that $T(n,1)=b_1(n)$, $T(n,2)=b_2(b_1(n))$ and $T(n,k)=T(n,k-1)+T(n,k-2)$ for $k>2$, then each natural number occurs exactly once in this infinite array.
- if $c(n)=T(c(\left\lfloor\frac{n}{2^{\operatorname{tr}(n)+1}}\right\rfloor+1),\operatorname{tr}(n)+1)$ for $n>1$ with $c(1)=1$ and $d(n)=[n=b_1(k)]2d(k)+[n=b_2(k)]\cdot(2d(k)+1)$ for $n>1$ with $d(1)=0$, then $c(n)$ is a permutation of natural numbers and $d(n)$ is a permutation of nonnegative integers such that $d(c(n))=n-1$.
Examples of such sequences:
- $b_1(n)$ is A007064, $b_2(n)$ is A007067, $T(n,k)$ is A035506, $d(n)$ is A200714
- $b_1(n)$ is A007066, $b_2(n)$ is A026355, $T(n,k)$ is A126714, $d(n)$ is A358654
- $b_1(n)$ is A003622, $b_2(n)$ is A022342, $T(n,k)$ is A035513, $d(n)$ is not in the OEIS
I conjecture that there are infinitely many pairs of sequences $b_1(n)$ and $b_2(n)$ with properties given above such that conditions for $T(n,k)$, $c(n)$ and $d(n)$ holds.
Is there a way to prove it?
