Let $\operatorname{tr}(n)$ be A007814, number of trailing zeros in the binary representation of $n$.

Also, let $\operatorname{ntr}(n)$ be A086784, number of non-trailing zeros in the binary representation of $n$.

Finally $$f(n)=2^{n-1}+n, f(0)=0$$

I conjecture that positive number $k$ has only one partition into parts with same binary weight as a binary weight of $k$ if and only if

• second leftmost bit of $k+1$ equals $0$ and $\operatorname{tr}(k+1)\geqslant f(\operatorname{ntr}(k+1))$
• second leftmost bit of $k+1$ equals $1$ and $\operatorname{tr}(k+1)\geqslant 2^{\operatorname{ntr}(k+1)}$

Here binary weight is A000120, number of $1$'s in the binary representation of $n$.

See also A091891 (number of partitions of $n$ into parts which are a sum of exactly as many distinct powers of $2$ as $n$ has $1$'s in its binary representation) and A091892 (numbers $k$ having only one partition into parts which are a sum of exactly as many distinct powers of $2$ as there are $1$'s in the binary representation of $k$).

Given conjecture was verified up to max term in a b-file for A091892 with no counterexamples.

Is there a way to prove it?