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.
Here is the PARI/GP prog to check it numerically:
tr(n) = valuation(n, 2)
ntr(n) = logint(n, 2) - hammingweight(n) - valuation(n, 2) + 1
f(n) = if(n == 0, 0, 2^(n - 1) + n)
my(z=1); for(k=1, 2053, while(!(if(bittest(z, logint(z, 2) - 1), tr(z) >= 2^ntr(z), tr(z) >= f(ntr(z)))), z++); print([k, z-1]); z++;);
Is there a way to prove it? Is there a way to compute more data (as much as possible) to additionally verify it?
Please leave the answers to the second question in the comments. Thank you in advance!