Let $a(n)$ be A110978 i.e. odd integers that are nonprime, such that there exist two factors of each number that when multiplied together in binary base, do not ever require the use of a "carry" bit.
Let $\operatorname{tr}(n)$ be A007814 i.e. number of trailing zeros in the binary expansion of $n$.
Let $b(n)$ be an integer sequence such that we start with $A=n$ and then consecutively apply $A:=A+\min(2^{\operatorname{tr}(A)},\frac{A}{2^{\operatorname{tr}(A)}})$ while $A<2n$.
Let $c(n)$ be an integer sequence of odd numbers $k$ such that
$$ b\left(\frac{k-1}{2}\right)=k $$
I conjecture that $c(n)$ is a subsequence of $a(n)$.
Here is the PARI/GP program to check it numerically:
f(n, m) = my(v1, v2); v1 = Vecrev(binary(n)); v2 = Vecrev(binary(m)); v3 = []; v4 = []; v5 = select(x -> (x > 0), v1, 1); for(i = 1, #v2, if(v2[i], my(v6); v6 = vector(#v5, j, v5[j]+i-1); v3 = concat(v3, v6); v4 = setunion(v4, v6))); vecsort(v3) == v4
b(n) = my(A = n); while(A < 2*n, my(B = 2^valuation(A, 2)); A += min(B, A/B)); A
isok1(n) = if(n%2, my(A = 2, v1); v1 = divisors(n); while(!(A>=#v1), if(f(v1[A], n/v1[A]), break); A++); A < #v1)
isok2(n) = if(n%2, b((n-1)/2)==n)
my(z=1); for(k=1,299, while(!(isok2(z)), z++); print(isok1(z)); z++;);
Is there a way to prove it?
