Let $\operatorname{wt}(n)$ be A000120, i.e. the number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $a(n)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)\operatorname{mod}2=\operatorname{wt}(k+1)\operatorname{mod}2=1$: the sequence begins with $$1, 7, 13, 21, 25, 31, 37, 41, 49, 55, 61, 69, 73, 81, 87, 93, 97, 103, 109, 117$$
Here $a(n)$ is A157971, odious twin locations: first members of pairs of consecutive odious numbers.
Let $b(n)=\frac{a(n)-1}{2}$.
Let $c(n)$ be the sequence of numbers $\frac{b(k)}{3}$ such that $3|b(k)$: the sequence begins with
$$1, 2, 4, 5, 6, 8, 9, 10, 12, 16, 17, 18, 20, 21, 22, 24, 25, 26, 30, 32$$
Conjecture: $c(n)$ is A095775, $\frac{n}{2}$ when $2$A003160$(n) = n$.
Here A003160 is $$d(n) = n-d(d(n-1))-d(d(n-2)), d(1) = d(2) = 1$$
Is there a way to prove it?
