Is it possible to find a closed expression for the following sum:
$\sum_{k=1}^{\infty}\frac{1}{a_k^2}$
where $a_k$ is the $k$-th number with its most significant '$1$' at an odd digit place?
A number of this kind is the number $5$, because $5$ in binary is $101_2$, which has its most significant '$1$' at place three. A list of these numbers is given by sequence A053738.
Your sequence $a_k$ consists of the numbers $2^{2j}$ to $2^{2j+1}-1$, for nonnegative integers $j$. Thus your sum is
$$ \sum_{j=0}^\infty \sum_{i=2^{2j}}^{2^{2j+1}-1} \frac{1}{i^2} = \sum_{j=0}^\infty \left(\Psi(1,2^{2j}) -\Psi \left( 1,2^{2j+1} \right) \right) = \sum_{k=0}^\infty (-1)^k \Psi(1, 2^k)$$
in Maple's notation. I don't know of a closed form for the last sum. Numerically it is approximately $1.1939512459529255006$. The Inverse Symbolic Calculator returns nothing.