If $q$ is a prime number I write $$\mathfrak a(q) = \binom{2^{q}}{2}-1$$ Surely $\mathfrak a(q) =2^{q-1}(2^q-1)-1$. I would like to answer the following claim
For every prime $q>2$ there exist an $n$ with $1\le n \le 5$ such that $3^nq\text{ }|\text{ }\mathfrak a(q)$
This is based on a hunch and computer checks. Also I suspect that the Wagstaff primes can occur as the largest prime factors of $\mathfrak a(q)$. The first prime to fail is $29$. In particular I am going to guess that
For each prime $q$ there exist a Wagstaff prime $p$ such that $p\text{ }|\text{ }\mathfrak a(q)$.