Prime numbers which solve $2^s=1\pmod p$ Here we define those primes $p$ for which $\operatorname{ord}_p(2)=s$, where $s$ is the minimum of the set $S$ of all divisors $d\mid p-1$ such that $2^d-1\geq p$.
For example: for $p=7$, $s=3$, $7\mid 2^3-1$ thus $\operatorname{ord}_p(2)=s=3$ ($7$ is such a prime).
Questions: how many such primes are there? 
Are such primes interesting?
Thanks.

If $p$ is a prime, one less than a multiple of $8$, and one more than twice a prime, then it satisfies the condition. This explains $7$; the next such example is $23$. It is generally believed, but not proved, that there are infinitely many primes satisfying the conditions I have given. 
There are primes that satisfy your condition but not mine, e.g., $17$. 
"Interesting" is a subjective term. I found them interesting enough to spend a few minutes writing up this answer. 
EDIT: It seems I can't comment on tomerg's answer, so I'll put my comment here. 
@tomerg, yes, I said there are primes of your kind that are not of my kind, and I gave $17$ as an example. $73$ is also an example. You wanted to know how many of your primes there are, and I have given a good reason to believe (but not a proof) that there are infinitely many. I hope someone else can build on what I've done, and give a complete answer. But this may be difficult, so I've done what I can. 
