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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.

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    JFYI: I've [opened a bug report](http://meta.stackexchange.com/questions/87685/migrating-from-meta-to-main-site-loses-user-information) about tomerg losing ownership to this question after the migration. This is somewhat unexpected.2011-04-17

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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.

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    Now that I think about it, $73$ isn't an example, is it? Don't we get $s=9$, but $8$ is a divisor of $p-1$, and $2^8-1\ge73$?2011-04-18