Of course there are a lot of special factorization properties of long strings of 1's. So it may not be so simple as all that. But the number of rotational factors between length 23 and 1031 matches the naive prediction surprisingly well. (The naive prediction is that the number of primes out to length n should be roughly log(n)/log(10), and the number in any interval should be the difference of those two. From 23 to 1031 we'd predict 1.65 and actually had 2.)

As we all know, (10**n-1)/9 is surely not a prime when n is composed, because (10**(k*l)-1)/9 = ((10**(k*l)-1)/(10**k-1)) * ((10**k - 1)/9). Thus, they do have special factoring properties, but this doesn't mean that they are either more often or less often primes then other numbers.

Indeed, I'd argue they're more likely to be primes: numbers of the form M(a, p) = (a^p - 1)/(a - 1) form an extended class of Mersenne numbers, and in particular will not share a factor with any M(a, q), q < p, and draw their factors from the restricted set {p, <2kp + 1>}.

I don't know if there's a way to adapt the Lucas-Lehmer test to check directly for divisors in this extended class.

Hugo

If so, there are an infinite number of (trivial) repunit rotational primes, but there may not be any more nontrivial ones.