Re: Rotationally Prime Numbers Revisited

[ambrus]

A068652 is the sequence of rotational primes. It references A003459, the sequence of absolute primes, the numbers of which all the permutations are primes. That sequence contains the numbers 1111111111111111111, and 11111111111111111111111. There's also a link to this page: MATHEWS: Circular, Permutable, Truncatable and Deletable Primes, which lists all known rotational primes. According to that list, you probably won't find any more rotational primes.

[tall_man]

According to the Mathews page, all repunit primes (primes consisting of the digit "1" only) are rotational primes -- and that there are conjectured to be an infinite number of them.

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

[tilly]

Fascinating. Contrary to what I said at Re: Re (tilly) 1: Rotationally Prime Numbers, when I consider numbers that are strings of 1's, the prime number theorem naively suggests that there should be an infinite number of rotational primes of that form. (This differs from most numbers in that there is only 1 number in the rotation - the odds of 1 number being prime is far different than the odds of n of them being prime.)

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.)

[ambrus]

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.

[hv]

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