A repunit is a number consisting solely of ones (such as 11
or 11111).
Let us call p(n) a 10base integer represented by a
string of n ones, e.g. p(1)=1, p(5)=11111 etc.
Most of the repunit numbers are composite.
2, 19,23,317 are
the first four indices of prime repunits.
Prove: For a prime repunit p(n) to be prime, n has to
be prime.
Suppose n is composite, and equals a*b (a and b both > 1)
Then p(n) can be subdivided into a groups of b digits each. Each group is p(b) times an appropriate power of 10, so p(n) = p(b) * [1 + 10^b + 10^2b + ... + 10^(a1)b] (there are a terms in the second factor)
So, p(n) for composite n is also composite unless either p(b) or the second factor is 1. Well, p(b) can't be 1 because b > 1 (see above) and the second factor is also only 1 if a = 1, which also contradicts the initial assumption.)
So all p(n) when n is composite are composite.
That doesn't mean that p(n) is necessarily prime when n is prime, but it does mean there are no cases where n is composite and p(n) is prime, and hence that for p(n) to be prime, n must also be prime.

Posted by Paul
on 20110110 22:23:15 