Distribution of decimal repunit primes The prime number theorem describes the distribution of prime numbers in positive integers. Is there a similar theorem describing the distribution of primes among positive integers of the form $\frac{10^n - 1}{9}$? 
If so, what techniques are used to arrive at the theorem?
 A: Acording to The Prime Glosary only 5 repunit primes are known, although it is conjectured that there is an infinite number of them. So the answer to your question is no, nothing is known about the distribution of repunit primes.
A: Before the Prime Number Theorem was proved, mathematicians used calculations and heuristics to determine that there 'should' be about one prime every $\log x$ numbers around $x$. We can work similarly: suppose a repunit is just like other numbers of its size which are relatively prime to 10. Even numbers are twice as likely to be prime as other numbers of their size, and non-multiples of 5 are 5/4 times more likely to be prime. Since the k-th repunit is $(10^k-1)/9,$ its logarithm is around $k\log10$ and so it 'should' be prime with 'probability' around
$$
2\cdot\frac{5}{4}\cdot\frac{1}{k\log10}=\frac{5}{2\log10}\cdot\frac1k\approx\frac{1.0857}{k}.
$$
Taking the integral, this would suggest that the number of $k$-digit repunit primes with $k\le n$ is roughly Poisson distributed with $\lambda\approx1.0857\log n,$ or that the number of repunit primes up to $x$ is roughly Poisson distributed with $\lambda\approx1.0857\log\log x.$
There are more complications since $k$ must be prime for the $k$-th repunit to be prime, but ultimately the expected order of growth doesn't change and Caldwell compares their growth to $\log\log x$ here:
http://primes.utm.edu/glossary/page.php?sort=Repunit
Of course all of this is highly conjectural -- counting primes in sparse forms is very hard to do in general.
