How many prime divisors does (10^263-1)/9 have?

Your number is 111111....1 (263 1's). If a number consists of a sequence of n 1's and n = st where s is prime, then 1111...1 (s 1's) divides n. Your n = 263 is a prime number, so it has no st decomposition. So a sequence of 263 1's is a prime number. However, this argument is heuristic and does not satisfy my notions of rigour.

Sorry to answer my own question, but I found the answer online, here:

http://stdkmd.com/nrr/repunit/tm.cgi?p=3

And the answer is that (10^263-1)/9 has 3 prime divisors, so it is neither prime nor semi-prime.