TITLE: Primality of the number of obtained by concatenating the n consecutive digits
QUESTION [5 upvotes]: Let $f_n$ be the number obtained by concatenating the first $n$ numbers (in base 10). For example
$f_1 = 1, f_3 = 123$ and $f_{13} = 12345678910111213.$
Now if $n$ is even or divisible by $5$ then so is $f_n$ and if $n(n+1) \equiv 0 \pmod{3}$ then $f_n$ is divisible by $3.$ My question is

Is $f_n$ always composite?

Verifying the question with a computer I get that it holds for $n$ up to $1000$ but I don't see how to prove it.
It does not seem to be case that $f_n$ always has small prime divisors as for example $f_7 = 127 \cdot 9721.$
Hence I don't see any simple approach to this question.

REPLY [2 votes]: As in the middle of November $2015$,there are no primes in the first $269293$ terms (!!), and it's very likely also that there are no primes in the first $300000$ terms. See the great Smarandache probable prime search managed by Serge Batalov. Update: As of 15 december 2015, there are no primes in the first $344869$ terms (the great Smarandache Prprime search).