# Does $19,199,1999,\dotsc$ contain infinitely many prime numbers?

Are there infinitely many primes of the form $F_n =2\times10^n-1$? That is, does this sequence,

$$19,199,1999,\dotsc$$

contain infinitely many prime numbers?

I think about Dirichlet's theorem on arithmetic progressions, but the problem is difficult to start.

Edit: $F_n$ is prime for $n=1, 2, 3, 5, 7, 26, 27,\dotsc, 55347$ (A002957) and $n=1059002$ (Kamada's tables). $F_{n}$ for $n=6m+4$ is divisible by $7$.

• Actually, Dirichlet's method won't suffice here. You need some strong sieve-theoretic tools like the ones provided by Hooley. But then a major problem on such analysis is described in here – Balarka Sen Jan 19 '14 at 15:46
• It will help if you ask your question as a question. Are you asking if it's true there are infinitely many primes of the given form? Are you asking if it's known? Or are you asking if there's an easy proof? As is, you seem to be making an assertion, rather than asking a question. – Barry Cipra Jan 19 '14 at 16:12
• @ziangchen, excellent, thank you! – Barry Cipra Jan 19 '14 at 16:22
• There is statistical evidence that there are infinitely many primes of the form $2*10^n-1$, but a proof seems to be out of reach. The situation is similar to the generalized mersenne numbers. – Peter Jan 20 '14 at 14:02
• 1 2 3 5 7 26 27 53 147 236 248 386 401 546 785 1325 1755 are the first few numbers n, such that $2*10^n-1$ is prime. – Peter Jan 20 '14 at 14:13