# Are there infinitely many primes of the form $12345678901234567890\dots$

Related to this question,

What is the smallest prime number made of sequential number?

are there infinitely many primes of the following form (OEIS A057137)?

$1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789, 1234567890, 12345678901, 123456789012, 1234567890123, 12345678901234, 123456789012345, 1234567890123456, 12345678901234567, 123456789012345678, 1234567890123456789, \dots$

The first five such primes are fairly easy to find via brute force with the aid of a computer and have the following lengths:

$171, 277, 367, 561, 567$

The sixth term comes after a bit of a gap with length $18881$ (see OEIS A120819), which takes a significant amount of computation time to reach, even just testing integers ending in $1$ and $7$.

Are there infinitely many such primes?

Edit: For such numbers ending in $1$, we can write

$$x_n = 10^{10n} + 2345678901 \cdot \frac{10^{10n}-1}{10^{10}-1}$$

and for such numbers ending in $7$, we can write

$$y_n = 1234567 \cdot 10^{10n} + 8901234567\cdot \frac{10^{10n}-1}{10^{10}-1}$$

• A possible approach towards solution: This sequence contains a sub-sequence that can be expressed using a recurrence relation of $a_1=1$ and $a_{n+1}=10000000000a_n+2345678901$. Try to convert this recurrence relation into an explicit formula, then perhaps use Dirichlet theorem (though I doubt that it will work, since the explicit formula is not going to be a polynomial I suppose). – barak manos Oct 4 '16 at 5:40
• If the answers were known, I'd expect they'd be mentioned at the OEIS page. – Gerry Myerson Oct 4 '16 at 6:24
• This might be hard... as @barakmanos says we can look at a recurrence $a_{n+1} = r a_n + C, a_0 = 1$ which has solution $a_{n} = r^n + Cr^{n-1} + \cdots + Cr + C$. Whether or not that solution has infinitely many primes with $C = 1, r = 2$ is the Mersenne prime conjecture. – Chris Jones Nov 20 '16 at 20:52
• At first $1234567=127*9721$,$8901234567=3*3*29*1291*26417$ and $2345678901=3*3*71*419*8761$. – Takahiro Waki Jan 30 '17 at 14:58
• @Pickle the recursion I wrote is not an arithmetic progression except for $r=1$. – Chris Jones Apr 5 '18 at 20:04