Infinitely many primes of the form $\underbrace{11\dots 1}_{k \text{ times}}\dots \underbrace{11\dots 1}_{k \text{ times}}$

Question

[Analytic Number Theory - Florian Luca and Jean Marie De Koninck, chapter 14, question 14.9]

Prove that for each positive integer $k$, there are infinitely many primes which when written in base $10$ look like $$\underbrace{11\dots 1}_{k \text{ times}}\dots \underbrace{11\dots 1}_{k \text{ times}}$$ where there are some unspecified digits in the middle.

I am sure that we need to construct a clever arithmetic progression and somehow use Dirichlet's Theorem. But, I can't figure out how.

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This post shows that there are infinitely many digits starting with a given string, and this one shows that there are infinitely many ending in $k$ $1$'s.

Answer 1

To elaborate a little on the comments: Let $$a = \underbrace{11\cdots 1}_{k \text{ times}}$$ and $$d = 1\underbrace{0\cdots 0}_{k \text{ times}}.$$ Then, by Dirichlet's theorem, there are infinitely many primes of the form $a + nd$ which then have the desired trailing ones. To get the initial ones, let $$x = a * 10^n = \underbrace{11\cdots 1}_{k \text{ times}}\;\;\underbrace{00\cdots 0}_{n \text{ times}}$$ for some large $n$ to be decided. Set $c = 1 + 10^{-k}$; then any $y$ with $x \leq y < cx$ must start with $k$ $1$'s as well. Letting $\pi_{d,a}(x)$ be the number of primes $p \leq x$ of the form $a + nd$, our goal is to show that $$\pi_{d,a}(cx) - \pi_{d,a}(x) \geq 1.\tag{*}\label{*} $$ Then there is one prime of the desired form. And by varying $n$, we will get infinitely many.

To show $(*)$, we use the prime number theorem for arithmetic progressions, which states that for any given $\epsilon$ and all sufficiently large $x$,

$$(1- \epsilon) \frac {L(x)}{\phi (d)} < \pi_{d,a}(x) < (1+ \epsilon) {\frac {L(x)}{\phi (d)}}.$$

where $L(x) = x / \log(x)$. Hence $$\pi_{d,a}(cx) - \pi_{d,a}(x) > \frac{L(cx) - L(x)}{\phi(d)} - 2 \epsilon \frac{L(cx)}{\phi(d)}$$

and so we are done if we can show that for some $\epsilon$ depending only on $k$ and $d$, that for all sufficiently large $x$ $L(cx) - L(x) > \phi(d) + 2 \epsilon L(cx)$.