Prove there are no prime numbers in the sequence $a_n=10017,100117,1001117,10011117, \dots$

It is as simple as $$a_{n+1}=10a_n-53$$

Use induction in order to complete the (excellent) hint by @Michael.

First, show that this is true for $n=1$:

$a_1=53\cdot189$

Second, assume that this is true for $n$:

$a_n=53k$

Third, prove that this is true for $n+1$:

$a_{n+1}=$

$10\cdot\color\red{a_n}-53=$

$10\cdot\color\red{53k}-53=$

$530k-53=$

$53(10k-1)$

Please note that the assumption is used only in the part marked red.

The sequence is given by $$ a_n = 10^{n+3}+10\cdot \frac{10^n-1}{9}+7 $$ Then $$ 9a_n = 9\cdot10^{n+3}+10\cdot (10^n-1)+63 = 9010\cdot 10^n+53 = 53\cdot(170 \cdot 10^n+1) $$ Therefore, $53$ divides $9a_n$. Since $53$ does not divide $9$, we have that $53$ divides $a_n$, by Euclid's lemma. (We don't even need to use that $53$ is prime, just that $9$ and $53$ are coprime.)