Prove there are no prime numbers in the sequence $a_n=10017,100117,1001117,10011117, \dots$ Define a sequence as $a_n=10017,100117,1001117,10011117$. (The $nth$ term has $n$ ones after the two zeroes.)
I conjecture that there are no prime numbers in the sequence. I used wolfram to find the first few factorisations:
$10017=3^3 \cdot 7 \cdot 53$
$100117=53\cdot 1889$
$1001117=13 \cdot 53\cdot1453$ and so on.
I've noticed the early terms all have a factor of $53$, so the problem can be restated as showing that all numbers of this form have a factor of $53$. However, I wouldn't know how to prove a statement like this. Nor am I sure that all of the terms do have a factor of $53$. 
I began by writing the $nth$ term of the sequence as 
$a_n=10^{n+3}+10^n+10^{n-1}+10^{n-2}+10^{n-3}+\cdots+10^3+10^2+10^1+7$ but cannot continue the proof.
 A: 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.
A: 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.)
A: It is as simple as $$a_{n+1}=10a_n-53$$
A: Another way to find the inductive relationship already cited, from a character manipulation point of view:
Consider any number in the sequence, $a_n$.  To create the next number, you must:


*

*Subtract $17$, leaving a number terminating in two zeroes;

*Divide by $10$, dropping one of the terminal zeroes;

*Add $1$, changing the remaining terminal zero to a $1$;

*Multiply by $100$, sticking a terminal double zero back on;

*Add $17$, converting the terminal double zero back to $17$


Expressing this procedure algebraically, and simplifying:
$$a_{n+1}=\left (\frac{a_n-17}{10}+1 \right ) \times 100+17=10a_n-53$$
