How many primes in the series?

The sequence can be given recursively by: $a_{0}=13$ , $a_{n+1}=100\cdot a_{n}+273$

Since $273$ is divisible by $13$ , every term of $a$ is divisible by $13$ so there are $\boxed{0}$ primes in the sequence.

My solution is the following: To find a pattern, I compared the divisors in the first three numbers (I wrote java code to know the divisords) and then found out, that they all shared 13 as a divisor. I then tried to prove that all numbers in the secuence had 13 as a divisor, and to do that, I prefered to show that all the diferences were divisible by 13. (In $a+b=c$ , if two variables are divisible by some number, then the third one is divisible aswell. In this case it´s $1573+(157573-1573)=157573$ , $157573+(15757573-157573)=15757573$ ... and so on, and as we know that 1573 is divisible by 13, we just have to proof that the diferences are divisible.) To proof that the differences are divisible, I also tried to find patterns. The differences were the following $156000, 15600000, 1560000000...$ . I then used the 13 divisibility rule (https://www.quora.com/What-is-the-divisibility-rule-of-13) and found out that all diferences converged in 156, which is divisible by 13.