Proving $a(n)$ is not prime for all $n\ge 8$ Let $a(n)$ be the number formed by concatenating $n+1$ nines, $n$ zeroes and a one. I noticed the following pattern for the first 9 terms:
\begin{array}{|c|c|c|}
\hline
n & a(n)& \text{Is prime} \\ \hline
0& 91 & \color{red}{\text{False}} \\ \hline
1& 9901 & \color{blue}{\text{True}} \\ \hline
2& 999001 & \color{red}{\text{False}} \\ \hline
3& 99990001 & \color{blue}{\text{True}} \\ \hline
4& 9999900001 & \color{red}{\text{False}} \\ \hline
5& 999999000001 & \color{blue}{\text{True}} \\ \hline
6& 99999990000001 & \color{red}{\text{False}} \\ \hline
7& 9999999900000001 & \color{blue}{\text{True}} \\ \hline
8& 999999999000000001 & \color{red}{\text{False}} \\ \hline
\end{array}
Interestingly, the pattern fails for $n>9$, and it seems that all the next terms are not prime (checked up to $n=1000$ with Python). So how would one prove that for $n\ge8$ all terms are not prime?

The smallest such number for which factordb does not know a prime factor is here. According to my calculations with PFGW there is no further prime of the form $$(10^n-1)\cdot 10^n+1$$ up to $n=13\ 000$ , hence another prime of this form must have at least $26\ 000$ digits.
Small forced factors apparently do not exist and I think neither algebraic or aurifeuillean factors. In this case, there are probably infinitely many such primes, but already to find the next can be an enormous task.
