smc

The $n$th term of this sequence is $$\sum _{k=n}^{2n}10^k+\sum _{k=0}^{n}10^k=10^n\sum _{k=0}^{n}10^k+\sum _{k=0}^{n}10^k=\left(10^n+1\right)\sum _{k=0}^{n}10^k.$$ It follows that the terms are $$\begin{array}{rl}121& =11\cdot 11,\\ 11211& =101\cdot 111,\\ 1112111& =1001\cdot 1111,\\ & ⋮\end{array}$$ Therefore, there are $\boxed{0}$ prime numbers in this sequence.