OK, so we can show that for $n>0$ $$a_n = 10^{10^{10^n}}+10^{10^n}+10^n-1$$
is divisible by $10^{2^k}+1$ where $2^k$ is the greatest power of $2$ that divides $n$, ie. $n=2^kd$, with $d$ odd
So, for $n$ odd, $k=0, 2^k=1$ and $n$ is divisible by $11$; for $n \equiv 2 \bmod 4$ , $k=1, 2^k=2$ and $n$ is divisible by $101$; for $n \equiv 4 \bmod 8$ , $k=2, 2^k=4$ and $n$ is divisible by $10001$, etc.
as a separate demonstration for the $n$ odd case, examining powers of $10, \bmod 11$:
$$\begin{array}{c|c|c}
\hline x & 10^x \bmod 11 \\
\hline 1 & 10 \\
\hline 2 & 1 \\
\hline 3 & 10 \\
\hline etc \\
\end{array}$$
Note that the higher power components are $10^\text{even}$
So when $n$ is odd, $a_n \equiv (1+1+10-1)\bmod 11 \equiv 0 \bmod 11$
For the general case, obviously $10^{2^k}-1 \equiv -2 \bmod (10^{2^k}+1)$ and we can see that $(10^{2^k}+1) \mid (10^{2(2^k)}-1)$ and also $(10^{2^k}+1) \mid (10^{2j(2^k)}-1)$ for $j\geq 1$,
So for example, $\overbrace{99999999999999}^{\text{14 9's}}\equiv -2 \bmod 101$
Now the two larger powers both have a high enough power of two in the exponent that they are each $\equiv 1 \bmod 10^{2^k}$ as above. So these add back $1$ each to the modular sum and the end result is that a suitable divisor can be found depending on the factors of $2$ in $n$.
So all $a_n$ are composite.
