How many are prime?

$\large a_2, a_{19}, a_{23}, a_{317},a_{1031}$ are the only ones known to be primes.

We already know $a_2 = 11$ is prime.

For each other case note that if $n = ij$ for $i,j \neq 1$ , then $b_i = (10^i-1)/9$ and $a_n/b_i=10...010...01$ with $j$ 1's and $i-1$ zeros in between each 1. Since 91=7(13), 837=3(279), and 951=3(319). The only case that is prime is $a_2$ .

This solution is similar to Jaleb's solution but I decided to post it anyways to provide an alternate method.

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$r = \underbrace{111111111\ldots 1}_{n \text{ number of 1's}}$

To solve this problem, we can use the fact that if $n$ is composite then the above number is composite. But let's prove the statement before.

Let $n$ be equal to $ab$ and $s$ be equal to $\underbrace { 111\dots 1 }_{a \text{ number of 1's}}$ . $a$ needs to be greater than 2 while $b$ needs to be greater than 1.

We can rewrite $r$ as

$s\cdot { 10 }^{ n-a }+s\cdot { 10 }^{ n-2a }+\dots +s$

Since $s$ is common in every term, $r$ is not prime.

As 91, 837 and 951 are all composite, they are out of the list.

Now we are left with 11. Through analysis, it's the only prime.

If $n$ is composite, obviously that $a_n$ is not a prime. The inverse here doesn't apply.

Proof: Let $a_n$ be written as the polynomial function for $f:\mathbb{N} \rightarrow \mathbb{N}$ , $f(x) =1+x+...+ x^{n-2} +x^{n-1}$ (later, we substitute $x=10$ ).

From the polynomial, create partitions by $m, m>1$ of $f(x)$ in order, e,g. $f(x) = 1+x+...+x^{n-1}=(1+x+ ... +x^{m-1}) +(x^{m} + ... ) + ... + x^{n-1}$

Let $B$ as the sets of the partitions and let the partitions become somewhat as a subset of $b_1, b_2, ..., b_j$ . $f(x)$ will be factorable if $B' \in \emptyset$ . Thus, there will be $\frac{n}{m}$ subsets, which $m|n$ , from $b_1, b_2, ..., b_{\frac{n}{m}}$ . Exactly, the GCF of the subsets is $b_1$ . We may factor $f(x)$ as $f(x) = (1+x+...+x^{m-1})(1+x^m+...+x^{n-m})$

Setting to $x=10$ . $f(10)$ consists of at least two factors, so it won't be prime. In conclusion, for $n$ composite and a subset partitions of $m$ , which is $b_1, b_2, ..., b_j \in B$ , and $B' \in \emptyset$ , $a_n = f(10)$ is not prime.

For $n$ prime, the inverse of above is broken. Easy to prove that for $n = 2$ , we have $11$ which is a prime, but $n=3$ , we have $111$ which is divisible by $3$ .

a2 is a prime number=11 all of these type of numbers such as 111,1111,11111,... except 1 and 11 are composite numbers. so a2 is the only prime number in the options.