Only the $1$st term of this sequence is prime; all other terms are composite.

Denote the $n$th term of this sequence by $a_n$. By mathematical induction, the following facts can be established:

- $a_1$ is divisible by $37$. $a_{n+3} = 555 \cdot 10^{n+1} + a_n$, and $555 = 3 \cdot 5 \cdot 37$. So $a_1, a_4, a_7, \dots$ are all multiples of $37$.

- $a_2 = 537$ is divisible by $3$. $a_{n+3} = 555 \cdot 10^{n+1} + a_n$, and $555 = 3 \cdot 5 \cdot 37$. So $a_2, a_5, a_8, \dots$ are all multiples of $3$.

- $a_3 = 5537 = 7 \cdot 791$ is a multiple of $7$. $a_{n+6} = 555555 \cdot 10^{n+1} + a_n$, and $555555 = 555 \cdot 1001$ is a multiple of $7$. So $a_3, a_9, a_{15}, \dots$ are all multiples of $7$.

- $a_6 = 5555537 = 13 \cdot 427349$ is a multiple of $13$. $a_{n+6} = 555555 \cdot 10^{n+1} + a_n$, and $555555 = 555 \cdot 1001$ is a multiple of $13$, so $a_6, a_{12}, a_{18}, \dots$ are all multiples of $13$.

In summary, only $a_1 = 37$ is prime; all other terms are divisible by one of $7, 13, 37$, and are therefore composite.