If there exists $n$ that $p_n\neq 37$, we get that $37$ divides all of $a_{n+1},a_n,\frac{a_n}{p_n}$. So $37\mid p_n\implies p_n=37$. Contradiction. So, $p_n=37$ for all $n\in \mathbb{Z}^+$. For each $n\in \mathbb{Z}^+$, let $a_n=37b_n$. We have $b_{n+1}=b_n-1+\frac{b_n}{37}\implies 37\mid b_n$. For each $n\in \mathbb{Z}^+$, let $a_n=37^2c_n$. We have $37c_{n+1}=37c_n+(c_n-1)\implies 37\mid c_n-1$. For each $n\in \mathbb{Z}^+$, let $a_n=37^2(37d_n+1)$. We have $37d_{n+1}=38d_n$ which implies $37^{\ell} \mid d_n$ for all $n\in \mathbb{Z}^+$ for all $\ell \in \mathbb{Z}^+$. So, $d_n=0$ for all $n\in \mathbb{Z}^+$. Hence $a_1=37^2$.