Let $a_1 = 2$ and for all natural number n, define $a_{n+1}= a_{n}(a_{n}+1)$. Then as $n\rightarrow \infty$, the number of prime factors of $a_{n}$

Question

Let $a_1 = 2$ and for all natural number n, define $a_{n+1}= a_{n}(a_{n}+1)$. Then as $n\rightarrow \infty$, the number of prime factors of $a_{n}$:

goes to infinity.
goes to a finite limit.
oscillates boundedly.
oscillates unboundedly.

Answer 1

$$a_{n+1}=a_n(a_n +1)$$

$$=a_{n-1}(a_{n-1}+1)(a_n+1)$$

$$=a_1(a_1+1)(a_2+1)...(a_n+1)$$

Now notice that $a_{k+1}+1=a_k(a_k+1)+1$, which means that $a_{k+1}+1\equiv1$(mod $a_k+1$), $a_{k+1}+1$ and $a_k+1$ are coprime.