Showing from a recurrence that n-th term has at least n primefactors

Question

Let the recurrence be $$ a_{n+1}=a_n(a_n+1) $$

How can I show that the n-th Term has at least n primefactors?

And how does it conclude from that statement that there are infinite many primenumbers?

Anyone any ideas?

Answer 1

If we start with $a_1 > 1$, then since $\gcd(a_1, a_1 + 1) = 1$, $a_2 = a_1(a_1 + 1)$ has at least 2 distinct prime factors. Assume $a_n$ has at least $n$ distinct prime factors, then since $\gcd(a_n, a_n + 1) = 1$, $a_{n+1} = a_n(a_n + 1)$ has at least $n + 1$ distinct prime factors. Thus by induction, $a_n$ has at least $n$ distinct prime factors for all $n$.

It should be clear that there are infinitely many primes now, since we can take $n$ arbitrarily large.

Answer 2

In MSE question 3009367 "Another Proof of Euclid's Theorem (infinite number of primes)?" theorem 2 has the equation

$a_2 = \sum_{i=1}^{a_1+1} a_1 = \sum_{i=1}^{a_1} (a_1+1) \tag{4}$

which is essentially the recursion $a_{n+1} = a_n(a_n+1).$ Later is this statement

Also, if we set $a_1$ to $1$ in theorem 2 we get the researched OEIS sequence A007018.

A comment to this question by Bill Dubuque states

The generated sequence is OEIS A007018. Note: adding 1 yields Sylvester sequence $a_{n+1}=a_n^2-a_n+1=$ OEIS A000058. See the OEIS notes for other connections.

As noted in a comment to A00058

An infinite coprime sequence defined by recursion.

The significance of this is that any infinite coprime sequence of integers implies that the primes are infinite since each term of A000058 has at least one prime factor different than any other terms of A000058. Since each term of A007018 is the partial product of the terms of A000058, this proves that $a_n = A007018(n)$ has at least $n$ prime factors.