# Does there exist $a_0$, such that $\{a_n\}_{n=0}^{\infty}$ is unbounded?

Suppose $$\{a_n\}_{n=0}^{\infty}$$ is a sequence, defined by the recurrence relation

$$a_{n+1} = \phi(a_n) + \sigma(a_n) - a_n,$$

where $$\sigma$$ denotes the divisor sum function and $$\phi$$ is Euler's totient function. Does there exist $$a_0$$ such that the corresponding $$\{a_n\}_{n=0}^{\infty}$$ is unbounded?

As $$\phi(a_n) + \sigma(a_n) \geq 2a_n$$ (see here: https://math.stackexchange.com/questions/2888880/is-phin-sigman-geq-2n-always-true), every sequence of this type is monotonically non-decreasing. This means that it is bounded iff it contains an element $$a_n$$ such that $$\phi(a_n) + \sigma(a_n) = 2a_n$$. We know, that to satisfy this equation, $$a_n$$ must either be $$1$$ or prime (see: https://math.stackexchange.com/questions/1215261/find-all-positive-integers-n-such-that-phin-sigman-2n/1215337#1215337). Thus, the question is equivalent to: "Does every such sequence $$\{a_n\}_{n=0}^{\infty}$$ with $$a_0 \geq 2$$ contain a prime element?".

This question was first asked on math.stackexchange.com: https://math.stackexchange.com/questions/2889876/does-there-exist-a-0-such-that-a-n-n-0-infty-is-unbounded

The existing answers and comments contain a lot of information on this topic, that may be useful in solving the question. However, no full answer was provided yet. So I decided to re-ask this question here.

Any help will be appreciated.

• Consider when the expression is odd. Gerhard "That Might Get You Somewhere" Paseman, 2018.12.15. – Gerhard Paseman Dec 15 '18 at 18:43