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.

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

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