Equation $x=\phi(x)+\phi(x+1)-1$

Does anybody have any ideas how to solve the equation $x=\phi(x)+\phi(x+1)-1$, where $x$ is a natural number and $\phi$ is Euler's totient function?

I failed even to figure out whether this equation has finite number of solutions or infinite.

Any help will be appreciated.

• It seems nicer to write it as y=phi(y)+phi(y-1). What have you tried? Can you deduce some divisibility properties of y? Does this arise as part of a research problem? Gerhard "Has More Questions And Answers" Paseman, 2017.06.29. Jun 29 '17 at 17:50
• It appears that there is a lot of solutions, the number of which appears to grow logarithmically. Jun 29 '17 at 17:55
• My belief is that the solutions of the equation, A067798 from the OEIS, are the same that the solutions of $$x=\phi(\phi(x)+\phi(x+1)-1)+\phi(\phi(x)+\phi(x+1))-1.$$ I've tested it for the segment of integers $1\leq x\leq 10^6$. I have not studied subsequent compositions. Oct 17 '19 at 14:41

This is OEIS Sequence A067798. Nothing else seems to be known about it; at any rate OEIS gives no references to the literature, only a link to a list of further such $x$ from Giovanni Resta that extends it from the 43rd solution, $722015$, to the 76th, $1103806594815$, which is just a a bit larger than $2^{40}$. (Thanks to Robert Israel for noting this link in his comment.)
It seems very hard to do anything with the problem other than make a few elementary observations (e.g. all $x>2$ will be odd), do some heuristics (biases mod $3$, $4$, $5$, and other small moduli; expected asymptotics), and use yet more computing power to find a handful of further solutions.
• @GerhardPaseman That would require $\varphi(2^{2^n}-1) = 2^{2^n-1}$. For $n \ge 6$, $2^{2^n}-1$ is divisible by $641$, so $\varphi(2^{2^n}-1)$ divisible by $\varphi(641) = 640$ which is not a power of $2$. So $2^{2^n}$ is never a solution for $n \ge 6$. Jun 29 '17 at 19:31