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Let $V(x)$ count the number of $n\leq x$ such that $\phi(m)=n$ is solvable. Does $V(2x)/V(x)\to 2$? Is there an asymptotic formula for $V(x)$?
Pillai [Pi29] proved $V(x)=o(x)$. Erdős [Er35b] proved $V(x)=x(\log x)^{-1+o(1)}$.

The behaviour of $V(x)$ is now almost completely understood. Maier and Pomerance [MaPo88] proved \[V(x)=\frac{x}{\log x}e^{(C+o(1))(\log\log\log x)^2},\] for some explicit constant $C>0$. Ford [Fo98] improved this to \[V(x)\asymp\frac{x}{\log x}e^{C_1(\log\log\log x-\log\log\log\log x)^2+C_2\log\log\log x-C_3\log\log\log\log x}\] for some explicit constants $C_1,C_2,C_3>0$. Unfortunately this falls just short of an asymptotic formula for $V(x)$ and determining whether $V(2x)/V(x)\to 2$.

In [Er79e] Erdős asks further to estimate the number of $n\leq x$ such that the smallest solution to $\phi(m)=n$ satisfies $kx<m\leq (k+1)x$.

See also [417].

Additional thanks to: Kevin Ford