For the convenience of MO readers, here is a transcript of James Davis's letter to Conway, as reported in the Numberphile video mentioned in James Davis's answer.

I've enjoyed your popular work for years, even as a non-mathematician. Thanks for making life slightly more enjoyable.

Concerning the attached problem 5, your conjecture that "every number eventually climbs to a prime" is incorrect. $13532385396179$ factors to $13 \cdot 53^2 \cdot 3853 \cdot 96179$ and thus climbs to itself.

Other numbers (obviously) climb to $13532385396179$ and "terminate" on this composite also. ($13^{532385396179}$, or $13 \cdot 53238539^{6179}$ etc).

I arrived at this counter-example after first being heartened that some 2-step cycles, $f(f(n)) = n$ occur quickly when the problem is done in base 2 or base 4, and $3^3$ maps to itself in base 8.

In searching for a number satisfying $f(n) = n$ in base 10, I restricted myself to numbers where the largest prime factor was to the first power, eventually simplifying it to looking for an $x$ satisfying $x\cdot p = f(x)\cdot 10^y + p$ where $p$ is the largest prime factor of $n$. $f(x)$ should be smaller than $x$ and constrains $y$. $n = x\cdot p = f(x)\cdot 10^y + p$

The rub is that $f(x)/(x-1)$ must terminate and have a prime number decimal expansion. But this doesn't require $f(x)$ to be prime or $(x-1)$ to be a power of ten, since they can have common factors(!) Requiring that $(x-1)$ has only factors of $10$ and common factors with $f(x)$ motivates looking for $x$ of the form $x=m\cdot 10^y+1$, where $y$ should probably be around $\lceil \log_{10}(m)\rceil$. With $y=5$, astonishingly, $m=1407$ was found almost instantly! Then $x=140700001$, $f(x)=135323853$, $p=f(x)/(x-1) \cdot 10^5 = 961789$, and $n=13532385396179$. So $f(13532385396179) = 13532385396179$

Thanks again for the problem, I've certainly enjoyed it (despite spending an embarrasingly long time playing with it).

For the convenience of MO readers, here is a transcript of James Davis's letter to Conway, as reported in the Numberphile video mentioned in James Davis's answer.

I've enjoyed your popular work for years, even as a non-mathematician. Thanks for making life slightly more enjoyable.

Concerning the attached problem 5, your conjecture that "every number eventually climbs to a prime" is incorrect. $13532385396179$ factors to $13 \cdot 53^2 \cdot 3853 \cdot 96179$ and thus climbs to itself.

Other numbers (obviously) climb to $13532385396179$ and "terminate" on this composite also. ($13^{532385396179}$, or $13 \cdot 53238539^{6179}$ etc).

I arrived at this counter-example after first being heartened that some 2-step cycles, $f(f(n)) = n$ occur quickly when the problem is done in base 2 or base 4, and $3^3$ maps to itself in base 8.

In searching for a number satisfying $f(n) = n$ in base 10, I restricted myself to numbers where the largest prime factor was to the first power, eventually simplifying it to looking for an $x$ satisfying $x\cdot p = f(x)\cdot 10^y + p$ where $p$ is the largest prime factor of $n$. $f(x)$ should be smaller than $x$ and constrains $y$. $n = x\cdot p = f(x)\cdot 10^y + p$

The rub is that $f(x)/(x-1)$ must terminate and have a prime number decimal expansion. But this doesn't require $f(x)$ to be prime or $(x-1)$ to be a power of ten, since they can have common factors(!) Requiring that $(x-1)$ has only factors of $10$ and common factors with $f(x)$ motivates looking for $x$ of the form $x=m\cdot 10^y+1$, where $y$ should probably be around $\lceil \log_{10}(m)\rceil$. With $y=5$, astonishingly, $m=1407$ was found almost instantly! Then $x=140700001$, $f(x)=135323853$, $p=f(x)/(x-1) \cdot 10^5 = 961789$, and $n=13532385396179$. So $f(13532385396179) = 13532385396179$

Thanks again for the problem, I've certainly enjoyed it (despite spending an embarrassingly long time playing with it).