Implications of the disproof of the "climb-to-a-prime" conjecture

Question

Now that James Davis has found a counter example, 13532385396179, to John Conway's climb-to-a-prime conjecture, I would be interested to learn whether this has any implications of interest in number theory.

Answer 1

Hans said that Conway's point in asking it was that there exist problems easy to state but impossible to prove. The point I took away was that there exist problems that look so hard, nobody has tried anything easy.

Read the letter to Conway on Numberphile if you want to see how it was easy. I'll ask Conway if it has any other implications (if I get the chance). I kinda doubt it, but who knows. The idea isn't really limited to base 10. The problem and short search that worked generalizes to other bases. I worked in smaller bases at first.

Edit: If anyone does think of a mathematical usefulness to this, please let me know!

Answer 2

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).