The script successfully finds the 15-digit value with 11 steps but I have yet to find a 12 stepper, having run with up to 27-digit values. Output from a 26- and 27-digit run below:-
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As you can see, the above run finds all the steps up to 11, just with a series of 1s prepended and the whole run took almost 9 hours on a 2012 MacBook Pro 2.3GHz quad core i7. It may be that there are no 12-steppers at all, I don't have the maths to tell,...
It was mentioned in the video, and I am quoting the similar idea from the Wolfram MathWorld Multiplicative Persistence article: "There is no number <10^(233) with multiplicative persistence >11". You're going to have to go a lot higher than 27 digits if you want to find the elusive 12-stepper.
I tried talking the lowest 11-stepper (277777788888899), then permuting its digits, and listing the factors of each of those permutations (keeping the single-digit factors separate, then lumping what's left if it's not), trying to find one or more permutations that is soley made up of single-digit factors -- because if there's a group of only-single-digit factors that make up a 11-stepper, then making a 12-stepper is as simple as concatenating those digits. -- Actually, I remembered that I started with a 10-stepper, because I wanted to see if I could proof-of-concept it to go from the 10-stepper to a known 11-stepper. I only made it about a million permutations through. If I had started with the 11-stepper, that would have been almost enough, because there are only 15! / 6! / 6! / 2! permutations of 277777788888899, which is 1.3million permutations. But since I was using the 10-stepper 4996238671872 => 1223466778899, which has fewer repeating digits, so is 13!/2!/2!/2!/2!/2! = 195million permutations.
Looks like I'll have to find some spare CPU cycles to try the 11-stepper, too.
