But you know that'll fail?

Yes. I believed I was eventually going to have to inject additional 1s into the 11-stepper to generate enough single-digit factors to get at least 233 digits in the 12-stepper. And I knew the chances were that the first 11-stepper, even with 1s, might not ever be able to generate a 12-stepper. Since the conjecture is that there are no 12-steppers -- though there isn't a solid mathematical reason, other than "233 digits wasn't enough for 12-step, when 15 digits was sufficient for 11-step, so it's not likely to have any 12-steppers" -- the chances are small than any amateur (like us me) will find a 12-stepper. (Though mathematics does have some strange outliers, like the Monster Group being so far out there, or the sixth platonic-solid-analog in 4d, when all other dimensions have fewer.)

update: I am an amateur at mathematics. I should not have spoken for anyone else. Sorry, LanX.

I have a math degree and know if many cases even nowadays were "amateurs" find good solutions.

Don't wanna go to much into details but in order to calculate the likelihood of a solution (which is not alien to number theory). You'd need to calculate the density of possible products of single digits numbers in a number range.(easily done with the sieve approach)

Since the number of 11 step solutions becomes infinite by just adding more 1s it's probably not that unlikely to find a solution with several hundreds or thousands digits.*

Otherwise you'd need to prove why it's impossible. ( Which could be done by showing that the density becomes 0)

Cheers Rolf
_{(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery FootballPerl is like chess, only without the dice}

update

*) or at least prove it's existence.

cases even nowadays where "amateurs" find good solutions.

Good point. Even Numberphile and related YouTube channels have highlighted the answer-hidden-in-the-comments on superpermutations (4)(3)(2)(1). And as @standupmaths likes recommending, give it a go, and make a "Parker Square" (5)(6) of it.

Since the number of 11 step solutions becomes infinite by just adding more 1s it's probably not that unlikely to find a solution with several hundreds or thousands digits.*

*: given enough computing power. :-) I mean, adding one 1 will multiply the number of permutations by 16, and from there it will keep on increasing by factors of (15+i)/i . (edit: remove stray factorial, which belied my statement)

You'd need to calculate the density of possible products of single digits numbers in a number range.(easily done with the sieve approach)

...

Otherwise you'd need to prove why it's impossible. ( Which could be done by showing that the density becomes 0)

Hmm. I'll have to think about this some more. I can imagine how I'd start generating the potentials -- 2**$i * 3**$j * 7**$k -- but without generating all the potentials below a threshold, I cannot see any way to generate just potentials of say 200-300 digits. (for any, using the iterators from Higher Order Perl -- section 6.4 especially -- would be how I would start... and brings this reply back into being remotely perl-related :-) ).

Since the number of 11 step solutions becomes infinite by just adding more 1s

I'm also wondering whether there are other base n=11 solutions that aren't permutations of 277777788888899 and however many 1s you insert. A003001 only lists 277777788888899 (since that's the only minimal one), and the other sites I've seen only list that and some of its permutations -- unless I've just not noticed one that looks similar enough to that one that I didn't notice it was separate. OEIS has Numbers with multiplicative persistence value lists for 1-9, but doesn't list the persistence 10 or 11 numbers. But A003001 has a table that (if I understand it correctly) implies there's a second family of persistence=11.

edit 2: found it. A046150 shows 99999999998777772, which would be the family 27777789999999999.