-4
$\begingroup$

tried to ask on the math reddit but got deleted due to my account being new.

Is the record for highest multiplicative persistence found still 11? As I may have just found a number with persistence of 12.

Feel like it might be a bug, but whenever I run the code on any numbers with known persistence, it always outputs expected result. Wondering if anybody here knows if this is significant, or if others have already done this. Thanks

$\endgroup$
16
  • 6
    $\begingroup$ What is "multiplicative persistence"? Why would it be a bug if running code (what code?) on numbers with known persistence outputs the expected result? \\ Also, MO is for research-level mathematics, not as a replacement for being unable to post to r/math. You might have better luck at our sister site MSE, once you clarify your terms. (Also also—unless you just want to advertise, which is not appropriate here or on MSE—once you define what multiplicative persistence is, you should give the number you think brings it to 12.) $\endgroup$
    – LSpice
    Commented Sep 28, 2023 at 23:54
  • 1
    $\begingroup$ Still no idea if this comment is notifying you as a reply, or just a standard comment, but I pushed enter trying to go to next line earlier, I meant to explain in my original reply, my bad. ^ Edited first comment. $\endgroup$
    – mwt2212
    Commented Sep 28, 2023 at 23:59
  • 2
    $\begingroup$ Multiplicative Persistence is defined here: mathworld.wolfram.com/…. I would be shocked that it has a finite bound, it just appears that it takes very very large numbers to move the persistence needle. In this case the wolfram article suggests that there is no number with less than 233 digits and this individual if they are to be believed says they could not find a number with less than 400 digits: $\endgroup$ Commented Sep 29, 2023 at 0:09
  • 1
    $\begingroup$ Yeah I've seen that post. Number I found has 500 digits ^. Still don't know how to send direct replies @SidharthGhoshal $\endgroup$
    – mwt2212
    Commented Sep 29, 2023 at 0:11
  • 2
    $\begingroup$ @SidharthGhoshal, re, I think the arXiv folks would probably frown on the use of the arXiv as a proof checker / "challenge" site. $\endgroup$
    – LSpice
    Commented Sep 29, 2023 at 0:56

1 Answer 1

12
$\begingroup$

First, to your direct question: "Is the record still 11", the answer is "yes", according to a variety of sources (MathWorld, Wikipedia, OEIS A003001). If anyone has found a new record, at least it is not in these sources yet.

Now for some general advice.

Finding a number that has high multiplicative persistence is a prime example of "hard to find, easy to verify". Given a (say) 500-digit number, it only takes a few lines of code, and an eyeblink of computer time, to compute its persistence.

My advice in such situations is:

If you think you have succeeded in the difficult "find" task, surely you should spend a reasonable effort in the relatively easy "verify" task before announcing your result.

So, if you think you have found such a small number with persistence 12, surely you should first verify your claim with some simple computer code. If that seems to succeed, perhaps double-check using another implementation, perhaps in a different language, perhaps on a different computer. After a reasonable amount of verification succeeds -- OK, perhaps then it is time to think about telling others.

The comments in OEIS claim various lower bounds on the size of a number of persistence 12: a comment in 2019 claims $10^{20585}$, and a comment in 2023 claims $2.67 \times 10^{30000}$. A 500-digit number would contradict both, so all the more reason to double-check.

For completeness, here is straightforward SageMath code to verify such claims. It can be run online in this SageMathCell.

def digits(n):
    d = []
    while n:
        n, remainder = divmod(n, 10)
        d.append(remainder)
    return d
        
def multiplicative_persistence(n):
    i = 0
    print("%d: %d" % (i,n))
    while n>=10:
        n = prod(digits(n))
        i += 1
        print("%d: %d" % (i,n))
    print("It has multiplicative persistence %d\n" % (i))

multiplicative_persistence(277777788888899)
multiplicative_persistence(10^500-1)

The output from these numbers is:

0: 277777788888899
1: 4996238671872
2: 438939648
3: 4478976
4: 338688
5: 27648
6: 2688
7: 768
8: 336
9: 54
10: 20
11: 0
It has multiplicative persistence 11

0: 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
1: 1322070819480806636890455259752144365965422032752148167664920368226828597346704899540778313850608061963909777696872582355950954582100618911865342725257953674027620225198320803878014774228964841274390400117588618041128947815623094438061566173054086674490506178125480344405547054397038895817465368254916136220830268563778582290228416398307887896918556404084898937609373242171846359938695516765018940588109060426089671438864102814350385648747165832010614366132173102768902855220001
2: 0
It has multiplicative persistence 2
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .