# [seqfan] What I've found so far about natural numbers not sharing digits with their squares

WayneMV waynemv at gmail.com
Tue Oct 16 04:24:57 CEST 2018

```I've been investigating integers that do not share any (base 10) digits in common with their squares, or higher powers. I haven't yet generated an interesting enough (in my opinion) sequences from this yet, but if I come up with any I will submit them. I was working on this myself for a couple weeks before seeing another post in the group recently that mentioned a similar property. My only questions are general ones: What more can be said? And what can be proved? Also, who might want to help me work on optimizing my search methods to be more efficient? (If so, private message me.)

The following is the highest example I have so far found where N has five distinct digits. But my brute force search method wasn't very systematic, so I suspect higher examples could be found with only a little more effort:

N=519686196115698
N^2=270073742433203723023002027204

An example where N has three distinct digits:

N=94249994994992499999999999999
N^2=8883061556556111300100075056061500010010015000000000000001

An example where N has only two distinct digits:

N=44944999944499999999944949499999999999999999944
N^2=2020053020011105003075301510561110605500003025523710256216000000006165656000000000000000003136

My original question would have been whether this can be proven to be any upper limit to how large of N can work. But I started on this I realized that the square of an arbitrarily long series of 3s easily provably only ever contains the digits 1,0,8, and 9:

N  =33333333333333333333333333333333333333333333333
N^2=1111111111111111111111111111111111111111111111088888888888888888888888888888888888888888888889

And the last digit can be 4 or 8:
N=33333333333333333333333333333333333333333333338
N^2=1111111111111111111111111111111111111111111111422222222222222222222222222222222222222222222244

N=33333333333333333333333333333333333333333333334
N^2=1111111111111111111111111111111111111111111111155555555555555555555555555555555555555555555556

Other minor modifications to the series of 3s are likely possible.

QUESTION: Might there be a maximum value that works for N vs N^2 in the cases where N doesn't contain any 3s? (That is, after some point, does every working value of N contain at least one 3? Or might it be shown that some pattern of 2s, 4s, and 9s continues arbitrarily long?)

I also haven't found any regular patterns for cubes, that would guarantee that arbitrarily large values work. The largest cube I have so far found that works is:

N=8989989888898999899898888
N^3=726570247462175445177573435176557560741123542602751424712547043304461763072

For 7th and 11th powers, these may well be the maximum possible values that work. I've not found any higher working values under 5 million.

31333^7=29649259480984782858578077020877
12^11=743008370688

QUESTION: Do any of you know a method for actually proving that conjecture?

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