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

[M. F. Hasler]

On Mon, Oct 15, 2018 at 10:25 PM WayneMV <waynemv at gmail.com> wrote:

> I've been investigating integers that do not share any (base 10) digits in
> common with their squares, or higher powers.


reference : oeis.org/A029783

The following is the highest example I have so far found where N has five
> distinct digits.
> N=519686196115698, N^2=270073742433203723023002027204


I think you could submit the sequence(s) of such solutions with 5 or even 4
distinct digits,
since there are (?wild guess?) only sporadic solutions. (I didn't really
look closer at this.)

An example where N has three distinct digits:
> N=94249994994992499999999999999 (...)


There are simple infinite three and two digit sequences of the form
s(n;a,b) = a*R(n)+b
(where R(n) = (10^n-1)/9 is the repunit of length n)
e.g.: 3...34  and 3...38 and 6...67 (for 2 digits), and
6...659, 6...672 and 6....673 for 3 digits.


> 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?)


The examples with 6's above show the contrary, e.g.,
( 6*R(n)+1 )^2 = 4*R(2n) + 4*R(n) + 1 for all n >= 1.

I also haven't found any regular patterns for cubes, that would guarantee
> that arbitrarily large values work.


reference : oeis.org/A029785


> The largest cube I have so far found that works is:
> N=8989989888898999899898888


I found the 29 digit term 80800000008880080808880080088 using PARI :
is(n)=!#setintersect(Set(digits(n)),Set(digits(n^3)))
{for(n=1,oo,print1("/*"n+1" digits*/"); P=vector(n+1,k,10^k)*8\10;
for(L=max(n\2-2,2),n\2+1,
forvec(v=vector(L,i,[1+(i==L)*n,1+(i>1)*n]),is(vecsum(vecextract(P,v)))||next;
print1(vecsum(vecextract(P,v))",");next(3),2)))}

This searches for n-digit numbers with digit '8' in L locations (here L
between n/2-2 and n/2+1) among which the first and last digit, and 0's
elsewhere. The "next(3)" skips to n+1 when a term has been found.
To search for numbers with digits 0,4 and 8 you can multiply the base-3
digits by 4:
for(n=1,oo, is(t=fromdigits(digits(n,3)*4)) && print1(t","))
This approach, easily generalized to other cases, has the drawback that you
cannot restrict the numbers created to have a minimum number of digits 0.


> 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?


I don't. It could be difficult, since simpler similar questions are still
open.

Maximilian