[seqfan] Re: An arithmetic conjecture

David Wilson dwilson at gambitcomm.com
Tue Mar 17 15:46:44 CET 2009

Statistically, you can argue that the number of powers of 2 having no 
base-10 digit 0 is finite. This argument rests on the conjecture that 
the base-10 digits of large powers of 2 are more or less uniformly 
distributed, which is borne out empirically. For very large powers of 2, 
with n digits, the expected number of zeroes will be n/10 with a fairly 
narrow variance, and the probability of no zeroes will be vanishingly small.

Dan Hoey appears to have tested powers of 2 up to around 750 million 
digits. Such a number should have about 75 million zeroes in it, the 
likelihood it would have none is incomprehensibly small. Indeed, if we 
compute the likelihood that this or any larger power of 2 omit a zero, 
the probability is unfathomably tiny. So while we do not have 
mathematical certainty that 2^86 is the largest power of 2 without a 
zero, we have the next best thing. Conway called such a statement a 

The statement

For all m > 26 there exists some k > 0 such that floor(2^m / 3^k) = 3 (mod 6).

is similar. It can be restated as

For m > 26, 2^m, written in base 3, has the form x0y where x is an even 
base-3 number and y is any base-3 string.

It's slightly different, but the same reasoning applies to conclude that 
the number of 2^m not of the desired form is finite. If you test all 
powers of 2 up to, say, 1000 base-3 digits and find no exceptions with m 
 > 26, you can be extremely confident that there is no such m (I'm 
assuming 26 is accurate).

I have a conjecture along these lines:

If two bases a >= 2 and b >= 2 and two sets of integers A and B where

    a and b are not powers of the same integer (e.g, a = 4, b = 8 is 
    A and B are infinite,
    A and B have limit density 0 over the integers,
    A is a-automatic (the base-a representations of the elements of A 
form a regular language) and B is b-automatic.

Then A and B have finite intersection.


Let a = 2, b = 10, A = powers of 2, B = numbers with no 0 in their 
base-10 numerals.

This example easily conforms to the first three conditions.
A is 2-automatic, with its base-2 numerals forming the regular language 10*.
B is 10-automatic, with is base-10 numerals forming the regular language 

My conjecture implies that A and B have finite intersection, that is, 
there are a finite number of powers of 2 without zeroes in their base-10 

My conjecture also implies your conjecture.

I understand Knuth's original intent of separating content from 
presentation, that authors should be concerned with content, publishers 
with format. I think this model was generally accurate when Knuth 
authored TeX, however, it is becoming less and less true. The age of 
paper is coming to an end. Modern editing programs allow authors to 
control both content and presentation of their work, and publishers 
rarely sully their hands with typesetting, they demand photo-ready 
documents from the author.

I really think LaTeX should get with the times. If, say, OpenOffice.org 
understood LaTeX, I would probably long ago have submitted several 
papers to the JIS in the required format.

Jim Nastos wrote:
> On Mon, Mar 16, 2009 at 9:20 AM, Rainer Rosenthal <r.rosenthal at web.de> wrote:
>> David Wilson wrote:
>>> DW> To my knowledge, no one knows how to prove your statement.
>>> DW> Without getting heavily into it, this problem belongs to a class
>>> of problems
>>> DW> with problems like:
>>> DW> Does every sufficiently large power of 2 include the digit 0 in base 10?
>> I'd like to ask my preferred book UPINT, but I am not sure
>> where I could find an answer (or at least related questions).
>   The Google Books copy of UPINT has the section "F24: Some Decimal
> Digit Problems:"
> where it says: "The decimal rep of 2^n contains no zero digit for
> n=1,2,3,4,5,6,7,8,9,13,14,15,16,18,19, ... 86.
> Shall we ever know if this sequence is complete? Dan Hoey has checked
> that there are no others with n <= 2.5billion "
> JN
>> The actual problem, posed by Peter Luschny, was: what can
>> be said about the remainders of floor(2^m/3^k) modulo 6?
>> His problem restated:
>>    ================= Conjecture ===================
>>    For all m > 26 there exists some k > 0 such that
>>             floor(2^m / 3^k) = 3 (mod 6).
>>    ================================================
>> Could someone please be so kind as to give me a pointer
>> to UPINT (Unsolved Problems in Number Theory)? I have
>> the third edition available at home.
>> Best regards,
>> Rainer
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