3
$\begingroup$

Inspired by this question, is there some conjecture stating that $$ \limsup_{n \to \infty} \frac{d_j(2^n)}{dc(2^n)} = \frac{1}{10} $$ where $d_j(m)$ counts the number of $j$s in the digits of $m$, and $dc(m)$ is just the number of digits in $m$?

Or is this false for obvious reasons? If so, what is this limit? Is the limit same as the $\liminf$?

This can of course be stated in a more general form, where we use another number than 2, and another basis than 10. There are of course some trivial combinations where the corresponding statement is clearly false, for example, if we count binary digits in the above problem (base 2).

This is probably very hard to solve, since it is closely related to normal numbers.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ I'm sure it's true, and I'm sure that far weaker statements are out of reach. So far as I know, there isn't even a proof that every sufficiently large power of 2 has a zero somewhere in its base-10 expression. $\endgroup$
    – Gerry Myerson
    Commented Jun 27, 2014 at 22:55
  • $\begingroup$ I might be missing something simple, but doesn't the fact that the sequence $(2^n)$ follows Benford's law in base $10$ give you some things like this? $\endgroup$
    – Bill Mance
    Commented Jun 28, 2014 at 1:39
  • $\begingroup$ Reference that $2^n$ follows Benford's law? $\endgroup$
    – Gerald Edgar
    Commented Jun 28, 2014 at 2:09
  • $\begingroup$ @BillMance: Benford's law (which is simply the equidistribution of $n\log_{10} 2$ in ${\Bbb R}/{\Bbb Z}$) will tell you that given any initial pattern of digits there are many $n\le N$ with $2^n$ beginning with that pattern. It does not say anything about the decimal digits of $2^n$ for every value of $n$, which is the problem raised here (and in Gerry Myerson's comment). $\endgroup$
    – Lucia
    Commented Jun 28, 2014 at 2:50
  • $\begingroup$ @Lucia oops I see. I misread it as "there are infinitely many powers of 2 with a zero in their base 10 expansion". GeraldEdgar: Lucia answered your question, but I want to add further this useful resource benfordonline.net $\endgroup$
    – Bill Mance
    Commented Jun 28, 2014 at 3:29

1 Answer 1

Reset to default
2
$\begingroup$

At least as of 2012 this should still be an open question. If it were known to be true, then the number

$$ .2\ 4\ 8\ 16\ 32\ 64\ \cdots $$

would be known to be simply normal in base $10$ (of course a strengthening would give normality in base $10$). Pillai conjectured that this number is normal in 1939 and according to Yann Bugeaud's book "Distribution modulo one and diophantine approximation", published in 2012, it is still open.

Improve this answer
$\endgroup$

You must log in to answer this question.

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