Posted by: Gary Ernest Davis on: April 23, 2011

An old problem asks if is the highest power of whose base representation: , does *not*

For instance, the next powers of do indeed contain the digit :

Instead of asking the existential question “Is there a zero in the base representation of for ?” we can ask the more quantitative question: “How many zeros are there in the base representation of .

Here are plots of the number of zeros in the base representation of for up to respectively:

We see, progressively, what seems to be oscillations around a straight line.

What could this straight line be?

And how large are the oscillations?

The number of digits of isÂ , where is the integer part of .

Neither the first nor last digit of is , so there are digits of that can be .

If we assume that all digits are equally distributed among these places then we expect to see about zeros in base representation of for large enough

So we take as a linear model for the number of zeros in base representation of the quantity

To get a visual feel for how this fits let’s plot the actual number of zeros along with this linear model:

We now have a statistical model for the number of zeros, in the base representation of :

where we will refer to as the “error” (meaning how much differs from the linear model).

How does the error behave?

In particular, how is it distributed?

Below is a histogram of the error for , together with a normal curve of the same mean and standard deviation:

We can see the the empirical distribution of the error is “peakier” than the normal distribution, and this is borne out by the kurtosis which is , substantially greater than a kurtosis of for the normal distribution.

Plotting these histograms for various values of indicates that the variance of the distribution of the errors varies linearly with , or, what amounts to the same thing, the standard deviation varies with .

What we really seek is a bound on the errors sufficient to show that for large enough .

What does computation suggest?

Below is a plot of the size of the error divided by .

It appears that even for quite small :

If this does hold good then

So now we have a more precise statement: prove that for .

Maybe techniques of higher order Fourier analysis (Higher-Order_Fourier_Tao) could be brought to bear on this problem since they are useful in proving other equidistribution theorems? Then again, just because Weyl’s polynomial equidistributionÂ theorem can be interpreted as a higher-order Fourier analysis application may have nothing at all to do with equidistribution of digits in .

Tags: equidistribution, powers of 2

2^22 ends in …04.

@Xan: “The tens digit can also never by 0.”

2^23 = 8,388,608

Did I misinterpret what you said?

1 | Xan Gregg

April 24, 2011 at 1:28 pm

The tens digit can also never by 0. (You can reason it out or just look at the 2^n mod 100 cycle.) I don’t think it affects the slope of your model, but it would affect the intercept. Maybe that would make your histogram be better centered on the normal curve.