Sequence of digits of powers of two

Question

Elementary number theory tells us a lot about the final digits of the powers of two, and ergodic theory (more specifically the theory of equidistribution of points in the orbit of an irrational rotation map) tells us a lot about the initial digits. But when $n$ is large, most digits of $2^n$ are neither initial digits nor final digits. Do we currently have any tools that would let us draw nontrivial conclusions about the statistical properties of the sequence obtained by concatenating the decimal expansions of the successive powers of two (https://oeis.org/A000455)? It’s easy to show that every digit string occurs infinitely often, but do we know (for instance) that asymptotically each decimal digit occurs 1/10 of the time?

In a similar spirit: There is no reason to think that, from some point onward, sequence A000455 agrees with sequence A000796 (the sequence of successive digits of pi), but I do not see a way to disprove this unlikely hypothesis. Can anyone show that the sequences disagree infinitely often?

Answer 1

The digits of power of $2$ follow Benford's law, not just the first digit of it, but a general distribution of all digits.

Leading digits of power of 2 Benford's Law first couple of digits

Therefore we can talk about the arrays of digits with a specific distribution, which are progressing in size.

The number of digits that $2^n$ has is $\left \lceil \log_{10}(2^n)\right \rceil=\left \lceil n \log_{10}(2) \right \rceil$

This means that the total number of digits that concatenating the decimal expansions of $2^x$ has is about

$$\frac{x(x+1)\log_{10}(2)}{2}$$

This all together gives a specific probability that combines the probability of a digit at position $m$ being $n^{th}$ digit of some $2^k$ and then we can argue about having a particular digit at that position.

If the length of the concatenated expansion is $L$ we can find the highest power of $2$ it contains using $$\frac{n(n+1)\log_10(2)}{2}=L$$

Now we have the first digit for each of $n$ powers, we have the second digit for $n-\left \lceil \log_2(10) \right \rceil$, third digit for $n-\left \lceil \log_2(10) \right \rceil - \left \lceil \log_2(100) \right \rceil$ and so on. If we take any digit from the concatenated value with the uniform distribution, this is telling us about the distribution of a digit being the first, the second, the third and so on.

Now combining it with the distribution of digits in we have our desired distribution.

In we can see that the farther we are from the first digit, the distribution becomes all closer to $\frac{1}{10}$. The selection of the first, the second, the third... digit does not affect this conclusion as long as the effect of the position within some power of $2^k$ is infinitesimally smaller than the uniformity of probability of digit being $0$,$1$,$2$,$3$,$4$...,$9$, i.e. being $\frac{1}{10}$

From , the error term of $n^th$ position not conforming to the uniform probability of $\frac{1}{10}$ is about $\frac{1}{10^n}$. Notice that the number of powers with $m$ digits is $3$ or $4$ so this does not affect the final distribution since they all have the same distribution within the same number of digits.

We can then take purely concatenated $1$,$2$,$3$,$4$,$5$... digits of the same distribution without repeating any of it.

It follows, since the expected error, the measure of how much we diverge from the uniform distribution of digits, is then

$$\lim_{n \to \infty}\sum_{k=0}^{n-1}\frac{1}{n-k}\frac{1}{10^k}=0$$

that the distribution of digits is $\frac{1}{10}$ at infinity. Which is to say that we are dealing with the uniform distribution. However, we are reaching it. It is not true that the distribution is uniform for any fixed length.