# First digits of the iterated powers of 2

I wanted to show that the first digits of $$(2^{2^j})_{j=1}^\infty$$ are not periodic. By the standard Dirichlet trick I can show that any array of digits forms the initial digits of some number of the form $$2^m$$. Is there a way to deduce the former fact from the latter?

The sequence of leading digits of the sequence $$(2^{2^j})_j$$ is probably not periodic, as suggested by Benford's Law.

Notice the first digit of $$2^{2^j}$$ is $$k$$ if, and only if, for some natural number $$m$$, $$k\cdot 10^m\le 2^{2^j}<(k+1)\cdot10^m$$ $$m+\log_{10}(k)\le 2^j\log_{10} 2 $$\log_{10}(k)\le \{2^j\log_{10}(2)\}<\log_{10}(k+1)$$ where $$\{x\}$$ denotes the fractional part of the real number $$x$$.

It is known that the sequence $$(\{b^j\alpha\})_j$$ is equidistributed over $$[0, 1]$$ if, and only if, $$\alpha$$ is normal in base $$b$$. It is also known that the set of real numbers that are not normal in base $$b$$ has measure zero. Therefore, $$\log_{10}(2)$$ is probably normal in base $$2$$ and the sequence $$(2^j\log_{10}(2))_j$$ is probably equidistributed over $$[0, 1]$$ (be aware that showing a number is normal is very hard).

If this is the case, the digit $$k$$ appears as the leading digit of $$2^{2^j}$$ with the frequency equal to the length of the interval $$[\log_{10}(k),~\log_{10}(k+1)]$$, i.e., $$\log_{10}(k+1)-\log_{10}(k) = \log_{10}\left(1+\dfrac1k\right)$$, which is irrational. Of course, if the sequence were periodic, the frequency of every digit should be rational.

If we compute the decimal logarithm of $$2^{2^n}$$, or $$2^n \log_{10}(2)$$, and convert to binary digits, we find that the first digit of $$2^{2^n}$$ sequence is directly linked with the binary representation of $$\log_{10}(2)$$. For example:
$$log_{10}(2) = 0.30102999..._{10} = 0.010011010001000001001101..._2$$
$$log_{10}(2^{2^7}) = 128 \log_{10}(2) = 100110.10001000001001101..._2$$
Since $$\log_{10}(2)$$ can be shown to be irrational quite easily, its binary expression can never repeat. If the sequence were periodic, then it would suggest that the binary representation of $$\log_{10}(2)$$ is periodic as well.