First digits of the iterated powers of 2

Question

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?

Answer 1

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 $m$, $$k\cdot 10^m\le 2^{2^j}<(k+1)\cdot10^m$$ $$m+\log_{10}(k)\le 2^j\log_{10} 2<m+\log_{10}(k+1)$$ $$\log_{10}(k)\le \{2^j\log_{10}(2)\}<\log_{10}(k+1)$$ 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.

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Useful links

• Analysis of the leading digits of the sequence $(2^j)_j$, by Numberphile.
• Your sequence is A364837 on OEIS.
• This is a related question.

Answer 2

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:

$log_{10}(2) = 0.30102999..._{10} = 0.010011010001000001001101..._2$

$log_{10}(2^{2^7}) = 128 \log_{10}(2) = 100110.10001000001001101..._2$

When multiplying by a power of 2, in binary we are simply shifting bits.

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.