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Terence Tao in his brilliant book Solving Mathematical Problems: a Personal Perspective states (page 17):

It is highly probable (though not proven!) that the digit-sum of $2^{n}$ is approximately $(4.5 \log_{10} 2)n≈1.355n$ for large $n$.

This problem sounds very interesting to me! Do you know something more about this problem? What is its exact formulation (replacing word approximately with a limit)? Is it still not proven?

I found only this which does not satisfy me enough.

Thank you very much for your answers and have a nice week!

Joseph
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  • it seems inherantly illogical, as n tends towards infinity, the amount of digits increases by more than 10 each time, which means that it will eventually be way more than 1.355, unless there is some dodgy illogical proof – Alex Robinson Dec 05 '16 at 20:28
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    Intuitively, this is just the statement that the digits are uniformly random. After all, the expected value of a randomly selected digit is $\frac 92=4.5$. As $2^n$ has about $n\log_{10}2$ digits, the result follows at once. Of course, it's not even remotely obvious that the digits of $2^n$ are uniformly random. After all, the units place isn't uniform (it can't be odd!) nor can the leading digits be uniform. – lulu Dec 05 '16 at 20:33
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    You can often find more information by looking for the sequence in the OEIS. In this case $2,4,8,7,5,10,11,13$ (e.g., $2+5+6=13$ for $2^8$) leads to https://oeis.org/A001370 which gives a few references. – Barry Cipra Dec 05 '16 at 20:38
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    "[…] He asked whether the number $0.248163264128 ...$ formed of the increasing sequence of powers of 2 is normal to base ten. This question remains open." in Yann Bugeaud, Distribution Modulo One and Diophantine Approximation, §4.7, p. 97. – Watson Dec 05 '16 at 20:38
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    @lulu: I think your comment should be the answer to this question. – Ross Millikan Dec 05 '16 at 21:10
  • @RossMillikan Duly posted. – lulu Dec 05 '16 at 21:17
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    @lulu, it's perhaps worth noting that although the ones digit, which falls into the repeating cycle $2,4,8,6$ of period $4$, is not uniform, the tens digit, which falls into a repeating cycle of length $20$, is uniform (two of each digit). So is the hundreds digit, which falls into a repeating cycle of length $100$, with exactly ten of each digit. I'm not sure if this keeps up, but it's kind of curious. (It also wouldn't hurt if someone doublechecked my counts.) – Barry Cipra Dec 05 '16 at 21:21
  • @BarryCipra Interesting. Surprising...worth a look. – lulu Dec 05 '16 at 21:23
  • Thanks to IBM! You asked my question, Watson. Lulu also helped me to understand the issue. – Joseph Dec 05 '16 at 21:41

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As discussed in the comments:

Informally, this is a statement about the distribution of the digits in $2^n$. If we imagined that they were distributed uniformly, then the claim would follow at once: The average value of a randomly selected digit is $\frac 92=4.5$ and the number of digits in $2^n$ is $\lceil n\log_{10} 2\rceil$.

Of course, it isn't at all clear that this assumption is justified, nor is it clear how one might go about proving it. Indeed, the units place is obviously not uniform (it must be even) nor is the lead digit uniform (the low digits are favored disproportionately, see, e.g., this). Of course, a couple of digits at the ends do not have any material impact on the overall distribution of the units.

Community
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lulu
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  • Just to repeat the gist of my comment below the OP, with a bit more detail, the seqence of digits in the tens position, starting at $4,8,16,32,64$, is $$0,0,1,3,6,2,5,1,2,4,9,9,8,6,3,7,4,8,7,5$$ and then repeats those $20$ terms. Each digit appears exactly twice. Something similar occurs with the $100$-term repeat in the hundreds position, starting at $8,16,32,64,128$. – Barry Cipra Dec 05 '16 at 21:40