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If I convert a very large power of 2 into decimal, the digits looks pretty random. For example, count the decimal digits of a very large $2^{10^6}$, we have:

1 - 30354

2 - 30047

3 - 30193

4 - 30230

5 - 30174

6 - 30103

7 - 29840

8 - 29096

9 - 30007

0 - 30186

That's a high likehood for a sample from uniform distribution.

What would be an explanation for this phenomenon? Can we prove that, when integer $n \to \infty$, the decimal digit frequency of $2^n$ converge to uniform distribution?

user1950580
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    It is easy enough to show that the fractional part of $\log_{10}(2^n)$ tends to uniformly distributed, but extending that to an average over all the digits doesn't look entirely trivial. – hmakholm left over Monica May 16 '17 at 08:33
  • have you tried any other tests of randomness? Such as the run test? You could test for runs of odd and even digits. – Cato May 16 '17 at 09:23
  • Besides the sure question that is complex, I just read it 3 times and I can not understand anything ... if you make an attempt to explain it only with numbers, there is a remote possibility that can bring you something. – Guillemus Callelus May 16 '17 at 11:14
  • @GuillemusCallelus It's extremely simple, write down $2^{10^6} = 9900656229 ...$, you will notice there's almost equal chance for each digit in 0-9 to appear. The question is why this happens. However I would expect a proof being far more complicated. – user1950580 May 16 '17 at 11:50
  • The calculations you refer to exceed the capacity of my programs (the Maxima software obviously gives me overflow)andof course that of an "extremely simple" calculator. Giving for certain the data that you provide, roughly, you would say that the greater the degree of the exponent ... greater manipulation and RANDOM SENSATION there is (mainly because of the huge amounts of figures and the difficulty of its verification) ... As you can see and feel very much, I am far from being able to affirm or deny whether it is a uniform distribution or not when it tends to infinity and much less its proof. – Guillemus Callelus May 16 '17 at 13:03
  • @GuillemusCallelus Just for providing reference, it's provable the digits of $\pi$ in decimal would be uniform due to $\pi$ being a normal number. Since we are generating digit sequence from a deterministic algorithm which can be ran on a Turning Machine (think about concat all $2^n$ digits), the similar definition can be used. – user1950580 May 16 '17 at 13:17
  • What an interesting question! Also interesting: for lots of n, some digits of 2^n appear the same number of time, more often than would happen for a random sequence of $0,\ldots,9$ of similar length. – orangeskid Jan 05 '24 at 18:27
  • Maybe when $m$ tends to infinity then the decimal digit frequency of most $m$-digit numbers is close to uniform? – Alex Ravsky Jan 05 '24 at 18:48
  • I'd guess it has something to do with the fact that the set of $2^n$ is dense in the 5-adics, so you're necessarily filling out digits in base 5 which ends up filling out digits in base 10 after combining with the Chinese remainder theorem. – Merosity Jan 05 '24 at 19:37
  • I think this is an application of (generalized) Benford's law. That is, the first few digits follow Benford's law which rapidly converges to uniform for digits after the third place. With so many digits in your number it appears uniform by simply counting. You'd see this behavior if you just counted the appearances of numbers in say the fifth or sixth spot – Gregory Jan 05 '24 at 19:46
  • @Merosity:$10=2\cdot 5$, that might be exploited . but this seems to work for all power sequences $(a^n)_n$, and bases $b$ (provided $\log_b a$ is not rational). – orangeskid Jan 05 '24 at 23:24
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    The question asks for an explanation where none is required. It would be a non-random-looking pattern in the digits that would require an explanation. If you pick a uniformly random number with $n$ digits, you get $n$ independent uniformly random digits. Unless there's some reason to expect a pattern in the digits, one would expect the digits of powers of $2$ to be just like the digits of any other number, and thus to exhibit the statistical features that uniformly randomly drawn numbers exhibit. So there's no there there. – joriki Jan 05 '24 at 23:33
  • @joriki It OP's question, after the question about an explanation, there is a question about a proof. Although the decimal digit frequency of powers of $2$ seems to be generic, there remains a problem to prove this. This reminds me the situation with normal numbers: "While a general proof can be given that almost all real numbers are normal (meaning that the set of non-normal numbers has Lebesgue measure zero), this proof is not constructive, and only a few specific numbers have been shown to be normal". – Alex Ravsky Jan 06 '24 at 04:20
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    @AlexRavsky: Yes, the question about a proof, while very ambitious, is a real question that could potentially be answered (though I personally don't know of any other example of such a fundamental innovation originating on math.SE). I was only commenting on the question about an explanation. – joriki Jan 06 '24 at 08:41
  • Related: https://math.stackexchange.com/questions/1902684/distribution-of-digits-in-powers-of-2 – leonbloy Jan 08 '24 at 17:54

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