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I asked this question a few hours ago on MathStackExchange and there it received some attention but we still do not have a proof so I decided to ask it here also, in an unchanged form, and here it is:

Digit sums of numbers $3^m$ in base $10$ for $m=1,2,...,50$ are:

$3,9,9,9,9,18,18,18,27,27,27,18,27,45,36,27,27,45,36,45,27,45,54,54,63,63,81,72,72,63,81,63,72,99,81,81,90,90,81,90,99,90,108,90,99,108,126,117,108,144$.

Ratios $\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$ for $m=1,2,...,49$ to three decimal places are:

$0.333,1.000,1.000,1.000,0.500,1.000,1.000,0.666,1.000,1.000,1.500,0.666,0.600,1.250,1.333,1.000,0.600,1.250,0.800,1.666,0.600,0.833,1.000,0.857,1.000,0.777,1.125,1.000,1.142,0.777,1.285,0.875,0.727,1.222,1.000,0.900,1.000,1.111,0.900,0.909,1.100,0.833,1.200,0.909,0.916,0.857,1.076,1.083,0.750$

Does there exist limit of the sequence $a(m)=\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$?

I cannot resist to note some kind of chebyshevness of this question (if there is one) because we know that Chebyshev proved that if limit in the prime number theorem exists then it must be equal to $1$. It could be that this is also the case here.

I also welcome any computational effort and results obtained from such an experimental work if the proof is out of reach.

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If the limit exists, it of course equals 1. Indeed, $A(m):=a(m)a(m+1)\dots a(2m-1)=ds_{10}(3^m)/ds_{10}(3^{2m})\in [\frac1{9m},9m]$. But if $\lim a(m)\ne 1$, then $A(m)$ is either exponentially large or exponentially small for large $m$.

Probabilistic speculations suggest that that $ds_{10}(3^m)=(\frac92 \log_{10} 3+o(1))m$, that would imply that your limit equals 1. But I am afraid that it is too hard to prove.

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  • $\begingroup$ I think there is a recent ArXiv post by David Radcliffe on digit sums which might resolve this. I'll check it out and report back. Gerhard "From The IRC #Math Days" Paseman, 2017.09.21. $\endgroup$
    – Gerhard Paseman
    Commented Sep 21, 2017 at 20:34
  • $\begingroup$ Is ir reasonable to expect that there will appear some fluctuations that will prevent the existence of limit? $\endgroup$
    – user114642
    Commented Sep 22, 2017 at 1:16
  • $\begingroup$ if the digits are less or more random, than the expectation of having too large or too small sum of digits is exponentially small in $m^2$. Since such a series converges very quickly, Borel--Cantelli heuristics says that we should expect this limit to exist. $\endgroup$
    – Fedor Petrov
    Commented Sep 22, 2017 at 16:49
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Experimentally, the sequence converges to $1,$ at the logarithmic rate suggested in Fedor's answer. Here is the graph for the first 20000 numbers:

enter image description here

Now, when we fit the actual $ds_{10}(3^m),$ we get the following suggestive graph:

enter image description here

Whose slope is pretty close to Fedor's "probabilistic" value. HOWEVER convergence is slow - empirically, Fedor's $o(1)$ term is actually of order $1/\log m.$

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