TITLE: Does this sequence have a limit?
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: So I programmed the different plots for base 3-10 for 1000 points, and for almost all of them you can see that it basically converges to 1, but in base 9 it has extremely weird behavior. This may be due to my programming, but it does seem to suggest that in base 9 it doesn't converge to 1.
The plots can be seen here: https://imgur.com/a/QvNlA.
The code can be seen here: https://gist.github.com/anonymous/a4888cc09178dc8a967596e085dbd165
EDIT:
So I have an idea that it seems to do this oscillation when the basis is some power of the sequence number. So for example, if we consider the sequence $\frac{ds_{4}(2^m)}{ds_{4}(2^{m+1})}$ (here the base is $2^2$), we see the oscillation, and for the sequence $\frac{ds_{8}(2^m)}{ds_{8}(2^{m+1})}$ (here the base is $2^3$), we can see the oscillation again. Below are the sequences respectively:  

EDIT 2:
We can in fact show this oscillation. Notice that we can rewrite $3^m = a_0 + a_1 3^j + a_2 3^{2j} + \cdots + a_n 3^{nj}$. We can then rewrite our $ds_{3^j}$ function to be $ds_{3^j}(3^m) = \sum_{i=0}^n a_i$.
So let's think about the case for $3^2$, or base 9. Let $\mu(x) = ds_{3^2}(x)$ for notational simplicity. Then we will go through some examples. We have then $3^1 = 3$ base 9, so $\mu(3) = 3$. Then $3^2 = 1 * 3^2$, so we have $\mu(9) = 1$. Notice $3^3 = 3 * 3^2$, and so $\mu(27) = 3$. So now we start to generalize. If $m$ is even, then we have that $\mu(3^m) = 1$, and if $m$ is odd then we have $\mu(3^m) = 3$. This is pretty clear to see (notice that if it's even then we have that then we have $(3^2)^k$ and so we have $a_k = 1$ and $a_0, \ldots, a_{k-1} = 0$, and if its odd then we have it's of the form $3 * (3^2)^k$ and so we have $a_k = 3$ and $a_0, \ldots, a_{k-1} = 0$) but it gives us this nice oscillation between $1/3$ and $3$, since if $\mu(3^m)$ is even then we have $\mu(3^{m+1})$ is odd and vice versa. We can in fact show that this will hold for any power, and that if we have $3^{n+1}$ as our base then our number will remain at $1/3$ for $n$ steps before shifting to $3^n$. This is something that we see in the case of the series $2^m$ in base 8; notice that in the second graph I posted it seems to linger at $1/2$ for two steps before shifting up to $4$.
Moreover, we can notice that if our base is of the form $3^j$ for $j >1$, then in fact your sequence does not converge.
EDIT 3: Asking a friend, he says that it is impossible for this to converge to 1 in any base since there will always be jumps down to 0. I don't have a formal proof for this, but this makes sense.