# Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?

For any $m\in\mathbb N$, let $S(m)$ be the digit sum of $m$ in the decimal system.

For example, $S(1234)=1+2+3+4=10, S(2^5)=S(32)=5$.

Question 1 :Is the following true? $$\lim_{n\to\infty}S(3^n)=\infty.$$

Question 2 :How about $S(m^n)$ for $m\ge 4$ except some trivial cases?

Remark : This question has been asked previously on math.SE without receiving any complete answers, where spin proved that $\lim_{n\to\infty} \sup S(m^n )=\infty$ when $m$ is not a power of $10$.

https://math.stackexchange.com/questions/501019/letting-sm-be-the-digit-sum-of-m-then-lim-n-to-inftys3n-infty

Motivation : I've got the following : $$\lim_{n\to\infty}S(2^n)=\infty.$$

Proof : The point of this proof is that there exists a non-zero number between the ${m+1}^{th}$ digit and ${4m}^{th}$ digit.

If $$2^n=A\cdot{10}^{4m}+B, B\lt {10}^m, 0\lt A,$$ then $2^n\ge {10}^{4m}\gt 2^{4m}$ leads $n\gt 4m$. Hence, the left side can be divided by $2^{4m}$. Also, $B$ must be divided by $2^{4m}$ because ${10}^{4m}=2^{4m}\cdot 5^{4m}$. However, since $$B\lt {10}^m\lt {16}^m=2^{4m},$$ $B$ can not be divided by $2^{4m}$ if $B\not=0$. If $B=0$, then the right side can be divided by $5$ but the left side cannot be divided by $5$. Hence, we now know that there is a non-zero number between the ${m+1}^{th}$ digit and ${4m}^{th}$ digit. Since $2^n$ cannot be divided by $5$, the first digit is not $0$. There exists non-zero number between the second digit and the fourth digit. Again, there exists non-zero number between $5^{th}$ digit and ${16}^{th}$ digit. By the same argument as above, if $2^n$ has more than $4^k$ digits, then $S(n)\ge {k+1}$. Hence, $$n\log {2}\ge 4^k-1\ \ \Rightarrow \ \ S(n)\ge k+1.$$ Now we know that $$\lim_{n\to\infty}S(2^n)=\infty$$ as desired. Now the proof is completed.

However, I've been facing difficulty for the $m=3$ case. I've got $\lim\sup S(3^n)=\infty$.

Proof : Suppose that $3^n$ has $m$ digits. Letting $l=\varphi({10}^m)+n$, then $$3^l-3^n=3^n(3^{\varphi({10}^m)}-1).$$ Since this can be divided by ${10}^m$, we know that the last $m$ digits of $3^l$ are equal to those of $3^n$. Hence, we get $\lim\sup S(3^n)=\infty$.

However, I can't get $\lim_{n\to\infty}\inf S(3^n)$. Can anyone help?

• I like your argument. Note you can replace 4m by ceil(mlog_2 10), giving a slightly tighter location of nonzero digits. For 3, maybe you can argue that a large string of 0's near the least significant bits gives that many nonzeros further up. – The Masked Avenger Sep 26 '13 at 15:28
• @TheMaskedAvenger: Thank you for pointing it out. – mathlove Sep 26 '13 at 15:35
• Btw this sequence is A004166 in the OEIS oeis.org/A004166. It has no much information however; maybe you could add there the link to this question... – Pietro Majer Sep 26 '13 at 18:40
• See this paper of Blecksmith, Filaseta, and Nicol for some closely related results: math.sc.edu/~filaseta/papers/blockpaper.pdf – so-called friend Don Sep 27 '13 at 1:01

This follows from W. M. Schmidt's Subspace theorem, which is a deep theorem in diophantine approximations generalizing Roth's to several variables. A full account of this theorem and its proof, as well as some of its striking applications, can be found in chapter 7 of Heights in Diophantine Geometry by Bombieri and Gubler. The following result, the finiteness of the number of non-degenerate solutions to the so-called "$S$-unit equation," is a straightforward application of Schmidt's theorem. (See Theorem 7.4.2 in [HIDG]):

Let $S$ be a finite set of prime numbers, and fix $n \in \mathbb{N}$. Consider $\mathcal{X}$ the set of solutions to $x_1 + \cdots + x_n = 1$ in rational numbers $x_i$ of the form $\pm \prod_{p \in S} p^{a_p}$, $a_p \in \mathbb{Z}$, such that no proper subsum of $x_1+\cdots+x_n$ vanishes. Then $\mathcal{X}$ is a finite set.

This implies your question immediately upon considering $S := \{2,3,5,7\}$.

However, the proof of the Subspace theorem is not effective, and this only shows $S(3^n) \to +\infty$ without any lower estimate on the rate of growth. An effective lower bound on $S(3^n)$ (going to infinity with $n$) is available through Baker's method; it is due to Stewart, and the google search led me to the old MathOverflow post linked to in my comment below. I will just copy the references which Gerry Myerson supplied there:

C. L. Stewart, On the representation of an integer in two different bases, J Reine Angew Math 319 (1980) 63-72, MR 81j:10012; H G Senge, E G Straus, PV-numbers and sets of multiplicity, Proceedings of the Washington State University Conference on Number Theory (1971) 55-67, MR 47 #8452.

• You are welcome! And actually, upon looking at a paper of Stewart from 1980, I realize that I was wrong about effectivity: an explicit lower bound is actually possible with Baker's method. I will edit. – Vesselin Dimitrov Sep 26 '13 at 15:54
• Actually, see here: mathoverflow.net/questions/38971/… – Vesselin Dimitrov Sep 26 '13 at 15:56
• Again, thank you for great information. – mathlove Sep 26 '13 at 16:23
• The only problem with copying my references is that you copied my typo. I can't fix it in the comments at 38971, but I did fix it here. Incidentally, Senge & Straus also published their result in another paper of the same title, Period. Math. Hungar. 3 (1973) 93–100, MR0340185 (49 #4941). – Gerry Myerson Sep 26 '13 at 23:26
• @WillSawin: Your are right, it is enough to take $S = \{2,3,5\}$. (I was thinking of accomodating the prime factors of every digit in $\{1,\ldots,9\}$, but this is unnecessary since e.g. we may split $7 \cdot 10^k \cdot 3^{-m}$ into $7$ new $\{2,3,5\}$-unit variables $= 10^k \cdot 3^{-m}$.) – Vesselin Dimitrov Jan 11 '14 at 19:52

There is more simple agrument which solves the problem: powers $m^n$ can start with arbitrary string of digits.

I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace S(m^n) ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now by contrary, suppose $Max$ $SS(m^{n_1})=M$ for some $n_1\in N$, and just look to this sequence of integer:

$\overline {m^{n_1}1}, \overline {m^{n_1}11}, \overline {m^{n_1}111}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^{t_i}$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.