# Can this heuristic argument be useful to prove: Sum of digits of $a^b$ equals $ab$ conjecture?

I've been trying to write a proof for the following conjecture (from this question):

Let $$s\left(a^{b}\right)$$ denote the sum of the digits of $$a^{b}$$ in base $$10$$. Then the only integer values $$a$$,$$b>1$$ that satisfy $$s\left(a^{b}\right)=ab$$ are $$(2,2),(3,3),(3,6),(3,9)$$ and $$(3,27)$$.

I found what I think is a heuristic argument, but i'm not sure if it can be useful in proving the conjecture.

Let $$d\left(n\right)$$ denote the number of digits of integer $$n$$ in base $$10$$:

$$d\left(n\right)=1+\left\lfloor \log_{10}n\right\rfloor$$

Let $$s\left(n\right)$$ denote the digit sum of integer $$n$$ in base $$10$$.

Now from the conjecture, take for example the case $$a=2$$. I've been looking at the following sum: $$\sum_{n=1}^{b}\frac{s\left(2^{n}\right)}{\sum_{k=1}^{n}d\left(2^{k}\right)}$$
The plot of that sum for $$1\leq b\leq20000$$ looks like that:

And now the same plot, with in orange, $$9\log b$$:

The difference between the $$2$$ curves quickly converges to a value $$c$$, and we see that: $$\lim_{b\rightarrow\infty}\left(9\log b-\sum_{n=1}^{b}\frac{s\left(2^{n}\right)}{\sum_{k=1}^{n}d\left(2^{k}\right)}\right)=c\approx12.721\ldots$$ From this, we can also conclude that: $$\frac{s\left(2^{b}\right)}{\sum_{k=1}^{b}d\left(2^{k}\right)}\sim9\log\left(\frac{b-1}{b}\right)\sim\frac{9}{b}$$ And since: $$d\left(n\right)=1+\left\lfloor \log_{10}n\right\rfloor \approx1+\log_{10}n$$ Then: $$\sum_{k=1}^{b}d\left(2^{k}\right)\approx\frac{b^{2}\log_{10}2}{2}$$ And: $$s\left(2^{b}\right)\sim\left(\frac{9}{b}\right)\left(\frac{b^{2}\log_{10}2}{2}\right)\sim\left(\frac{9}{2}\right)b\log_{10}2s\left(2^{b}\right)\sim1.3546\times b$$ The same applies to other values of $$a$$, so more generally: $$s\left(a^{b}\right)\sim\left(\frac{9}{2}\right)b\log_{10}a$$ Looking at plots of $$s(a^b)$$ for each values of $$a$$ from $$2$$ to $$8$$, we can see this asymptotic relation seems to be very accurate.

Now I have $$2$$ questions:

1: Does the asymptotic relation above is correct, or is there some errors in my reasoning?

2: Since $$a>\left(\frac{9}{2}\right)\log_{10}a$$, does an asymptotic relation like that is enough to prove $$s\left(a^{b}\right), for sufficiently large $$b$$?

Any help or advice would be appreciated.

• I don't know why you are interested in the sums in $$\sum_{n=1}^{b}\frac{s\left(2^{n}\right)}{\sum_{k=1}^{n}d\left(2^{k}\right)}$$ What you are really interested in is a particular $n$. Sep 26, 2019 at 15:54

Your relation $$s\left(a^{b}\right)\sim\left(\frac{9}{2}\right)b\log_{10}a$$ is a very reasonable heuristic. It assumes that the digits of $$a^b$$ are reasonably evenly distributed. Unfortunately the upper bound is $$9b\log_{10}a$$ (even ignoring the $$1$$) and $$9\log_{10}a \gt a$$ unless $$a=9$$. We cannot use this to set a hard upper limit on the $$b$$ that we would have to check for a given $$a$$. You can say that examples with large $$b$$ are unlikely because they would require the digits of $$a^b$$ to be larger than expected