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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:

enter image description here

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

enter image description here

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 $s\left(a^{b}\right)<ab$, for sufficiently large $b$?

Any help or advice would be appreciated.

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  • $\begingroup$ 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 $n$. $\endgroup$ Sep 26, 2019 at 15:54

1 Answer 1

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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

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