Sum of digits of $a^b$ equals $ab$

The following conjecture is one I have made today with the aid of computer software.

Conjecture:

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

Remarks:

• The number of digits of an integer $n$ is $1+\lfloor\log_{10}n\rfloor$ where $\lfloor\cdot\rfloor$ denotes the floor function. This means that $s(a^b)<9+9\lfloor b\log_{10} a\rfloor$ as each digit takes a value of at most $9$. From this plot it is evident that the equality will never hold for $a\ge 9$, after checking the first $21$ values of $b$. Therefore it suffices to consider $2\le a\le 8$.

• I have excluded the case $b=1$ as it is trivial - it forces $a$ to be a single-digit integer.

• In PARI/GP the code is given by mfun(b)={for(i=2,8,for(j=2,b,if(sumdigits(i^j,10)==i*j,print(i," ",j))));} and tests up to $b\le 10^6$ verify the above conjecture.

Aside:

• The equation $s(ab)=a+b$ is much simpler to solve. From here, the upper bound $a\le 23$ is apparent and for these values of $a$, we can form the upper bound $b\le 22$ as there will be no solutions when the red line lies above the blue lines. This $23\times22$ grid can be computed through software and it is found that the only solutions are $$(a,b)=(2,2),(3,6),(6,3).$$ The similarity in the first two solutions with the conjectured solutions may only be coincidental.

Advances on this will be appreciated.

I like to crack numbers but this time I'll pass. It's highly unlikely that there is any other solution. I have created a few plots of function:

$$f(a)=s(a^b)-ab$$

...for $a=2,3,4,5,6,7,8$ and $b\in [1,4000]$. All these plots look the same. The function slides towards negative infinity in a pretty linear fashion, with very little variation from the straight line. If you accept bets, I can bet a house that there are no other solutions except small ones that you already listed.

Mathematica code:

For[a=2,a<=8,a++,DiscretePlot[Total[IntegerDigits[a^b,10]]-a*b,{b,2,4000},Filling->f,Joined->False] // Print]

$a=2$

$a=3$

$a=4$

$a=5$

$a=6$

$a=7$

$a=8$

Here is a heuristic argument (not a proof sadly, but too long for a comment).

The sequences $2^n,3^n,\dots,9^n$ are known to follow Benford's law, i.e., their first digit has a probability $\log_{10}\left(1+\frac{1}{d}\right)$ of being $d$; their second digit has a probability $\log_{10}\left(1+\frac{1}{10+d}\right)+\log_{10}\left(1+\frac{1}{20+d}\right)+\dots+\log_{10}\left(1+\frac{1}{90+d}\right)$ of being $d$, etc.

Here, "probability" is to be understood as asymptotic density. For instance, for the first digit of the sequence $2^n$, we have: $$\lim_{n\to\infty}\frac{|\{k\in\{1,\dots,n\}:2^k\text{ has $d$ as first digit}\}|}{n} = \log_{10}\left(1+\frac{1}{d}\right)$$ and more generally, for all $a \in \{2,\dots,9\}$: $$\lim_{n\to\infty}\frac{|\{k\in\{1,\dots,n\}:a^k\text{ has $d$ as $j$-th digit}\}|}{n} = \sum_{i=10^{j-2}}^{10^{j-1}-1}\log_{10}\left(1+\frac{1}{10i+d}\right)$$ and as the last quantity approaches $\frac 1 {10}$ as $j \to \infty$, the $j$-th digit of $a^n$ are almost equidistributed.

If we heuristically take these probabilities at face value, the sum of the digits of $a^n$ should match the sum of their expected values, and, apart from the first figures, this expected value is close to $4.5$. Thus, we should expect to have: $$s(a^n) \approx 4.5(1+\lfloor n\log_{10}a\rfloor) \approx 4.5 n\log_{10}a$$ and thus: $$s(a^n) - an \approx (\underbrace{4.5\log_{10} a - a}_{<0 \textrm{ for all } a})n \to_{n \to\infty} -\infty$$ This aligns with the graphs provided by @Oldboy. For instance, for $a=6$, we should expect a slope of $4.5\log_{10}6 - 6 \approx -2.50$, which matches the observed slope of $-10000/4000$.

Also not an answer, but it greatly reduces the set of possible solutions... We know for a fact that the difference between $n$ and $s\left(n\right)$ is always a multiple of $9$: $$ \left(n-s\left(n\right)\right)\mod9=0 $$ So if $s\left(a^{b}\right)=ab$, then: $$\left(a^{b}-ab\right)\mod9=0$$ Therefore, for $2\leq a \leq 8$, $b$ must be of the following forms: $$ b\begin{cases} 3n & \iff a=3~or~6\\ 9n+1 & \iff a=4~or~7\\ 18n+1~or~18n+2 & \iff a=2\\ 18n+1~or~18n+8 & \iff a=8\\ 18n+1~or~18n+14 & \iff a=5\\ \end{cases} $$