Integers $n$ for which the digit sum of $n$ exceeds the digit sum of $n^5$

Question

This question is strongly inspired by The smallest integer whose digit sum is larger than that of its cube? by Bernardo Recamán Santos.

The number $n=124499$ has digit sum $1+2+4+4+9+9=29$ while its fourth power $n^4=240250031031001002001$ has a (strictly) smaller digit sum $2+4+0+2+5+0+0+3+1+0+3+1+0+0+1+0+0+2+0+0+1=25$.

In general, for a fixed integer exponent $k \ge 2$ we can ask for the set of positive integers $n$ whose digit sum exceeds that of $n^k$. One finds:

$$\begin{array}{|r|l|} k & \text{values $n$ so that $\mathrm{digsum}(n)>\mathrm{digsum}(n^k)$} \\ \hline 2 & 39, 48, 49, 79, 149, 179, 249, 318, 348, 349, 389, 390, 399, \ldots \\ 3 & 587, 5870, 28847, 28885, 28887, 46877, 48784, 49468, 49639, \ldots \\ 4 & 124499, 1244990, 12449900, 124499000, 594959999, 1244990000, 1349969999, \ldots \\ 5 & ? \\ \hline \end{array}$$

An immediate observation is that if $n$ belongs to a row above, so does $10^j \cdot n$ since appending a bunch of zeroes does not change the digit sum of $n$ or its power $n^k$. An $n$ value not ending in $0$ can be thought of as "primitive".

From a comment in A122484 we know that in the $k=4$ case the formula:

$$f(i)=75\cdot 10^{2i}-4\cdot 10^{i+1}-1 \quad , \quad i=7,8,9,10,\ldots$$

miraculously produces infinitely many "primitive" $n$ values for $k=4$.

The question here is, do numbers like this exist for exponent $k=5$ and higher?

(Or: In $39,587,124499,\ldots$, what comes next?)

Even if heuristics seem to imply that it is improbable for high values of $k$ (the number of digits increases dramatically when going from $n$ to $n^k$), maybe there exist miraculous formulas in the style of $f(i)$ above which produce examples for $k=5$ or higher?

We note that if we leave base $10$ and go to much higher radixes $b$, then it becomes easy to find examples for $k=5$.

Answer 1

Such formulas for such numbers exist for all even $k$. The formula is not as magical as it seems. Consider

$$ 10^{ni}-m\sum_{s=0}^{n-1}10^{si}\;. $$

with fixed $m,n$ for all $i$. There are $n$ stretches of $O(i)$ nines in this number. Taking it to the $k$-th power yields

$$ \left(10^{ni}-m10^{(n-1)i}\right)^k+R\quad\text{with}\quad R\in o\left(10^{k(n-1)i}\right)\;. $$

For even $k$, the terms with the highest powers of $m$ in the remainder $R$ are all positive, so by making $m$ large enough, we can make all terms in $R$ positive. Then we only have $k/2$ negative terms in the leading power, and hence only $k/2$ stretches of $O(i)$ nines, and all other stretches of $O(i)$ repeating digits are zeros. The remaining constant digits contribute $O(1)$ to the digit sum, so for $n\gt k/2$ the larger number of $O(i)$ stretches of nines wins out for sufficient $i$.

Unfortunately this doesn't work for odd $k$, since $R$ then contains a mixture of signs. Nevertheless, one could systematically search for such formulas using the ansatz

$$ \sum_{s=0}^nc_s10^{si}\;, $$

expanding the $k$-th power and solving the linear program resulting from requiring that only a small number of terms are negative.