It is easy to see that for each perfect power $N = n^k$ it holds that

$$N-1 = (n-1)\sum_{0\leq i<k} n^{i}$$

It is not much harder to see that for each $x < N$ there are $a_i < n$ such that

$$x = \sum_{0\leq i<k} a_i\ n^{i}$$

What makes perfect powers perfect bases for the redundancy-free representation of numbers $x<N$ is that there is a perfect 1:1 correspondence between all $x<N$ and all k-tuples $\{a_i\}$ with $a_i<n$.

I wonder why these - as I find important - facts are not more often prominently mentioned, e.g. in the Wikipedia article on perfect powers.

  • 3
    $\begingroup$ Whenever we write a natural number in base $n$ we use this representation...I'd say it was extremely well known. $\endgroup$ – lulu Oct 11 '16 at 16:11
  • $\begingroup$ I know that. But this was not the question. You can use any number k for k-ary representation. But perfect powers are special because they allow redundancy-free representations. $\endgroup$ – Hans-Peter Stricker Oct 11 '16 at 16:12
  • $\begingroup$ Isn't this simply representation in radix $n$? What am I missing? And what's the question? “Why somebody didn't mention this?” is not a question that can receive an answer other than “they didn't deem it important”. $\endgroup$ – egreg Oct 11 '16 at 16:13
  • $\begingroup$ @egreg: I added a note on "redundancy-free" which probably makes perfect powers "important". $\endgroup$ – Hans-Peter Stricker Oct 11 '16 at 16:20
  • 1
    $\begingroup$ @HansStricker This has nothing to do with perfect powers; it's simply a property of representation in base $n$. $\endgroup$ – egreg Oct 11 '16 at 16:26

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