3
$\begingroup$

I read that the set of integers that can be written in the form $$n = a^2 + b^4 + c^6$$ is of zero density, since the sum of inverses of exponents $1/2+1/4+1/6$ is less than $1$. I do not understand the argument, is there a probabilistic/density way of seeing/writing this?

$\endgroup$
8
$\begingroup$

I don't know much about any theory behind this, but to me it looks like simple counting.

The number of squares less or equal than $n$ is $O(n^{1/2})$. (In fact, it is easily calculated as $\lfloor\sqrt{n}\rfloor$.) Similarly, the number of fourth powers in the same interval is $O(n^{1/4})$ and the number of sixth powers in the same interval are $O(n^{1/6})$. Now, make up all the sums of any square, any fourth power and any sixth power in that set. The number of those sums is at most $O(n^{1/2})O(n^{1/4})O(n^{1/6})=O(n^{1/2+1/4+1/6})$ (there may be repetitions!) and so the number of those sums less than equal than $n$ (which is bound from above by the previous number, and may be even smaller if some of the sums end up bigger than $n$) is also $O(n^{1/2+1/4+1/6})=o(n^1)=o(n)$, because $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}<1$.

As the density is defined as the limit of the fraction of $\{1,2,\ldots, n\}$ belonging to the set, when $n\to\infty$, we have that the density here is $\lim_{n\to\infty}\frac{o(n)}{n}=0$.

$\endgroup$
4
  • $\begingroup$ it is also the case that $n=a^2 + b^2$ has density zero. Sad but true. $\endgroup$
    – Will Jagy
    Dec 6 '21 at 21:20
  • $\begingroup$ @WillJagy A more refined argument can deduce that number of integers $\le N$ that can be expressed as sum of squares is asymptotic to $N(\log N)^{-1/2}$ multiplied by an explicit constant. $\endgroup$
    – TravorLZH
    Dec 7 '21 at 2:27
  • $\begingroup$ @TravorLZH right. It's in Le Veque, volume 2. I asked a question about primitive representations, I was surprised to learn that the implied constant changed but nothing else....having trouble finding it, I think I did ask it here or on MO but perhaps it was just in comments $\endgroup$
    – Will Jagy
    Dec 7 '21 at 3:06
  • $\begingroup$ @TravorLZH found it math.stackexchange.com/questions/1282550/… $\endgroup$
    – Will Jagy
    Dec 7 '21 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.