2
$\begingroup$

Let $a,b$ be two positive integers. Let the sequence $\{s_n\}_n$ be the set of all possible sums of squares $a^2+b^2$, such that they are in ascending order \begin{align*} & n=1 & s_1=1^2+1^2=2 \\ & n=2 & s_2=1^2+2^2=5 \\ & n=3 & s_3=2^2+1^2=5 \\ & n=4 & s_4=2^2+2^2=8 \end{align*} etc. Here $n$ serves as a counting index for the sequence and does not serve any other purpose. I have found that the number of identical values that each sum can have is equal to the number of lattice points lying on a quarter of circumference of a circle of radius $n$. For example, $s_2=s_3=5$ we have two values, but there are other cases, such as \begin{align*} & n=31 & s_{31}=1^2+7^2=50 \\ & n=32 & s_{32}=5^2+5^2=50 \\ & n=33 & s_{33}=7^2+1^2=50 \end{align*}

To the best of my knowledge, there are no references to the asymptotic form of $\{s_n\}_n$ as $n\to \infty$, if indeed it exists. A brute-force approach with $n\le10^6$ led to the expression $$ \{s_n\}_n\sim \beta n,\qquad\beta\approx1.276\ldots. $$ Therefore my question is: is there any known closed form of the asymptotics of this particular succession? A more profound and interesting question is, why does it seem to be linear?

Improve this question
$\endgroup$
8
  • $\begingroup$ I do not understand your definition. I assume that "ordered set" (as in your body) means "sequence" (as in your title), but what is the order? Is it by the value of the sum, or by the larger of the summands and then lexicographically, or …? What is a ‘degenerate’ value of a sum? $\endgroup$
    – LSpice
    Commented Nov 16, 2022 at 0:29
  • 2
    $\begingroup$ It seems to me that introducing the sequence doesn't actually do anything; you're just counting how many times $n$ is a sum of squares, which, by definition, is the number of lattice points on a circle of radius $\sqrt n$ (I think you may have forgotten the square root). So you are just asking about the asymptotics of that point count, right? $\endgroup$
    – LSpice
    Commented Nov 16, 2022 at 0:46
  • 6
    $\begingroup$ usually this is studied in the nearly equivalent form of asking for the number of (x,y) such that $x^2 + y^2 \leq N$. The answer is asymptotic to the area of this circle, which is $\pi N$. Since you're using a quarter-circle, we replace the factor $\pi$ by $\pi / 4$. Then the $n$-th sum of squares should be asymptotic to $(4/\pi) n$. Indeed your $\beta$ nearly equals $4/\pi = 1.273\ldots$. $\endgroup$
    – Noam D. Elkies
    Commented Nov 16, 2022 at 0:47
  • 1
    $\begingroup$ This is essentially "The Gauss Circle Problem," whhich has an enormous literature. en.wikipedia.org/wiki/Gauss_circle_problem will get you started, TheVal. But note that while you write "nonzero integers," it seems you really mean positive integers, since you're not counting $5=(-2)^2+1^2=2^2+(-1)^2=(-2)^2+(-1)^2$ and the like. $\endgroup$
    – Gerry Myerson
    Commented Nov 16, 2022 at 21:58
  • 1
    $\begingroup$ @GerryMyerson thank you I'll edit it $\endgroup$
    – TheVal
    Commented Nov 17, 2022 at 14:22

1 Answer 1

Reset to default
0
$\begingroup$

The answer has been provided by Noam D. Elkies, for which I'll copy here his comment.

Usually this is studied in the nearly equivalent form of asking for the number of $(x,y)$ such that $x^2+y^2\le N$.

The answer is asymptotic to the area of this circle, which is $\pi N$. Since you're using a quarter-circle, we replace the factor $\pi$ by $\pi/4$.

Then the $n$-th sum of squares should be asymptotic to $$ \frac{4}{\pi}n $$

Indeed your $\beta$ nearly equals $4/\pi=1.273\ldots$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .