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The following question is related to http://mathoverflow.net/questions/27413. For a positive integer $n$, let $i(D,n)$ be the number of integer points in the disk $x^2+y^2\leq n^2/\pi$, so $i(D,n) = n^2 + o(n^2)$. Let $F(x)=\sum_{n\geq 0} i(D,n)x^n$. Does $i(D,n)$ satisfy any of the following progressive weaker properties? (1) $i(D,n)$ is a polynomial for $n$ sufficiently large. (2) $F(x)$ is rational. (3) $F(x)$ is algebraic. (4) $F(x)$ is D-finite. These properties seem very unlikely, but how to prove it?

We can also ask whether the error term $i(D,n)-n^2$ has similar behavior as for the disk $x^2+y^2\leq n^2$. An incidental question (quite possibly hopeless) is whether $i(D,n)=n^2$ for infinitely many $n$. The values of $n\leq 500$ for which this happens are 1, 3, 9, 11, 35, 45, 57, 61, 109, 155, 159, 227, 275, 365, 379, 383, 471, 481. (By a simple symmetry argument is is clear that $n$ must be odd.)

edited Jan 20 2012 at 17:05

asked Jan 20 2012 at 16:16

Richard Stanley
13.3k●2●60●94

100% the answer is "NO" to each of your questions and 100% it will be pain to prove. – Anton Petrunin Jan 20 2012 at 16:35

1

Anton, I would expect that it equals $n^2$ for infinitely many $n$. It happens with naive probability about $\geq n^{−1/2−\varepsilon}$ (because error term in circle problem is conjectured and believed to be $i(D,n)−n^2=O(n^{1/2+\varepsilon})$ for any ε>0). The series $\sum n^{−1/2−\varepsilon}$ diverges, this suggest that mathematical expectation of number of $n$ such that $i(D,n)=n^2$ is infinite. – Fedor Petrov Jan 20 2012 at 16:46

3

Just an easy remark: if it is polynomial, then it equals $n^2+$const, since error term is proved to be less then linear in $n$. If the generating function is rational, then the function $i(D,n)-n^2$ is periodic (sublinear function with rational generating function is bounded, and if it is also inteegr-valued, then surely periodic). So, $i(D,nT)-T^2n^2$ is constant, where $T$ stays for period. – Fedor Petrov Jan 20 2012 at 17:00

1

The generating function counting solutions to $x^2 + y^2 = m$ is a modular form, so transcedental but satifying a differential equation. But a change of variable like $m = \lfloor n^2 / \pi \rfloor$ will surely destroy a structure that delicate. – Noam D. Elkies Jan 21 2012 at 0:55

Isn't it known that the error term in $i(D,n)$ is at least $C n^{1/2}$ for infinitely many $n$ ? Together with Fedor's argument, this would imply the generating function is not rational. – François Brunault Jan 21 2012 at 10:10

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Fedor Petrov · Answer 1 · 2012-01-20 22:50:06Z UTC

This is the answer only to 1). If $i(D,n)$ is polynomial, then from known relation $i(D,n)-n^2=o(n)$ we get $i(D,n)=n^2+C$. But $i(D,n)$ is always odd, while $n^2+C$ does change its parity. A contradiction.

Alas, such trick does not work even for disproving $i(D,Tn)=T^2n^2+C$ for given positive integer $T$ (which would imply that generating function is not rational, see comments).

Integer points in dilations of a disk of volume one

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