平面区域内的整点个数

葡萄糖

非线性区域x^2+y^2+2x+4y-4≤0内的整点个数
任意线性平面区域内的整点个数能否结合皮克公式得出？

葡萄糖

非线性区域$x^2+y^2+2x+4y-4≤0$内的整点个数

wayne

其实就是 \(x^2+y^2<=9\) 的整数解.
这个Mathematica软件有对应的函数 SquaresR

1. Sum[SquaresR[2, i], {i, 0, n}]

复制代码

公式就是 http://oeis.org/A057655 里的  1 + 4*{ [n/1] - [n/3] + [n/5] - [n/7] + ... }

葡萄糖

wayne 发表于 2014-2-3 20:34
其实就是 \(x^2+y^2

谢谢！吧主！
FORMULA       
$a(n) = 1 + 4*{ [n/1] - [n/3] + [n/5] - [n/7] + ... }$. - Gauss
$a(n) = 1 + 4*Sum_{ k = 0 .. [sqrt(n)] } [ sqrt(n-k^2) ]$. - Liouville (?)
a(n) - Pi*n = O(sqrt(n)) (Gauss). a(n) - Pi*n = O(n^c), c = 23/73 + epsilon ~ 0.3151 (Huxley). If a(n) - Pi*n = O(n^c) then c > 1/4 (Landau, Hardy). It is conjectured that a(n) - Pi*n = O(n^(1/4 + epsilon)) for all epsilon >0.
a(n) = A122510(2,n). [R. J. Mathar, Apr 21 2010]
$a(n) = 1 + sum((floor(1/(k+1)) + 4 * floor(cos(Pi * sqrt(k))^2) - 4 * floor(cos(Pi * sqrt(k/2))^2) + 8 * sum((floor(cos(Pi * sqrt(i))^2) * floor(cos(Pi * sqrt(k-i))^2)), i = 1..floor(k/2))), k = 1..n).$
  - Wesley Ivan Hurt, Jan 10 2013.

manthanein

https://en.wikipedia.org/wiki/Gauss_circle_problem

mathematica

圆内整点问题很难