from i to π

[CRGreathouse]

Here's a heuristic suggesting that my constant is right. Start by looking at the circle with radius sqrt(x). All lattice points in the circle are (u, v) with u^2 + v^2 < x. The circle has area \(\pi x\). Now look only at the first quadrant where u, v > 0, a quarter-circle of area \(\pi x/4\). But we need to split it in half to get only the values with 0 < u < v which has area \(\pi x/8.\) Notice that the number of lattice points inside this area is equal to its area plus an error term equal to its outside (perimeter/circumference/etc.) for a total of \(\pi x/8+O(\sqrt x)\). So far this is a rigorous count of the number of points, but the next step is just a heuristic. The density of squarefree numbers is \(6/\pi^2,\) so the expected number of 0 < u < v with gcd(u, v) = 1 and u^2 + v^2 < x is about \(\pi x/8\cdot6/\pi^2 = \frac{3}{4\pi}x\) as desired.

The last step can probably be made rigorous -- the area is pretty regular and the error bounds on squarefree density are small.

[bhelmes]

A peaceful day for you,

thank you for your efforts and explications.

i know where the problem comes from:

u should be odd and v even, or inverse !

so the tripple (u,v)=(1,3) is not mentionned in my list.

i learned that for primitiv pyth. tripples that this fact is necessary,
but i did not mentionned it, it was my fault.

I will research if the amount of pyth. triples or the amount of different roots of 1 deliver a better approximation of pi.

Nice greetings from the primes
Bernhard

[CRGreathouse]

Originally Posted by bhelmes

u should be odd and v even, or inverse !

so the tripple (u,v)=(1,3) is not mentionned in my list.

i learned that for primitiv pyth. tripples that this fact is necessary,
but i did not mentionned it, it was my fault.

If you're going to arbitrarily reject some values then you can adapt my method and see if it comes out as desired.