Primitive pythagorean triple conjecture with pi ($\pi$) An Ordered Primitive Pythagorean Triple $(a,b,c)$ is one in which $a \le b \le c$ are coprime and $a^2+b^2 = c^2$.
$f(n) = |\{(a,b,c)~|~ a^2+b^2=c^2,a\le b\le c,~c \le n\}|$.
Function $f(n)$ defines the number of all distinct Ordered Primitive Pythagorean Triple $(a,b,c)$ with $c \le n$. For example, 
$f(4) = 0$ 
$f(5) = 1$ with triple $(3,4,5)$
Conjecture: for any $\epsilon > 0$, there exists $N_0$ such that 
$\forall n \ge N_0, \frac{n}{f(n)} \in (2 \pi - \epsilon, 2 \pi + \epsilon)$.
Question:I'd like to ask is there anyone has proposed such conjecture or is there any proof (true or false) for this conjecture?
Here are the first several triples
n   f(n)    n/f(n)
5   1   5
13  2   6.5
17  3   5.66666666666667
25  4   6.25
[3, 4, 5]
[5, 12, 13]
[8, 15, 17]
[7, 24, 25]
 A: Wolfram MathWorld, in its article on Pythagorean triples, says,

Lehmer (1900) proved that the number of primitive solutions with
  hypotenuse less than N satisfies
$$\lim_{n \to \infty} {\Delta_p(N) \over N} = {1 \over 2\pi} =  0.1591549\cdots$$

and taking the reciprocal gives your result.  This paper can be found on Google Books, and your conjecture is given in the discussion on pp. 327-328.  The proof seems to rest on a lot of theoretical apparatus, though; I don't know of a "simple" proof.
A: A loose heuristic argument uses the sort-of-result:

Picking two random integer, the probability that they are relatively prime is $\frac{6}{\pi^2}$

This is a "sort-of" result because we are actually talking about density, not probability.
Every primitive triple can be written $u^2-v^2,2uv,u^2+v^2$ with $\gcd(u,v)=1$, $u-v$ is odd.
So $f(n)$ can be seen as the number of pairs $v<u$ with $u^2+v^2\leq n$ and $\gcd(u,v)$ and $u-v$ is odd.
But then $g(n)$, the number of pairs $u^2+v^2\leq n$ is roughly $\frac{\pi n}{4}$, since it is the number of lattice points in a quadrant of a circle of radius $\sqrt{n}$. Of those, approximately $\frac{6}{\pi^2} g(n)$ have $\gcd(u,v)=1$, and of those, about $\frac{2}{3}$ have $u-v$ odd (because we've already excluded the cases where $u,v$ are both even.) And half of those have $v<u$.
So, altogether, you get: $$f(n)\approx \frac{1}{2}\frac{2}{3}\frac{6}{\pi^2}\frac{\pi n}{4}=\frac{1}{2\pi} n$$
Making this all rigorous would require an effort to get bounds in the various estimates, but this is the fundamentals of why this $\frac{n}{f(n)}$ converges to $2\pi$ - these estimates turn out to be "good enough."
