If we define a $PPTCountingFunction(n)$ as a function that returns the number of PPF with $c < n$ and $a>b$, then up to first $n=100,000$ it is near linear and
$\dfrac{n}{PPTCountingFunction(n)}=2\pi$
I have several questions (third question is the most interesting to me):
(1) Is this also an asymptotic behavior of this function, or does it have some other slowly changing factors that are not showing up when n is small?
(2) Is there a clear reasoning for frequencies of PPT?
(3) Can we apply similar reasoning to estimate the frequency of primitive counterexamples to Euler's hypothesis for $n=4$ (solution s of $A^{4}+B^{4}+C^{4}=D^{4}$)?
Regarding (3). First solution appears at $95800^{4} + 414560^{4} + 217519^{4} = 422481^{4}$. This is the only solution with $D<2000000$. Another known solution (not necessarily second) is $2682440^{4} + 15365639^{4} + 18796760^{4} = 20615673^{4}$. I am curious if there is a point to look for a solution between these two.
The list of hypotenuses is A020882 in OEIS. The following analysis follows the third comment on that sequence, and provides a reasonability argument, though not a proof.
Counting primitive triples with hypotenuse at most $n$ is the same as counting pairs $(a,b)$ with $\gcd(a,b)=1$, $a$ and $b$ not both odd, and $a>b>0$ inside the circle of radius $\sqrt{n}$. The total number of pairs $(a,b)$ inside such a circle is $\approx \pi n$ (see here, for example). Only $\frac{1}{8}$ of these have $a>b>0$; of these, only $\frac{6}{\pi^2}$ of them are relatively prime. Finally, asking that $a$ and $b$ be not both odd reduces by another factor of $\frac{2}{3}$ (note that $a$, $b$ both even was excluded by the $\gcd$). So altogether, the number of qualifying points is $$n\cdot\pi\cdot\frac{1}{8}\cdot\frac{6}{\pi^2}\cdot\frac{2}{3} = \frac{n}{2\pi}.$$ Thus $\text{ppt}(n)\approx \frac{n}{2\pi}$, and your result follows.