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Conjecture: Let $\mu_x$ be the arithmetic mean of the ratio of the perimeter to the hypotenuse of all primitive Pythagorean triplets in which no side exceeds $x$; then,

$$ \lim_{x \to \infty}\mu_x = 1 + \frac{4}{\pi}$$

Based on data for $x \le 10^{11}$, the computed value agrees with the conjectured value up to $5$ decimal places.

Primitive Pythagorean triplets $a^2 = b^2 + c^2, \gcd(b,c) = 1$ are given by $a = r^2 + s^2$, $b = r^2 - s^2$ and $c = 2rs$ where $r > s$ are natural numbers. Let the $n$-th primitive triplet be the one formed by the $n$-th smallest pair in increasing order of $(r,s)$. It was proved in this post MSE that the arithmetic mean $\mu_n$ of the ratio of the perimeter to the hypotenuse of first $n$ primitive Pythagorean triplets approaches $\dfrac{\pi}{2} + \log 2$ as $n \to \infty$. However the above claim is still open.

Question: I am looking for a proof or disproof of this conjecture.

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    $\begingroup$ You have some reason for believing the limit exists and equals $1+(4/\pi)$? $\endgroup$ – Gerry Myerson Oct 7 at 5:45
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    $\begingroup$ @GerryMyerson Three reasons to suspect the existence. (1) Similar limit of the first $n$ triplets exists and equals $\pi/2 + \log 2$ (2) May be related to Sieprinski's result that the number of triplets in which no side exceeds $x$ is $\frac{4x\log x}{\pi}+ O(x)$ and (3) Experiment data for $x \le 10^{11}$ seems to agree though this is the weakest of the three reasons. $\endgroup$ – Nilos Oct 7 at 6:25
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This is right. Primitive Pythagorean triples are parametrized as $(u^2-v^2, 2uv, u^2+v^2)$ with $\mathrm{GCD}(u,v) = 1$ and $u+v \equiv 1 \bmod 2$. To have $0 < a,b < c \leq R^2$, we must have $0 < v < u$ and $u^2+v^2 \leq R^2$. The ratio of perimeter to hypotenuse is $\tfrac{(u^2-v^2)+(2uv)+(u^2+v^2)}{u^2+v^2} = \tfrac{2u(u+v)}{u^2+v^2}$.

Ignoring the $GCD$ and parity conditions, we want to compute $$\sum_{0<v<u,\ u^2+v^2\leq R^2} \frac{2u(u+v)}{u^2+v^2} \ \mbox{and} \ \sum_{0<v<u,\ u^2+v^2\leq R^2} 1.$$ Replacing these sums with integrals and changing to polar coordinates gives $$\int_{r=0}^R \int_{\theta=0}^{\pi/4} 2r ( \cos \theta) (\cos \theta+\sin \theta) \, dr \, d\theta = \frac{4+\pi}{8} R^2$$ and $$\int_{r=0}^R \int_{\theta=0}^{\pi/4} r \, dr \, d\theta = \frac{\pi}{8} R^2$$ so the ratio approaches $1+\tfrac{4}{\pi}$.

It remains to analyze replacing the sum by an integer, and the effect of the conditions $\operatorname{GCD}(u,v) = 1$ and $u+v \equiv 1 \bmod 2$. I'll sketch the argument.

The GCD condition can be represented as an inclusion-exclusion sum $$\sum_{1 \leq k \leq R} \mu(k) \sum_{0<kv<ku,\ (ku)^2+(kv)^2\leq R} \frac{2(ku)(ku+kv)}{(ku)^2+(kv)^2}.$$ The inner sum is exactly the one we discussed above, with $R$ replaces by $R/k$, so the integral approximation gives $\tfrac{4+\pi}{8} (R/k)^2$. Being more precise, the error in replacing the sum by an integral is $O(R/k)$. So the sum with GCD condition imposed is $\tfrac{4+\pi}{8} R^2 \sum_k \tfrac{\mu(k)}{k^2} + O(\sum_{k \leq R} R/k) = \tfrac{4+\pi}{8} \tfrac{6}{\pi^2} R^2 + O(R \log R)$. Likewise, the denominator is $\tfrac{\pi}{8} \tfrac{6}{\pi^2} R^2 + O(R \log R)$. So the limiting ratio is still the same. Imposing that $u+v \equiv 1 \bmod 2$ (once we have already imposed $\mathrm{GCD}(u,v)=1$) multiplies top and bottom by $2/3$.

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    $\begingroup$ If we relax the $GCD$ condition the I think the mean will still converge to the same value but at a slower rate $\endgroup$ – Nilos Oct 7 at 15:20
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    $\begingroup$ The analysis above definitely shows it converges to the same value. I would guess the convergence is faster, but I am not an analytic number theorist, so that's just my guess. $\endgroup$ – David E Speyer Oct 7 at 16:56
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Let $(a,b,c)$ be a Pythagorean triple (in standard notation with $c$ being the hypotenuse). In the usual coordinate system the point $(a,b)$ corresponds to this triple. Now taking the sum of $(a+b+c)/c$, when $0<a \leq R, 0<b\leq R$, and in particular for the longest side $0 < c \leq R$, then taking the limit $R \rightarrow \infty$ means that the sum can be replaced by an integral, (with some rescaling).

It seems to me this integral is in the limit (in polar coordinates) $$\frac{1}{\pi R^2/4} \int_{\phi=0}^{\pi/2}\int_{c=0}^R (1+\cos(\phi)+\sin(\phi))\, c\, d \phi \, d c=1+\frac{4}{\pi}$$

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    $\begingroup$ Why can you replace the sum by an integral when $R \rightarrow \infty$? $\endgroup$ – Joël Oct 7 at 13:07
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    $\begingroup$ This is morally right, but the OP wanted to restrict to the case where $c$ is an integer (and also $GCD(a,b,c)=1$). I was in the middle of writing out an answer where I replace $(a,b,c)$ by $(u^2-v^2, 2uv, u^2+v^2)$, and integrate in the $(u,v)$ coordinates, but I seem to have an error somewhere in it. $\endgroup$ – David E Speyer Oct 7 at 13:08
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    $\begingroup$ Your integral should be written as $$ \int_{\phi=0}^{\pi/2} \left( \int_{c=0}^R (1+\cos(\phi)+\sin(\phi))\, c\, dc\right)\, d \phi $$ rather than as $$ \int_{\phi=0}^{\pi/2} \left( \int_{c=0}^R (1+\cos(\phi)+\sin(\phi))\, c\, d \phi\right) \, d c $$ $\endgroup$ – Michael Hardy Oct 7 at 15:11

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