7
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Let $a_k<b_k<c_k$ be the $k$-th primitive Pythagorean triplet in ascending order of the hypotenuse $c_k$. Define

$$ l = \frac{b_1 + b_2 + b_3 + \cdots + b_k}{c_1 + c_2 + c_3 + \cdots + c_k}, \text{ } s = \frac{a_1 + a_2 + a_3 + \cdots + a_k}{c_1 + c_2 + c_3 + \cdots + c_k} $$

Question: What is the limiting value of $l$ and $s$?

The difference between this question and the related question: Part 2: Does the arithmetic mean of sides right triangles to the mean of their hypotenuse converge? is that here the triangles are in sequenced in ascending order of the hypotenuse $c_k$ where as in the related question, they are sequenced in ascending order of $r$ and $s$, and depending on the choice of sequencing, the limiting value differs.

SageMath Code

c  = 1
sa = 1
sb = 1
sc = 1
f  = 0
sx = 0
while(c <= 10^20):
    a = c - 1
    b = 3
    while(a > b):
        b = (c^2 - a^2)^0.5
        if(b%1 == 0):
            if(b <= a):
                if(gcd(a,b) == 1):
                    f  = f + 1
                    sa = sa + a
                    sb = sb + b
                    sc = sc + c
                    sx = sx + 1/c.n()
                    print(f,c, sa/sc.n(),sb/sc.n(),sx)
            else:
                break
        a = a - 1
    c = c + 1
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  • 1
    $\begingroup$ You should mention that all values except $l,s$ are integer numbers. And so pay attention to your code b = (c^2 - a^2)^0.5, of that ^0.5 $\endgroup$ – Ripi2 Oct 25 '19 at 16:57
4
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Quantities $x=a/c$ and $y=b/c$ are the legs of a pythagorean triangle having unit hypotenuse, hence they are the coordinates of points lying on a unit circle centred at the origin, and $x=\cos\theta$, $y=\sin\theta$ with $\pi/4<\theta<\pi/2$.

It is reasonable to think that, at least in the case of primitive triples, those points are spread evenly on that arc. In that case their average values are: $$ \langle x\rangle={\int_{\pi/4}^{\pi/2}\cos\theta\,d\theta\over\int_{\pi/4}^{\pi/2}d\theta}= {4-2\sqrt2\over\pi}\approx 0.372923, $$ $$ \langle y\rangle={\int_{\pi/4}^{\pi/2}\sin\theta\,d\theta\over\int_{\pi/4}^{\pi/2}d\theta}={2\sqrt2\over\pi}\approx 0.900316. $$ One should then justify that $\langle a\rangle/\langle c\rangle$ and $\langle a/c\rangle$ have the same limiting value, but that also seems very reasonable. I ran a simulation up to $k\approx800000$ and found the encouraging results: $$ s_k\approx0.373,\quad l_k\approx0.900. $$

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  • $\begingroup$ Finding a bug in a code can be difficult. It would be easier, in my opinion, if you could share a file with all the triples you found. However, my simulation was made with primitive triples only, while yours involves all triples: I'll try to add non-primitive triples to see if I can reproduce your results. $\endgroup$ – Aretino Oct 25 '19 at 8:24
  • $\begingroup$ My code is still running and I will run it for a few more days. But I have added the code in the post now. $\endgroup$ – Nilotpal Kanti Sinha Oct 25 '19 at 8:39
  • $\begingroup$ Your answer is correct for primitive triplets and agrees with my computations. $\endgroup$ – Nilotpal Kanti Sinha Oct 27 '19 at 17:39

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