Constant Hiding Behind Particular Pythagorean Triples

Asked May 31 '23 at 18:51

Active May 31 '23 at 19:11

Viewed 44 times

The question pertains to an apparent constant appearing behind Pythagorean Triples whose "legs" have a difference of 1.

To start, here are the first 5 triples that satisfy the above condition:

(20, 21, 29)

(119, 120, 169)

(696, 697, 985)

(4059, 4060, 5741)

(23660, 23661, 33461)

If you take the average of the first two legs, call it $a_n$, and divide $\frac{a_{n+1}}{a_n}$, you get the following decimals:

$\frac{a_1}{a_0}=\frac{119.5}{20.5}=5.82926829268$

$\frac{a_2}{a_1}=\frac{696.5}{119.5}=5.82926829268$

$\frac{a_3}{a_2}=\frac{4059.5}{696.5}=5.82842785355$

$\frac{a_4}{a_3}=\frac{23660.5}{4059.5}=5.8284271462$

To me, it clearly appears to be approaching some constant and I would like some help working out how to solve for the closed form of this constant. I also doubt this is a new observation but I had trouble finding literature on it so if someone could provide a link, I would appreciate it!

edited May 31 '23 at 19:11

Bill Dubuque

282,220

asked May 31 '23 at 18:51

Connor James

1

That's $3+2\sqrt 2$. See this question. – lulu May 31 '23 at 18:54
2

your sequence of numbers obeys $a_{n+2} = 6 a_{n+1} - a_n$ – Will Jagy May 31 '23 at 18:56
Perfect! This is exactly what I was looking for and now my question is answered! Thank you :) – Connor James May 31 '23 at 18:57
Why don't you count (3,4,5) as the first triple? – Daniel Schepler May 31 '23 at 20:51
The ratio approaches \begin{align} \cot^2(\dfrac{π}{8})\approx 5.8284271247461900976\ \ 3 + 2 \sqrt{2} \approx 5.8284271247461900976\ \ (1 + \sqrt{2})^2 \approx 5.8284271247461900976 \end{align} – poetasis Jun 01 '23 at 18:52

Constant Hiding Behind Particular Pythagorean Triples

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