What is the formula for this recursive pattern?

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Hey what is the formula for this recursive pattern?

http://img440.imageshack.us/img440/7528/recursion.jpg

It looks shocking & I would bet that there is a name for it.

Perhaps it's a theorem or something?
 
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What do you want a formula for?
The sides of the hypothenuses? The area of the triangles? The total area of the first n?
 
Sorry, the fact that each hypoteneuse is an interger higher than the last hypoteneuse & that the leg is the previous hypoeneuse.

It's just a crazy pattern that I've never seen before, I seen it in a book and it just looks awesome, is there some explanation?
 
You can find the lengths of successive hypotenuses by using the Pythagorean theorem. If that's what you mean.
 
Yeah I know, I do mean that, but it's the kind of thing a number theorist or somebody would make into a theorem (or probably already has).

It's just amazing, I mean look at the pattern that emerges. That thing could probably go on to infinity, sprouting smaller and smaller triangles - kind of like a fractal.

I would have assumed that it is some theorem, or has some special name or something.

Maybe I've just found my own personal Fibonacci sequence :smile:

___The_Pythagorasequence___
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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