24: (24,32,40), (18,24,30), (24,70,74), (7, 24, 25)...
840: see 24 example but multiply by 35

So it's not exactly 3.

Oh, you're limiting yourself to primitive pythagorean triples only? (so [6,8,10] is invalid because it is simply 2x[3,4,5] ?)

Also, is it three exactly, or at least three?

Hm... are you counting certain triples like [20,21,29]? Not sure if they're fitting in the current description.

Currently, you're using the following:

Even values of 'a'
b=a^2/4-1
c=a^2/4+1

Odd values of 'a'
b=(a^2-1)/2
c=(a^2+1)/2

which means you're assuming that for odd a, b and c are 1 apart; for even a, b and c are 2 apart. This works usually, but sometimes fails. (e.g. a=20, c-b=8, not 2)

As mentioned a bit earlier in this thread, the generalized Pythagorean triple can be written in the form (m^2-n^2,2mn,m^2+n^2). If you're looking at only primitive triples, then m,n should be coprime.

Note that the first leg is (m+n)(m-n), so factoring should make this task easy.

Using your definition: http://oeis.org/A111766

EDIT: Notice that there's a closed form!

Necro update!
So, if you look at the image attached, that is a formula I was working on for the past few years. I made a rough estimate of how many pages of notes I put into this problem and it was over a hundred Well, guess what? In the time that I have worked on it, I was able to prove the formula worked and that it extended to any function that you can find derivatives and integrals for (and since you can use Fourier analysis to find a function for just about anything, you are good for just about anything). However, I was still having doubts that it was original. Well, last week I was looking up something only slightly related when I noticed a particular sequence:

1
1/2
1/6
-1/30
1/42
-1/30

At this point, I got excited. I know these numbers! Then the next term: 5/66. I was ecstatic! These were the coefficients to my function and they are known as the Bernoulli numbers. See, the issue is that when I was writing down the numbers, I computed the coefficients, but when you look them up on the internet and whatnot, most people simply refer to them as B_n. Not very helpful for me. On top of that, my function is a generalisation of the Euler-Macluarin formula which cannot be google searched very easily (unless you know the name) and most people don't know of it or have never worked with it! So what have I done with it in the past few months?

I have made some pretty cool identities including one dealing with the coefficients and "e" (so now I have gone back and rewritten e in terms of the Bernoulli numbers). I now have two representations for sum(sin(x),x,1,n) which can be compared and I can derive more relationships (one of them uses imaginary numbers, the other uses sine, cosine, and tangent. We know where this is going Another proof of Euler's identity). But more importantly to me, I can apply this to analysing the Riemann Zeta function which I did a bunch of research on over the summer (and I've even presented about it a few times).

I also use this quite a bit in my statistics class and it has been very useful. For the curious, any polynomial with positive integer powers will result in a finite closed form expression. Any polynomial using exponentials, trig functions, or non-integer powers will have an series with infinitely many terms. Luckily, it converges pretty rapidly to the desired sum.

If you want to have fun, look at "identity 0" below. If you replace a with e, all of those ln(a) terms get cancelled to 1

Also, you can use this to derive that the "first Bernoulli numbers" divided by the "second Bernoulli numbers" (the only difference is that B₁ is negative in the first, positive in the second) will equal e

More randomness here Basically, I was trying to evaluate pi when I was in 8th grade, and I found some notes on it. After updating the syntax a little (I had not been exposed to limits at that point), I realised a neat little identity and I tried to prove it This reveals an interesting thing about the limit of sin(x) as x approaches zero at an exponential rate compared to the limit of 2^x as x approaches zero. Anyways, another neat fact that this revealed to me is that if you divide pi by the nested radical (see the attached image, at the bottom), it will converge to a power of 2. For example:

pi/sqrt(2) is about 2.22144   (2¹=2)
pi/sqrt(2-sqrt(2)) is about 4.104688612 (2²=4)
pi/sqrt(2-sqrt(2+sqrt(2))) is about 8.051636 (2³=8)
pi/sqrt(2-sqrt(2+sqrt(2+sqrt(2)))) is about 16.025731 (2⁴=16)
pi/sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(2))))) is about 32.01285466 (2⁵=32)

Anyways, take a look at the proof I devised it when I was trying to show that it did in fact converge

No, I took out the /2. 2^x/2=2^x*2^-1=2^x-1