Just looking for verification on this...
Here are my algorithms:
For ODDS:
$$a_n=1+2n$$
$$b_n=(a_n-1)(n+1)$$
$$c_n=b_n+1$$
For EVENS:
$$a_n=4+2n$$
$$b_n=(a_n/2+1)(n+1)$$
$$c_n=b_n+2$$
Here's the proof for the odds: $$a_n^2=(2n+1)(2n+1)=4n^2+4n+1$$ $$b_n=(1+2n-1)(n+1)=2n(n+1)=2(n^2+n)$$ $$b_n^2=(2(n^2+n))^2=4(n^2+n)(n^2+n)=4n^4+8n^3+4n^2$$ $$a_n^2+b_n^2=4n^4+8n^3+8n^2+4n+1$$ $$c_n=2(n^2+n)+1=2(n^2+n+1/2)$$ $$c_n^2=4(n^2+n+1/2)(n^2+n+1/2)=4n^4+8n^3+8n^2+4n+1$$
You can see $a_n^2+b_n^2=c_n^2$, which I think proves it's correct.
And here's the proof for the evens: $$a_n^2=(2n+4)(2n+4)=4n^2+16n+16$$ $$b_n=((2n+4)/2+1)(n+1)=(n+3)(n+1)=n^2+4n+3$$ $$b_n^2=(n^2+4n+3)^2=4(n^2+4n+3)(n^2+4n+3)=n^4+8n^3+22n^2+24n+9$$ $$a_n^2+b_n^2=n^4+8n^3+26n^2+40n+25$$ $$c_n=n^2+4n+5$$ $$c_n^2=(n^2+4n+5)(n^2+4n+5)=n^4+8n^3+26n^2+40n+25$$
Again, $a_n^2+b_n^2=c_n^2$, which I think proves both sets of equations will always result in a Pythagorean triple.