Thread: from i to π
View Single Post
Old 2016-09-06, 17:03   #14
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts
Default

Quote:
Originally Posted by CRGreathouse View Post
Ah, you think he meant c^2 = u^2 + v^2 rather than c = u^2 + v^2? That doesn't match either, though: I get
Quote:
Originally Posted by A008846
Not only the square of these numbers is equal to the sum of two nonzero squares, but the numbers themselves also are; this sequence is then a subsequence of A004431. - Jean-Christophe Hervé, Nov 10 2013
so c=v^2+u^2 works. for c<2^n. you still get the same numbers I think but the statement is about that hitting 2 pi in ratio with 2^n ( aka that ratio ((2^(n-1))/(number of prim. pythag triples) tends to pi as n tends to infinity)

Last fiddled with by science_man_88 on 2016-09-06 at 17:05
science_man_88 is offline   Reply With Quote