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%I #10 Jul 07 2016 23:48:48
%S 5,13,10,25,17,15,41,26,61,20,37,85,50,25,39,113,34,65,145,30,82,181,
%T 29,52,101,35,75,221,122,265,40,51,74,145,65,313,170,45,123,365,53,
%U 100,197,421,50,78,226,481,68,130,257,55,65,183,545,290,91,125,613,60,85,111
%N Consider all Pythagorean triangles A^2 + B^2 = C^2 with A<B<C; sequence gives values of C, sorted to correspond to increasing A (A009004(n)).
%C The corresponding sequence for primitive triples is A156679. For all triples, the ordered sequence of C values is A020882 (allowing repetitions) and A009003 (excluding repetitions).
%D Beiler, Albert H.: Recreations In The Theory Of Numbers, Chapter XIV, The Eternal Triangle, Dover Publications Inc., New York, 1964, pp. 104-134.
%D Sierpinski, W.; Pythagorean Triangles, Dover Publications, Inc., Mineola, New York, 2003.
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Right-angled Triangles and Pythagoras' Theorem</a>
%F a(n) = sqrt(A009004(n)^2 + A156681(n)^2).
%e As the first four Pythagorean triples (ordered by increasing A) are (3,4,5), (5,12,13), (6,8,10) and (7,24,25), then a(1)=5, a(2)=13, a(3)=10 and a(4)=25.
%t PythagoreanTriplets[n_]:=Module[{t={{3,4,5}},i=4,j=5},While[i<n,h=Sqrt[i^2+j^2];If[IntegerQ[h] && j<n,AppendTo[t,{i,j,h}]]; If[j<n,j++,i++;j=i+1]];t];k=20;data1=PythagoreanTriplets[2k^2+2k+1];data2=Select[data1,#[[1]]<=2k+1 &];#[[3]] &/@data2
%Y Cf. A009004, A156681, A156679, A020882, A009003.
%K easy,nice,nonn
%O 1,1
%A _Ant King_, Feb 17 2009
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