A156689 - OEIS

%I #13 Jul 07 2016 23:48:48

%S 1,2,3,3,4,5,5,6,7,7,8,9,6,9,10,11,11,12,13,10,13,14,15,15,12,16,17,

%T 14,17,18,15,19,19,20,21,18,21,22,23,15,23,24,21,25,22,25,26,27,27,24,

%U 28,29,21,26,29,30

%N Inradii of primitive Pythagorean triples a^2+b^2=c^2, 0<a<b<c with gcd(a,b)=1, and sorted to correspond to increasing a (given in A020884).

%C The inradius is given by r=1/2 (a+b-c)=ab/(a+b+c)=area/semiperimeter, and the inradii ordered by increasing r are in A020888.

%D Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.  Solution published in Vol. 16, Issue 2, November 2008, p. 32.

%D D. G. Rogers, Putting Pythagoras in the frame, Mathematics Today, The Institute of Mathematics and its Applications, Vol. 44, No. 3, June 2008, pp. 123-125.

%H Reinhard Zumkeller, <a href="/A156689/b156689.txt">Table of n, a(n) for n = 1..1000</a>

%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 A156689(n)=1/2 (A020884(n)+A156678(n)-A156679(n))

%e The eighth primitive Pythagorean triple ordered by increasing a is (13,84,85). As this has inradius 1/2 (13+84-85)=6, we have a(8)=6.

%t PrimitivePythagoreanTriplets[n_]:=Module[{t={{3,4,5}},i=4,j=5},While[i<n,If[GCD[i,j]==1,h=Sqrt[i^2+j^2];If[IntegerQ[h] && j<n,AppendTo[t,{i,j,h}]];];If[j<n,j+=2,i++;j=i+1]];t];k=30;data1=PrimitivePythagoreanTriplets[2k^2+2k+1];data2=Select[data1,#[[1]]<=2k+1 &];1/2(#[[1]]+#[[2]]-#[[3]]) &/@data2

%o (Haskell)

%o a156689 n = a156689_list !! (n-1)

%o a156689_list = f 1 1 where

%o    f u v | v > uu `div` 2        = f (u + 1) (u + 2)

%o          | gcd u v > 1 || w == 0 = f u (v + 2)

%o          | otherwise             = (u + v - w) `div` 2 : f u (v + 2)

%o          where uu = u ^ 2; w = a037213 (uu + v ^ 2)

%o -- _Reinhard Zumkeller_, Nov 09 2012

%Y Cf. A020884, A020888, A156678, A156679.

%Y Cf. A037213.

%K easy,nice,nonn

%O 1,2

%A _Ant King_, Feb 18 2009