A156689 - OEIS

A156689

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).

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, 14, 17, 18, 15, 19, 19, 20, 21, 18, 21, 22, 23, 15, 23, 24, 21, 25, 22, 25, 26, 27, 27, 24, 28, 29, 21, 26, 29, 30

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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.

REFERENCES

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. 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.

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Ron Knott, Right-angled Triangles and Pythagoras' Theorem

FORMULA

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

EXAMPLE

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.

MATHEMATICA

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

PROG

(Haskell)

a156689 n = a156689_list !! (n-1)

a156689_list = f 1 1 where

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

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

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

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

-- Reinhard Zumkeller, Nov 09 2012

CROSSREFS

Cf. A020884, A020888, A156678, A156679.

Cf. A037213.