|
| |
|
|
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
|
|
|
|
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
|
|
|
|
FORMULA
|
|
|
|
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)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nice,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
