login
A156682
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)).
5
5, 13, 10, 25, 17, 15, 41, 26, 61, 20, 37, 85, 50, 25, 39, 113, 34, 65, 145, 30, 82, 181, 29, 52, 101, 35, 75, 221, 122, 265, 40, 51, 74, 145, 65, 313, 170, 45, 123, 365, 53, 100, 197, 421, 50, 78, 226, 481, 68, 130, 257, 55, 65, 183, 545, 290, 91, 125, 613, 60, 85, 111
OFFSET
1,1
COMMENTS
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).
REFERENCES
Beiler, Albert H.: Recreations In The Theory Of Numbers, Chapter XIV, The Eternal Triangle, Dover Publications Inc., New York, 1964, pp. 104-134.
Sierpinski, W.; Pythagorean Triangles, Dover Publications, Inc., Mineola, New York, 2003.
FORMULA
a(n) = sqrt(A009004(n)^2 + A156681(n)^2).
EXAMPLE
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.
MATHEMATICA
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
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
Ant King, Feb 17 2009
STATUS
approved