A156682 - OEIS

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, 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

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

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.

LINKS

Table of n, a(n) for n=1..62.

Ron Knott, Right-angled Triangles and Pythagoras' Theorem

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

Cf. A009004, A156681, A156679, A020882, A009003.

Sequence in context: A210694 A298316 A299190 * A089534 A116981 A222756

Adjacent sequences: A156679 A156680 A156681 * A156683 A156684 A156685

KEYWORD

easy,nice,nonn

AUTHOR

Ant King, Feb 17 2009

STATUS

approved