A198460 Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =2, ordered by a and then b; sequence gives c values.

7, 6, 17, 12, 11, 31, 20, 17, 49, 16, 30, 71, 22, 42, 21, 33, 97, 29, 56, 27, 43, 127, 26, 37, 72, 161, 32, 46, 90, 31, 67, 199, 56, 110, 37, 81, 241, 36, 46, 67, 132, 59, 287, 42, 54, 79, 156, 41, 69, 113, 337, 92, 182, 47, 131, 391, 40, 46, 72, 106, 210, 449, 45, 52

Offset

1,1

Comments

The definition can be generalized to define Pythagorean k-triples a<=b<c where (a^2+b^2-c^2)/(c-a-b)=k, or where for some integer k, a(a+k) + b(b+k) = c(c+k).

If a, b and c form a Pythagorean k-triple, then na, nb and nc form a Pythagorean nk-triple.

A triangle is defined to be a Pythagorean k-triangle if its sides form a Pythagorean k-triple.

If a, b and c are the sides of a Pythagorean k-triangle ABC with a<=b<c, then cos(C) = -k/(a+b+c+k) which proves that such triangles must be obtuse when k>0 and acute when k<0. When k=0, the triangles are Pythagorean, as in the Beiler reference and Ron Knott’s link. For all k, the area of a Pythagorean k-triangle ABC with a<=b<c equals sqrt((2ab)^2-(k(a+b-c))^2))/4.

References

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Links

Table of n, a(n) for n=1..64.

Ron Knott, Pythagorean Triples and Online Calculators

Example

3*5 + 6*8 = 7*9
4*6 + 4*6 = 6*8
5*7 + 16*17 = 17*18
6*8 + 10*12 12*14
7*9 + 8*10 = 11*13
7*9 + 30*32 = 31*33

Prog

(True BASIC)
input k
for a = (abs(k)-k+4)/2 to  40
for b = a to (a^2+abs(k)*a+2)/2
  let t = a*(a+k)+b*(b+k)
   let c =int((-k+ (k^2+4*t)^.5)/2)
    if c*(c+k)=t then print a; b; c,
next b
print
next a
end

Crossrefs

Cf. A103606, A198454-A198469.

Sequence in context: A309624 A078323 A099255 * A215334 A298377 A299244

Adjacent sequences: A198457 A198458 A198459 * A198461 A198462 A198463

Keyword

nonn

Author

Charlie Marion, Nov 15 2011

Status

approved