Ron Knott, <a href="http://www.mcsmaths.surrey.ac.uk/Personalhosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>

editing

approved

Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =3, ordered by a and then b; sequence gives a values.

Ron Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>

Cf. A103606, A198454-A198472A198469.

approved

editing

_Charlie Marion (charliemath(AT)optonline.net), _, Nov 26 2011

proposed

approved

editing

proposed

allocated for Charlie Marion Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =3, ordered by a and then b; sequence gives a values.

2, 3, 4, 5, 6, 6, 6, 7, 7, 8, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 13, 14, 14, 15, 15, 15, 15, 15, 16, 17, 17, 17, 18, 18, 18, 18, 18, 19, 19, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 24, 25, 25, 25, 25, 26, 26, 27, 27, 27, 27, 27, 28

1,1

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.

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

Ron Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>

2*5 + 3*6 = 4*7

3*6 + 7*10 = 8*11

4*7 +12*15 = 13*16

5*8 + 18*21 = 19*22

6*9 = 6*9 = 9*12

6*9 = 11*14 = 13*16

(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

Cf. A103606, A198454-A198472.

allocated

nonn

Charlie Marion (charliemath(AT)optonline.net), Nov 26 2011

approved

editing

allocated for Charlie Marion

allocated

approved