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A198454
Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =1, ordered by a and then b; sequence gives a values.
12
2, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 23, 23, 23, 24
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 Knottā€™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.
EXAMPLE
2*3 + 2*3 = 3*4
3*4 + 5*6 = 6*7
4*5 + 9*10 = 10*11
5*6 + 6*7 = 8*9
5*6 + 14*15 = 15*16
6*7 + 9*10 = 11*12
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
KEYWORD
nonn
AUTHOR
Charlie Marion, Oct 26 2011
STATUS
approved