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A198465
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, b and c values in that order.
0
3, 3, 4, 4, 6, 7, 5, 10, 11, 6, 7, 9, 6, 15, 16, 7, 10, 12, 7, 21, 22, 8, 28, 29, 9, 11, 14, 9, 36, 37, 10, 14, 17, 10, 22, 24, 10, 45, 46, 11, 27, 29, 11, 55, 56, 12, 15, 19, 12, 21, 24, 12, 66, 67, 13, 18, 22, 13, 25, 28, 13, 78, 79, 14, 45, 47, 14, 91, 92
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). See A198453 for more about Pythagorean k-triples.
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
EXAMPLE
3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15
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, Dec 19 2011
STATUS
approved