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A198467
Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = -1, ordered by a and then b; sequence gives b values.
0
3, 6, 10, 7, 15, 10, 21, 28, 11, 36, 14, 22, 45, 27, 55, 15, 21, 66, 18, 25, 78, 45, 91, 15, 19, 34, 52, 105, 22, 39, 120, 136, 23, 50, 76, 153, 26, 56, 85, 171, 36, 46, 190, 27, 40, 51, 69, 210, 30, 36, 76, 115, 231, 126, 253, 31, 91, 276, 34, 58, 99, 300
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