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A198468
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Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = -1, ordered by a and then b; sequence gives c values.
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1
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4, 7, 11, 9, 16, 12, 22, 29, 14, 37, 17, 24, 46, 29, 56, 19, 24, 67, 22, 28, 79, 47, 92, 21, 24, 37, 54, 106, 27, 42, 121, 137, 29, 53, 78, 154, 32, 59, 87, 172, 41, 50, 191, 34, 45, 55, 72, 211, 37, 42, 79, 117, 232, 128, 254, 39, 94, 277, 42, 63, 102, 301
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OFFSET
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1,1
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COMMENTS
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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.
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
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LINKS
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EXAMPLE
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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