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A198455
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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 b values.
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2
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2, 5, 9, 6, 14, 9, 20, 27, 10, 35, 13, 21, 44, 26, 54, 14, 20, 65, 17, 24, 77, 44, 90, 14, 18, 33, 51, 104, 21, 38, 119, 135, 22, 49, 75, 152, 25, 55, 84, 170, 35, 45, 189, 26, 39, 50, 68, 209, 29, 35, 75, 114, 230, 125
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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).
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.
The definition amounts to saying that T_a+T_b=T_c where T_i denotes a triangular number (A000217). - N. J. A. Sloane, Apr 01 2020
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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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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
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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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