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A198461
Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = 3, ordered by a and then b; sequence gives a, b and c values in that order.
2
2, 3, 4, 3, 7, 8, 4, 12, 13, 5, 18, 19, 6, 6, 9, 6, 11, 13, 6, 25, 26, 7, 15, 17, 7, 33, 34, 8, 42, 43, 9, 10, 14, 9, 15, 18, 9, 52, 53, 10, 30, 32, 10, 63, 64, 11, 36, 38, 11, 75, 76, 12, 14, 19, 12, 19, 23, 12, 27, 30, 12, 88, 89, 13, 102, 103, 14, 57, 59, 14, 117, 118
OFFSET
1,1
COMMENTS
See A198453.
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
EXAMPLE
2*5 + 3*6 = 4*7
3*6 + 7*10 = 8*11
4*7 + 12*15 = 13*16
5*8 + 18*21 = 19*22
6*9 + 6*9 = 9*12
6*9 + 11*14 = 13*16
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,tabf
AUTHOR
Charlie Marion, Nov 26 2011
STATUS
approved