

A198459


Consider triples a<=b<c where (a^2+b^2c^2)/(cab) =2, ordered by a and then b; sequence gives b values.


0



6, 4, 16, 10, 8, 30, 18, 14, 48, 12, 28, 70, 18, 40, 16, 30, 96, 25, 54, 22, 40, 126, 20, 33, 70, 160, 26, 42, 88, 24, 64, 198, 52, 108, 30, 78, 240, 28, 40, 63, 130, 54, 286, 34, 48, 75, 154, 32, 64, 110, 336, 88, 180, 38, 128, 390, 28, 36, 66, 102, 208, 448, 33, 42
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

The definition can be generalized to define Pythagorean ktriples a<=b<c where (a^2+b^2c^2)/(cab)=k, or where for some integer k, a(a+k) + b(b+k) = c(c+k).
If a, b and c form a Pythagorean ktriple, then na, nb and nc form a Pythagorean nktriple.
A triangle is defined to be a Pythagorean ktriangle if its sides form a Pythagorean ktriple.
If a, b and c are the sides of a Pythagorean ktriangle 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 ktriangle ABC with a<=b<c equals sqrt((2ab)^2(k(a+bc))^2))/4.


REFERENCES

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104134.


LINKS



EXAMPLE

3*5 + 6*8 = 7*9
4*6 + 4*6 = 6*8
5*7 + 16*17 = 17*18
6*8 + 10*12 12*14
7*9 + 8*10 = 11*13
7*9 + 30*32 = 31*33


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



STATUS

approved



