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A277269
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Hypotenuses of Pythagorean triples, generated by a variation of Euclid's formula.
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1
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5, 10, 13, 17, 10, 25, 26, 29, 34, 41, 37, 20, 15, 26, 61, 50, 53, 58, 65, 74, 85, 65, 34, 73, 20, 89, 50, 113, 82, 85, 30, 97, 106, 39, 130, 145, 101, 52, 109, 58, 25, 68, 149, 82, 181, 122, 125, 130, 137, 146, 157, 170, 185, 202, 221, 145, 74, 51, 40, 169, 30, 75, 122, 265, 170, 173, 178, 185, 194, 205, 218, 233, 250, 269, 290, 313, 197, 100, 205, 106, 221, 116, 35, 130, 277, 148, 317, 170, 365, 226, 229
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OFFSET
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1,1
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COMMENTS
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Take two positive integers, x > y. As shown in the referenced faux art, you can form a vector using the integers as the coordinates, and repeat that vector and its equal-length normal so that you get exactly to the x-axis. Now you can mirror the pattern: take the same number of normal vectors but in the opposite direction. You get an isosceles triangle and the equal sides represent a Pythagorean triple.
Let s = gcd(x,y). This is the scaling factor -- you divide x and y by it and get coprime x and y. The symmetry axis goes from (0,0) to (xx,xy). The first normal goes from (xx,xy) to (xx+yy,0). The second normal goes from (xx,xy) to (xx-yy,xy+xy). So x^2+y^2 is the hypotenuse of the triangle with catheti x^2-y^2 and 2xy. Scale these with s and you get the triple corresponding to the parameters. In the examples the hypotenuse will be called P(x,y).
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LINKS
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Juhani Heino, Faux art showing the motivation of this.
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EXAMPLE
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Triangle with each row r going from P(r+1,1) to P(r+1,r):
P(2,1)=5;
P(3,1)=10, P(3,2)=13;
P(4,1)=17, P(4,2)=2*P(2,1)=10, P(4,3)=25;
P(5,1)=26, P(5,2)=29, P(5,3)=34, P(5,4)=41;
P(6,1)=37, P(6,2)=2*P(3,1)=20, P(6,3)=3*P(2,1)=15, P(6,4)=2*P(3,2)=26, P(6,5)=61;
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PROG
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(PARI)
p(x, y) = x^2 + y^2
out=""
for (row = 1, 15, for (col = 1, row, s=gcd(row+1, col); out = Str(out, s * p((row+1)/s, col/s), ", ") ))
print(out);
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CROSSREFS
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When results are ordered and doubles removed, we should get A009003.
A222946 is similar but omits non-primitive triples (gives 0 for them).
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KEYWORD
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AUTHOR
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STATUS
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approved
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