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A335418
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Largest odd side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.
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1
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19, 31, 41, 59, 71, 101, 109, 79, 149, 121, 139, 211, 209, 271, 191, 229, 179, 269, 341, 361, 241, 251, 419, 341, 311, 449, 319, 439, 451, 551, 389, 319, 421, 599, 541, 401, 659, 701, 509, 649, 569, 491, 781, 479, 631, 811, 671, 589, 499, 761, 929, 571, 859, 739
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OFFSET
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1,1
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COMMENTS
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If medians at A and B are perpendicular at the centroid G, then a^2 + b^2 = 5 * c^2 (see picture in Maths Challenge link).
For the corresponding primitive triples and miscellaneous properties, see A335034.
In each increasing triple (c,a,b), c is always the smallest odd side (A335036), but the largest odd side b' can be either the middle side a (A335347) or the largest side b (A335348) (see formulas and examples for explanations).
The largest odd side b' is not divisible by 3 or 5, and the odd prime factors of this odd side b' are all of the form 10*k +- 1.
The repetitions for 319, 341 ... correspond to largest odd sides for triangles with distinct perimeters (see examples).
This sequence is not increasing: a(7) = 109 for triangle with perimeter = 252 and a(8) = 79 for triangle with perimeter = 266; hence the largest odd side is not an increasing function of the perimeter of these triangles.
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LINKS
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FORMULA
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There exist two disjoint classes of such triangles, obtained with two distinct families of formulas: let u > v > 0, u and v with different parities, gcd(u,v) = 1; a' is the even side and b' the largest odd side.
--> 1st class of triangles: (a',b',c) = (2*(u^2-uv-v^2), u^2+4*u*v-v^2, u^2+v^2) with u/v > 3 and 5 doesn't divide u-3v.
If 3 < u/v < 3+sqrt(10) then a' < b' (largest odd) and the triple in increasing order is (c, a = a', b = b'),
if u/v > 3+sqrt(10) then a' > b' (largest odd) and the triple in increasing order is (c, a = b', b = a').
--> 2nd class of triangles: (a',b',c) = (2*(u^2+uv-v^2), -u^2+4*u*v+v^2, u^2+v^2) with 1 < u/v < 2 and 5 doesn't divide u-2v.
If 1 < u/v < (1+sqrt(10))/3 then a' < b' (largest odd) and the triple in increasing order is (c, a = a', b=b').
If (1+sqrt(10))/3 < u/v < 2 then a' > b' (largest odd) and the triple in increasing order is (c, a = b', b = a').
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EXAMPLE
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The triples (257, 319, 478) and (289, 319, 562) correspond to triangles with respective perimeters equal to 1054 and 1170, so a(27) = a(32) = 319.
-> For 1st class of triangles, u/v > 3:
(u,v) = (4,1), then 3 < u/v < 3+sqrt(10) and (c,a,b) = (c, a', b') = (17,22,31); the relation is 22^2 + 31^2 = 5 * 17^2 = 1445 with a(2) = 31 = b' = b.
(u,v) = (10,1), then u/v > 3+sqrt(10) and (c,a,b) = (c, b' ,a') = (101, 139, 178), the relation is 139^2 + 178^2 = 5 * 101^2 = 51005 with a(11) = 139 = b' = a.
-> For 2nd class, 1 < u/v < 2:
(u,v) = (3,2), then (1+sqrt(10))/3 < u/v < 2 and (c,a,b) = (c, b', a') = (13,19,22), the relation is 19^2 + 22^2 = 5 * 13^2 = 845 with a(1) = 19 = b' = a.
(u,v) = (4,3), then 1 < u/v < (1+sqrt(10))/3 and (c,a,b) = (c, a', b') = (25,38,41); the relation is 38^2 + 41^2 = 5 * 25^2 = 3125 with a(3) = 41 = b' = b.
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PROG
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(PARI) mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]); }
triples(nn) = {my(vm = List(), vt); for (u=1, nn, for (v=1, nn, if (gcd(u, v) == 1, vt = 0; if ((u/v > 3) && ((u-3*v) % 5), vt = [2*(u^2-u*v-v^2), u^2+4*u*v-v^2, u^2+v^2]); if ((u/v > 1) && (u/v < 2) && ((u-2*v) % 5), vt = [2*(u^2+u*v-v^2), -u^2+4*u*v+v^2, u^2+v^2]); if (gcd(vt) == 1, listput(vm, vt)); ); ); ); vecsort(apply(vecsort, Vec(vm)), mycmp); } \\ A335034
lista(nn) = my(w=triples(nn)); vector(#w, k, vecmax(select(x->(x%2), w[k]))); \\ Michel Marcus, Jun 11 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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