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A335348
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Largest side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.
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6
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22, 31, 41, 59, 71, 101, 109, 122, 149, 158, 178, 211, 209, 271, 262, 242, 302, 278, 341, 361, 358, 362, 419, 382, 418, 449, 478, 439, 451, 551, 482, 562, 502, 599, 541, 638, 659, 701, 638, 649, 622, 718, 781, 758, 662, 811, 698, 802, 902
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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), hence c is always the smallest side.
For the corresponding primitive triples and miscellaneous properties, see A335034; such a triangle with sides of integer lengths cannot be isosceles.
The largest side b with c < a < b is not divisible by 3, 4, or 5, and the odd prime factors of this largest side term b are all of the form 10*k +- 1.
In each increasing triple (c,a,b), c is the smallest odd side (A335036), but the largest side b can be either the even side (A335273) or the largest odd side (see formulas and examples for explanations).
This sequence is not increasing: a(12) = 211 for triangle with perimeter = 442 and a(13) = 209 for triangle with perimeter = 464; hence the largest 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' (even) < b' and the triple in increasing order is (c, a = a', b = b'),
if u/v > 3+sqrt(10) then a' (even) > b' 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' (even) < b' and the triple in increasing order is (c, a = a', b = b'),
If (1+sqrt(10))/3 < u/v < 2 then a' (even) > b' and the triple in increasing order is (c, a = b', b = a').
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EXAMPLE
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-> 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) = 178 = 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) = 22 = 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]); }
lista(nn) = {my(vm = List(), vt, w); 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)); ); ); ); w = vecsort(apply(vecsort, Vec(vm)); , mycmp); vector(#w, k, w[k][3]); } \\ Michel Marcus, Jun 06 2020
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CROSSREFS
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Cf. A081804 (similar, with hypotenuse for primitive Pythagorean triples).
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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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