A335418 Largest odd side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

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

Offset: 1

Author

Bernard Schott, Jun 11 2020

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• Annales Concours Général, Sujet Concours Général 2007.
• Maths Challenge, Perpendicular medians, Problem with picture.
• Index to sequences related to Olympiads and other Mathematical competitions.

Crossrefs

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335347 (middle side), A335348 (largest side), A335273 (even side), this sequence (largest odd side).

Programs

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

Formula

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').

Extensions

More terms from Michel Marcus, Jun 11 2020