A335347 Middle side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

19, 22, 38, 58, 58, 62, 82, 79, 118, 121, 139, 122, 142, 178, 191, 229, 179, 269, 218, 202, 241, 251, 262, 341, 311, 298, 319, 398, 398, 302, 389, 319, 421, 362, 458, 401, 418, 418, 509, 538, 569, 491, 422, 479, 631, 478, 671, 589, 499

Offset: 1

Author

Bernard Schott, Jun 03 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), 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 middle side a with c < a < b is not divisible by 3, 4, or 5, and the odd prime factors of this middle side term a are all of the form 10*k +- 1.

In each increasing triple (c,a,b), c is the smallest odd side (A335036), but the middle side a can be either the even side (A335273) or the largest odd side (see formulas and examples for explanations).

The consecutive repetitions for 58, 398, 418,... correspond to middle sides for triangles with distinct perimeters (see examples).

This sequence is not increasing: a(7) = 82 for triangle with perimeter = 252 and a(8) = 79 for triangle with perimeter = 266; hence the middle side is not an increasing function of the perimeter of these triangles.

Examples

			The triples (37, 58, 59) and (41, 58, 71) correspond to triangles with respective perimeters equal to 154 and 170, so a(4) = a(5) = 58.
--> 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) = 22 = a = a',
(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 = a = b'.
--> 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 = a = b'
(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) = 22 = a = a'.
		

Links

• Maths Challenge, Perpendicular medians, Problem with picture.

Crossrefs

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335348 (largest side), A335273 (even side).

Programs

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][2]); }  \\ Michel Marcus, Jun 03 2020

Formula

a(n) = A335034(3n-1).

a(n) = A335035(n) - A335036(n) - A335348(n).

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').