A335035 Ordered perimeters of primitive integer triangles with two perpendicular medians.

54, 70, 104, 154, 170, 216, 252, 266, 352, 368, 418, 442, 464, 594, 598, 620, 638, 720, 740, 748, 792, 810, 902, 952, 962, 988, 1054, 1102, 1118, 1134, 1148, 1170, 1216, 1274, 1316, 1376, 1426, 1484, 1512, 1564, 1568, 1598, 1600, 1638, 1702, 1710, 1802, 1836, 1862

Offset

1,1

Comments

The study of these integer triangles that have two perpendicular medians was proposed in the problem of Concours Général in 2007 in France (see link).

If medians drawn from A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2 (see Maths Challenge picture in link).

All terms are even because each triple is composed of one even side and two odd sides.

For the corresponding primitive triples and miscellaneous properties, see A335034.

Links

Table of n, a(n) for n=1..49.

Annales Concours Général, Sujet Concours Général 2007

Maths Challenge, Perpendicular medians, Problem with picture.

Formula

a(n) = A335036(n) + A335347(n) + A335348(n).

Example

a(1) = 13 + 19 + 22 = 54 with 19^2 + 22^2 = 5 * 13^2 = 845.

Prog

(PARI) lista(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, vecsum(vt))); ); ); ); vecsort(vm); } \\ Michel Marcus, May 27 2020

Crossrefs

Cf. A024364 (perimeters of primitive Pythagorean triangles).

Cf. A335034 (corresponding primitive triples), A335036 (smallest side), A335347 (middle side), A335348 (largest side), A335273 (even side).

Sequence in context: A025331 A025323 A157934 * A350886 A281920 A005129

Adjacent sequences: A335032 A335033 A335034 * A335036 A335037 A335038

Keyword

nonn

Author

Bernard Schott, May 27 2020

Status

approved