A335036 Smallest side c of the primitive triples (c,a,b) for integer triangles that have two perpendicular medians, ordered by increasing perimeter.
13, 17, 25, 37, 41, 53, 61, 65, 85, 89, 101, 109, 113, 145, 145, 149, 157, 173, 181, 185, 193, 197, 221, 229, 233, 241, 257, 265, 269, 281, 277, 289, 293, 313, 317, 337, 349, 365, 365, 377, 377, 389, 397, 401, 409, 421, 433, 445, 461
Offset: 1
Keywords
Examples
The triples (145, 178, 271) and (145, 191, 262) correspond to triangles with respective perimeters equal to 594 and 598, so a(14) = a(15) = 145. The triples (365, 418, 701) and (365, 509, 638) correspond to triangles with respective perimeters equal to 1484 and 1512, so a(38) = a(39) = 365.
Links
- Maths Challenge, Perpendicular medians, Problem with picture.
Crossrefs
Programs
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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][1]);} \\ Michel Marcus, May 28 2020
Formula
a(n) = A335034(3n-2).
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; if c is the smallest odd side, then:
1st class of triangles: c = u^2+v^2 with u/v > 3 and 5 doesn't divide u-3v,
2nd class of triangles: c = u^2+v^2 with 1 < u/v < 2 and 5 doesn't divide u-2v.
Comments