A335273 -id:A335273 - cp's OEIS frontend

A335034 Primitive triples, in nondecreasing order of perimeter, for integer-sided triangles with two perpendicular medians; each triple is in increasing order, and if perimeters coincide then increasing order of the smallest side.

13, 19, 22, 17, 22, 31, 25, 38, 41, 37, 58, 59, 41, 58, 71, 53, 62, 101, 61, 82, 109, 65, 79, 122, 85, 118, 149, 89, 121, 158, 101, 139, 178, 109, 122, 211, 113, 142, 209, 145, 178, 271, 145, 191, 262, 149, 229, 242, 157, 179, 302, 173, 269, 278, 181, 218, 341

Offset: 1

Bernard Schott, May 20 2020

The study of these integer triangles that have two perpendicular medians was proposed in the problem of Concours Général in 2007 in France (see links).
If medians drawn from A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2, hence c is always the smallest side.
Some theoretical results and geometrical properties:
-> 1/2 < a/b < 2 and 1 < a/c < 2 (also 1 < b/c < 2).
-> a, b, c are pairwise coprimes.
-> a et b have different parities, so c is always odd.
-> a and b are not divisible nor by 3 nor by 4 neither by 5.
-> The odd prime factors of the even term a' (a' = a or b) are all of the form 10*k +- 1 (see formula for a').
-> The prime factors of the largest odd side b' (b' = a or b) are all of the form 10*k +- 1 (see formula)
-> Consequence: in each increasing triple (c,a,b), c is always the smallest odd side, but a can be either the even side or the largest odd side (see formulas and examples for explanations).
-> cos(C) >= 4/5 (or tan(C) <= 3/4), hence C <= 36.86989...° = A235509 (see Maths Challenge link with picture).
-> CG = c (see Mathematics Stack Exchange link).
-> Area(ABC) = (2/3) * m_a * m_b with m_a (resp. m_b) is the length of median AA' (resp. BB') (see Mathematics Stack Exchange link).
-> cot(A) + cot(B) >= 2/3 (see IMO Compendium link and Doob reference). - Bernard Schott, Dec 02 2021

			For 1st class, u/v > 3: (u,v) = (4,1), then (c,a,b) = (c,a',b') = (17,22,31) is the second triple and 22^2 + 31^2 = 5 * 17^2 = 1445.
For 2nd class, 1 < u/v < 2: (u,v) = (3,2), then (c,a,b) = (c,b',a') = (13,19,22) is the first triple and 19^2 + 22^2 = 5 * 13^2 = 845.

Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 3, 1993, page 253, 1993.

Annales Concours Général, Corrigé Concours Général 2007.
Annales Concours Général, Sujet Concours Général 2007.
The IMO Compendium, Problem 3, 25-th Canadian Mathematical Olympiad 1993.
Maths Challenge, Perpendicular medians (Problem with picture).
Mathematics Stack Exchange, Perpendicular medians (Proves that AB = GC).
Mathematics Stack Exchange, Solving area of a triangle where medians are perpendicular.

Cf. A103606 (primitive Pythagorean triples), A235509 = arccos(4/5).

Cf. A335035 (corresponding perimeters), A335036 (smallest side and 1st trisection), A335347 (middle side and 2nd trisection), A335348 (largest side and 3rd trisection), A335273 (even side).

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); 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);} \\ Michel Marcus, May 21 2020

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.
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.
For n>=1, a(3n-2) = A335036(n), a(3n-1) = A335347(n), a(3n) = A335348(n).

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

Original entry on oeis.org

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

Bernard Schott, May 27 2020

nonn

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.

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

Annales Concours Général, Sujet Concours Général 2007
Maths Challenge, Perpendicular medians, Problem with picture.

Cf. A024364 (perimeters of primitive Pythagorean triangles).

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

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

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

A335036 Smallest side c of the primitive triples (c,a,b) for integer triangles that have two perpendicular medians, ordered by increasing perimeter.

Original entry on oeis.org

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

Bernard Schott, May 28 2020

nonn

If medians drawn from A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2, hence c is always the smallest odd side (see link Maths Challenge).
c = u^2 + v^2 for some u and v (see formula), so this sequence is subsequence of A004431.
For the corresponding primitive triples and miscellaneous properties, see A335034.
The repetitions for 145, 365, 377,... correspond to smallest sides for triangles with distinct perimeters (see examples).
This sequence is not increasing a(30) = 281 for triangle with perimeter = 1134 and a(31) = 277 for triangle with perimeter = 1148. The smallest side is not an increasing function of the perimeter of these triangles.

			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.

Maths Challenge, Perpendicular medians, Problem with picture.

Subsequence of A004431.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335347 (middle side), A335348 (largest side), A335273 (even side).

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

a(n) = A335034(3n-2).
a(n) = A335035(n) - A335347(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; 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.

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

Original entry on oeis.org

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

Bernard Schott, Jun 03 2020

nonn

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.

			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'.

Maths Challenge, Perpendicular medians, Problem with picture.

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

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

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').

A335348 Largest side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

Original entry on oeis.org

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

Offset: 1

Bernard Schott, Jun 06 2020

nonn

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.

			-> 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'.

Maths Challenge, Perpendicular medians, Problem with picture.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335347 (middle side), A335273 (even side).

Cf. A081804 (similar, with hypotenuse for primitive Pythagorean triples).

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

a(n) = A335034(3n).
a(n) = A335035(n) - A335036(n) - A335347(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').

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

Original entry on oeis.org

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

Bernard Schott, Jun 11 2020

nonn

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.

			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.

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

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

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

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').

More terms from Michel Marcus, Jun 11 2020

cp's OEIS Frontend

A335034 Primitive triples, in nondecreasing order of perimeter, for integer-sided triangles with two perpendicular medians; each triple is in increasing order, and if perimeters coincide then increasing order of the smallest side.

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

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A335036 Smallest side c of the primitive triples (c,a,b) for integer triangles that have two perpendicular medians, ordered by increasing perimeter.

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

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A335348 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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A335418 Largest odd side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

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