A335893 -id:A335893 - cp's OEIS frontend

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A336750 Triples for integer-sided triangles whose sides a < b < c are in arithmetic progression.

2, 3, 4, 3, 4, 5, 3, 5, 7, 4, 5, 6, 4, 6, 8, 5, 6, 7, 4, 7, 10, 5, 7, 9, 6, 7, 8, 5, 8, 11, 6, 8, 10, 7, 8, 9, 5, 9, 13, 6, 9, 12, 7, 9, 11, 8, 9, 10, 6, 10, 14, 7, 10, 13, 8, 10, 12, 9, 10, 11, 6, 11, 16, 7, 11, 15, 8, 11, 14, 9, 11, 13, 10, 11, 12, 7, 12, 17, 8, 12, 16

Offset: 1

Bernard Schott, Aug 03 2020

The triples are displayed in increasing order of perimeter, and if perimeters coincide then by increasing order of the smallest side, hence, each triple (a, b, c) is in increasing order.
Equivalently: triples of integer-sided triangles such that b = (a+c)/2 with a < c.
As the perimeter of these triangles = 3*b, the triples are also displayed in increasing order of middle side.
When a < b < c are in arithmetic progression with b - a = c - b = x, then 1 <= x <= floor((b-1)/2), hence, there exist for each side b >= 3, floor((b-1)/2) = A004526(b) triangles whose sides a < b < c are in arithmetic progression.
The only right integer-sided triangles such that a < b < c are in arithmetic progression correspond to the Pythagorean triples (3k, 4k, 5k) with k > 0.
There do not exist triangles whose sides a < b < c and angles A < B < C are both in arithmetic progression.
Three geometrical properties about these triangles, even if they are not integer-sided:
   1) tan(A/2) * tan(C/2) = 1/3,
   2) r = h_b/3, where r is the inradius and h_b the length of the altitude through B,
   3) The line (IG) is parallel to side (AC), where I is the incenter and G is the centroid of the triangle.

			The smallest such triangle is (2, 3, 4).
The only triangle with perimeter = 12 corresponds to the Pythagorean triple: (3, 4, 5).
There exist two triangles with perimeter = 15 corresponding to triples (3, 5, 7) and (4, 5, 6).
There exist also two triangles with perimeter = 18 corresponding to triples (4, 6, 8) and (5, 6, 7).
The table begins:
  2, 3, 4;
  3, 4, 5;
  3, 5, 7;
  4, 5, 6;
  4, 6, 8;
  5, 6, 7;
  4, 7, 10;
  5, 7, 9;
  6, 7, 8;

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-288 p. 120, André Desvigne.

Michel Marcus, Table of n, a(n) for n = 1..16650 (Rows 1 to 5550)
Wikipedia, Integer triangle.

Cf. A336751 (smallest side), A307136 (middle side), A336753 (largest side), A336754 (perimeter), A024164 (number of triangles with perimeter = n), A336755 (primitive triples), A336756 (perimeter of primitive triangles), A336757 (number of primitive triangles with perimeter = n).
Cf. A004526 (number of triangles with middle side = b).
Cf. A103605 (similar, with Pythagorean triples).
Cf. A335893 (similar, with A, B, C in arithmetic progression).

for b from 3 to 20 do
for a from b-floor((b-1)/2) to b-1 do
c := 2*b - a;
print(a,b,c);
end do;
end do;

Block[{nn = 12, a, b, c}, Reap[Do[Do[Sow@ {a, b, 2 b - a}, {a, b - Floor[(b - 1)/2], b - 1}], {b, 3, nn}]][[-1, 1]] ] // Flatten (* Michael De Vlieger, Oct 15 2020 *)

tabf(nn) = {for (b = 3, nn, for (a = b-floor((b-1)/2), b-1, my(c = 2*b - a); print(a, " ", b, " ", c);););} \\ Michel Marcus, Sep 08 2020

T(n,1) = A336751(n); T(n,2) = A307136(n); T(n,3) = A336753(n).

A336754(n) = T(n,1) + T(n,2) + T(n,3).

A335895 Middle side of primitive triples, in nondecreasing order, for integer-sided triangles whose angles A < B < C are in arithmetic progression.

Original entry on oeis.org

7, 7, 13, 13, 19, 19, 31, 31, 37, 37, 43, 43, 49, 49, 61, 61, 67, 67, 73, 73, 79, 79, 91, 91, 91, 91, 97, 97, 103, 103, 109, 109, 127, 127, 133, 133, 133, 133, 139, 139, 151, 151, 157, 157, 163, 163, 169, 169, 181, 181, 193, 193, 199, 199, 211, 211, 217, 217, 217, 217

Offset: 1

Bernard Schott, Jul 04 2020

nonn

Equivalently, lengths of the middle side b of primitive non-equilateral triangles that have an angle of Pi/3; indeed, this side is opposite to angle B = Pi/3.
Also solutions b of the Diophantine equation b^2 = a^2 - a*c + c^2 with a < b and gcd(a,b) = 1.
For the corresponding primitive triples and miscellaneous properties and references, see A335893.
When (a, b, c) is a triple or a solution, then (c-a, b, c) is another solution, so every b in the data is present an even number of times (see examples).
From Bernard Schott, Apr 02 2021: (Start)
Terms are primes of the form 6k+1, or products of primes of the form 6k+1. Three observations:
-> The lengths b are in A004611 \ {1} without repetition, 1 corresponds to the equilateral triangle (1, 1, 1).
-> Every term appears 2^k (k>0) times consecutively and the smallest term that appears 2^k times is precisely A121940(k); see examples.
-> The terms that appear precisely twice consecutively are in A133290. (End)

			b = 7 appears twice because A121940(1) = 7 and:
  7^2 = 3^2 - 3*8 + 8^2, with triple (3, 7, 8),
  7^2 = 5^2 - 5*8 + 8^2, with triple (5, 7, 8).
b = 91 appears four times because A121940(2) = 91 and:
  91^2 = 11^2 - 11*96 + 96^2, with triple (11, 91, 96),
  91^2 = 85^2 - 85*96 + 96^2, with triple (85, 91, 96),
  91^2 = 19^2 - 19*99 + 99^2, with triple (19, 91, 99),
  91^2 = 80^2 - 80*99 + 99^2, with triple (80, 91, 99).
b = 1729 appears eight times because A121940(3) = 1729 and the triples of these 2^3 = 8 triangles are (96, 1729, 1775), (1679, 1729, 1775), (249, 1729, 1840), (1591, 1729, 1840), (656, 1729, 1961), (1305, 1729, 1961), (799, 1729, 1984), (1185, 1729, 1984).

Cf. A004611, A121940, A133290.

Cf. A335893 (triples), A335894 (smallest side), this sequence (middle side), A335896 (largest side), A335897 (perimeter).

for b from 3 to 100 by 2 do
for a from 1 to b-1 do
c := (a+ sqrt(4*b^2-3*a^2))/2;
if igcd(a, b) = 1 and issqr(4*b^2-3*a^2) then print(b); end if;
end do;
end do;

lista(nn) = {forstep(b=1, nn, 2, for(a=1, b-1, if (gcd(a, b) == 1, my(d = 4*b^2 - 3*a^2); if (issquare(d), my(c = (a + sqrtint(d))/2); if (denominator(c)==1, print1(b, ", "));););););} \\ Michel Marcus, Jul 05 2020

a(n) = A335893(n, 2).

b is such that b^2 = a^2 - a*c + c^2 with gcd(a,b) = 1 and a < b.

A335894 Smallest side of integer-sided primitive triangles whose angles A < B < C are in arithmetic order.

Original entry on oeis.org

3, 5, 7, 8, 5, 16, 11, 24, 7, 33, 13, 35, 16, 39, 9, 56, 32, 45, 17, 63, 40, 51, 11, 85, 19, 80, 55, 57, 40, 77, 24, 95, 13, 120, 23, 120, 65, 88, 69, 91, 56, 115, 25, 143, 75, 112, 15, 161, 104, 105, 32, 175

Offset: 1

Bernard Schott, Jul 15 2020

nonn

The triples of sides (a,b,c) with a < b < c are in nondecreasing order of middle side, and if middle sides coincide, then by increasing order of the largest side, and when largest sides coincide, then by increasing order of the smallest side (see last example). This sequence lists the a's.
Equivalently, lengths of the smallest side a of primitive non-equilateral triangles that have an angle of Pi/3; indeed, this side is opposite to the smallest angle A.
Also, solutions a of the Diophantine equation b^2 = a^2 - a*c + c^2 with gcd(a,b) = 1 and a < b.
For the corresponding primitive triples and miscellaneous properties and references, see A335893.
When (a, b, c) is a triple with a < c/2, then (c-a, b, c) is the following triple because if b^2 = a^2 - a*c + c^2 then also b^2 = (c-a)^2 - (c-a)*c + c^2; hence, for each pair (b,c), there exist two distinct triangles whose smallest sides a_1 and a_2 satisfy a_1 + a_2 = c (see first example).

			For the pair b = 7, c = 8 the two corresponding values of a are 3 and 5 with 3 + 5 = 8 = c because:
    7^2 = 3^2 - 3*8 + 8^2, with triple (3, 7, 8),
    7^2 = 5^2 - 5*8 + 8^2, with triple (5, 7, 8).
For b = 91, there exist four corresponding values of a, two for b = 91 and c = 96 that are 11 and 85 with 11 + 85 = 96 = c, and two for b = 91 and c = 99 that are 19 and 80 with 19 + 80 = 99 = c; also these four smallest sides are ordered 11, 85, 19, 80 in the data because:
    91^2 = 11^2 - 11*96 + 96^2, with triple (11, 91, 96),
    91^2 = 85^2 - 85*96 + 96^2, with triple (85, 91, 96),
    91^2 = 19^2 - 19*99 + 99^2, with triple (19, 91, 99),
    91^2 = 80^2 - 80*99 + 99^2, with triple (80, 91, 99).

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.

Cf. A335893 (triples), A335895 (middle side), A335896 (largest side), A335897 (perimeter).

for b from 3 to 250 by 2 do
for c from b+1 to 6*b/5 do
a := (c - sqrt(4*b^2-3*c^2))/2;
if gcd(a,b)=1 and issqr(4*b^2-3*c^2) then print(a,c-a); end if;
end do;
end do;

lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print1(a, ", ", c-a, ", "); ); ); ); ); } \\ Michel Marcus, Jul 16 2020

a(n) = A335893(n, 1).

a is such that a^2 - c*a + c^2 - b^2 = 0 with gcd(a,b) = 1 and a < b.

A335896 Largest side of integer-sided primitive triangles whose angles A < B < C are in arithmetic order.

Original entry on oeis.org

8, 8, 15, 15, 21, 21, 35, 35, 40, 40, 48, 48, 55, 55, 65, 65, 77, 77, 80, 80, 91, 91, 96, 96, 99, 99, 112, 112, 117, 117, 119, 119, 133, 133, 143, 143, 153, 153, 160, 160, 171, 171, 168, 168, 187, 187, 176, 176, 209, 209, 207, 207, 221, 221, 224, 224, 225, 225

Offset: 1

Bernard Schott, Jul 10 2020

nonn

The triples of sides (a,b,c) with a < b < c are in nondecreasing order of middle side, and if middle sides coincide then by increasing order of the largest side. This sequence lists the c's.
Equivalently, lengths of the largest side c of primitive non-equilateral triangles that have an angle of Pi/3; indeed, this side is opposite to the largest angle C.
Also, solutions c of the Diophantine equation b^2 = a^2 - a*c + c^2 with gcd(a,b) = 1 and a < b.
For the corresponding primitive triples and miscellaneous properties and references, see A335893.
When (a, b, c) is a triple, then (c-a, b, c) is another triple, so every c in the data is twice consecutively present according to the corresponding pair (b, c) (see examples).
As B = Pi/3 and C runs from Pi/3 to 2*Pi/3, sin(C) gets a maximum when C = Pi/2 with sin(C) = 1, hence, from law of sinus, b/sin(B) = c/sin(C), c < b/sin(Pi/3) = b * 2/sqrt(3) < 6*b/5. This bound is used in PARI and Maple programs.
This sequence is not increasing. For example, a(41) = a(42) = 171 for triangle with middle side = 151 while a(43) = a(44) = 168 for triangle with middle side = 157.

			c = 8 appears twice because:
  7^2 = 3^2 - 3*8 + 8^2, with triple (3, 7, 8),
  7^2 = 5^2 - 5*8 + 8^2, with triple (5, 7, 8).
c = 96 and c = 99 each appear twice associated with b = 91 because:
  91^2 = 11^2 - 11*96 + 96^2, with triple (11, 91, 96),
  91^2 = 85^2 - 85*96 + 96^2, with triple (85, 91, 96),
  91^2 = 19^2 - 19*99 + 99^2, with triple (19, 91, 99),
  91^2 = 80^2 - 80*99 + 99^2, with triple (80, 91, 99).

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.

Wikipedia, Law of sines.

Cf. A089025 (terms in increasing order without repetition).

Cf. A335893 (triples), A335894 (smallest side), A335895 (middle side), this sequence (largest side), A335897 (perimeter).

for b from 3 to 250 by 2 do
for c from b+1 to 6*b/5 do
a := (c - sqrt(4*b^2-3*c^2))/2;
if gcd(a,b)=1 and issqr(4*b^2-3*c^2) then print(c,c); end if;
end do;
end do;

lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print(c, ", ", c, ", ");););););} \\ Michel Marcus, Jul 15 2020

a(n) = A335893(n, 3).

c satisfies c^2 - a*c + a^2 - b^2 = 0 with gcd(a,b) = 1 and a < b.

A335897 Perimeters of primitive integer-sided triangles whose angles A < B < C are in arithmetic order.

Original entry on oeis.org

18, 20, 35, 36, 45, 56, 77, 90, 84, 110, 104, 126, 120, 143, 135, 182, 176, 189, 170, 216, 210, 221, 198, 272, 209, 270, 264, 266, 260, 297, 252, 323, 273, 380, 299, 396, 351, 374, 368, 390, 378, 437, 350, 468, 425, 462, 360, 506, 494, 495, 432, 575, 476, 585, 464, 630

Offset: 1

Bernard Schott, Jul 17 2020

nonn

The triples of sides (a,b,c) of A335893 with a < b < c are in nondecreasing order of middle side, and if middle sides coincide, then by increasing order of the largest side, and when largest sides coincide, then by increasing order of the smallest side. This sequence lists the sums a+b+c (see last example).
Equivalently, perimeters of primitive non-equilateral triangles that have an angle of Pi/3.
Also, sum a+b+c of the solutions of the Diophantine equation b^2 = a^2 - b*c + c^2 with gcd(a,b) = 1 and a < b.
For the corresponding primitive triples and miscellaneous properties and references, see A335893.
This sequence is not increasing. For example, a(8) = 90 for triangle with middle side = 31 while a(9) = 84 for triangle with middle side = 37.

			For b = 7 and c = 8, the two corresponding triangles satisfy:
   7^2 = 3^2 - 3*8 + 8^2, with triple (3, 7, 8) and perimeter = 18,
   7^2 = 5^2 - 5*8 + 8^2, with triple (5, 7, 8) and perimeter = 20.
For b = 91, there exist four corresponding triangles, two for b = 91 and c = 96 and two for b = 91 and c = 99; the four corresponding perimeters are ordered 198, 272, 209, 270 in the data because:
   91^2 = 11^2 -11*96 +96^2, with triple (11, 91, 96) and perimeter 11+91+96 = 198,
   91^2 = 85^2 -85*96 +96^2, with triple (85, 91, 96) and perimeter 85+91+96 = 272,
   91^2 = 19^2 -19*99 +99^2, with triple (19, 91, 99) and perimeter 19+91+99 = 209,
   91^2 = 80^2 -80*99 +99^2, with triple (80, 91, 99) and perimeter 80+91+99 = 270.

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.

Cf. A335893 (triples), A335894 (smallest side), A335895 (middle side), A335896 (largest side).

for b from 3 to 250 by 2 do
for c from b+1 to 6*b/5 do
a := (c - sqrt(4*b^2-3*c^2))/2;
if gcd(a,b)=1 and issqr(4*b^2-3*c^2) then print(a+b+c,2*c-a+b); end if;
end do;
end do;

lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print1(a+b+c, ", ", 2*c-a+b, ", "); ); ); ); ); } \\ Michel Marcus, Jul 17 2020

a(n) = A335893(n, 1) + A335893(n, 2) + A335893(n, 3).

a(n) = A335894(n) + A335895(n) + A335896(n).

A343063 Primitive triples (a, b, c) for integer-sided triangles whose angle B = 2*C.

Original entry on oeis.org

5, 6, 4, 7, 12, 9, 9, 20, 16, 11, 30, 25, 13, 42, 36, 15, 56, 49, 16, 15, 9, 17, 72, 64, 19, 90, 81, 21, 110, 100, 23, 132, 121, 24, 35, 25, 25, 156, 144, 27, 182, 169, 29, 210, 196, 31, 240, 225, 32, 63, 49, 33, 28, 16, 33, 272, 256, 35, 306, 289, 37, 342, 324, 39, 40, 25, 39, 380, 361, 40, 99, 81, 41, 420, 400, 43, 462, 441

Offset: 1

Bernard Schott, Apr 04 2021

This sequence is inspired by the problem of French Baccalauréat Mathématiques at Lyon in 1937 (see link).
The triples (a, b, c) are displayed in increasing order of side a, and if sides a coincide then in increasing order of the side b.
If in triangle ABC, B = 2*C, then the corresponding metric relation between sides is a*c + c^2 = c * (a + c) = b^2.
This metric relation is equivalent to a = m^2 - k^2, b = m * k, c = k^2, gcd(m,k) = 1 and k < m < 2k; hence every c is a square number.
When A <> 45° and A <> 72°, table below shows there exist these 3 possible inequalities: c < b < a, c < a < b, a < c < b.
   ------------------------------------------------------------------------
   | A | 180 |   decr.  |   72    |   decr.   |   45    |   decr.   |  0  |
   ------------------------------------------------------------------------
   | B |  0  |   incr.  |   72    |   incr.   |   90    |   incr.   | 120 |
   ------------------------------------------------------------------------
   | C |  0  |   incr.  |   36    |   incr.   |   45    |   incr.   |  60 |
   ------------------------------------------------------------------------
   | < | No | c < b < a | c < b=a | c < a < b | c=a < b | a < c < b |  No |
   ------------------------------------------------------------------------
where 'No' means there is no such corresponding triangle.
If (A,B,C) = (72,72,36) then a = b = c * (1+sqrt(5))/2 and isosceles ABC is not an integer-sided triangle.
If (A,B,C) = (45,90,45) then ABC is isosceles rectangle in B, so a = c with b = a*sqrt(2) and ABC is not an integer-sided triangle.

			The smallest such triangle is (5, 6, 4), of type c < a < b with 4*(5+4) = 6^2.
The 2nd triple is (7, 12, 9) of type a < c < b with 9*(7+9) = 16^2.
The 7th triple (16, 15, 9) is the first of type c < b < a with 9*(16+9) = 15^2.
The table begins:
   5,  6,  4;
   7, 12,  9;
   9, 20, 16;
  11, 30, 25,
  13, 42, 36;
  15, 56, 49;
  16, 15,  9;
  17, 72, 64;
  ...

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-336 p. 178, André Desvigne.

APMEP, Baccalauréat Mathématiques, Lyon, Septembre 1937.

Cf. A335893 (similar for A < B < C in arithmetic progression).

Cf. A343064 (side a), A343065 (side b), A343066 (side c), A343067 (perimeter).

for a from 2 to 60 do
for c from 3 to floor(a^2/2) do
d := c*(a+c);
if igcd(a,sqrt(d),c)=1 and issqr(d) and abs(a-c)

A339856 Primitive triples for integer-sided triangles whose sides a < b < c form a geometric progression.

Original entry on oeis.org

4, 6, 9, 9, 12, 16, 16, 20, 25, 25, 30, 36, 25, 35, 49, 25, 40, 64, 36, 42, 49, 49, 56, 64, 49, 63, 81, 49, 70, 100, 49, 77, 121, 64, 72, 81, 64, 88, 121, 81, 90, 100, 81, 99, 121, 81, 117, 169, 81, 126, 196, 100, 110, 121, 100, 130, 169, 121, 132, 144, 121, 143, 169

Offset: 1

Bernard Schott, Dec 19 2020

These triangles are called "geometric triangles" in Project Euler problem 370 (see link).
The triples are displayed in increasing lexicographic order (a, b, c).
Equivalently: triples of integer-sided triangles such that b^2 = a*c with a < c and gcd(a, c) = 1.
When a < b < c are in geometric progression with b = a*q, c = b*q, q is the constant, then 1 < q < (1+sqrt(5))/2 = phi = A001622 = 1.6180... (this bound is used in Maple code).
For each triple (a, b, c), there exists (r, s), 0 < r < s such that a = r^2, b = r*s, c = s^2, q = s/r.
Angle C < 90 degrees if 1 < q < sqrt(phi) and angle C > 90 degrees if sqrt(phi) < q < phi with sqrt(phi) = A139339 = 1.2720...
For k >= 2, each triple (a, b, c) of the form (k^2, k*(k+1), (k+1)^2) is (A008133(3k+1), A008133(3k+2), A008133(3k+3)).
Three geometrical properties about these triangles:
  1) The sinus satisfy sin^2(B) = sin(A) * sin(C) with sin(A) < sin(B) < sin(C) that form a geometric progression.
  2) The heights satisfy h_b^2 = h_a * h_c with h_c < h_b < h_a that form a geometric progression.
  3) b^2 = 2 * R * h_b, with R = circumradius of the triangle ABC.

			The smallest such triangle is (4, 6, 9) with 4*9 = 6^2.
There exist four triangles with small side = 49 corresponding to triples (49, 56, 64), (49, 63, 81), (49, 70, 100) and (49, 77, 121).
The table begins:
   4,  6,  9;
   9, 12, 16;
  16, 20, 25;
  25, 30, 36;
  25, 35, 49;
  25, 40, 64;
  36, 42, 49;
  ...

David A. Corneth, Table of n, a(n) for n = 1..10002
Project Euler, Problem 370: Geometric Triangles.

Cf. A339857 (smallest side), A339858 (middle side), A339859 (largest side), A339860 (perimeter).
Cf. A336755 (similar for sides in arithmetic progression).
Cf. A335893 (similar for angles in arithmetic progression).
Cf. A001622 (phi), A139339 (sqrt(phi)), A008133.

for a from 1 to 300 do
for b from a+1 to floor((1+sqrt(5))/2 * a) do
for c from b+1 to floor((1+sqrt(5))/2 * b) do
k:=a*c;
if k=b^2 and igcd(a,b,c)=1 then print(a,b,c); end if;
end do;
end do;
end do;

lista(nn) = {my(phi = (1+sqrt(5))/2); for (a=1, nn, for (b=a+1, floor(a*phi), for (c=b+1, floor(b*phi), if ((a*c == b^2) && (gcd([a,b,c])==1), print([a,b,c])););););} \\ Michel Marcus, Dec 25 2020

upto(n) = my(res=List(), phi = (sqrt(5)+1) / 2); for(i = 2, sqrtint(n), for(j = i+1, (i*phi)\1, if(gcd(i, j)==1, listput(res, [i^2, i*j, j^2])))); concat(Vec(res)) \\ David A. Corneth, Dec 25 2020

Data corrected by David A. Corneth, Dec 25 2020

A357277 Largest side c of primitive triples, in nondecreasing order, for integer-sided triangles with angles A < B < C = 2*Pi/3 = 120 degrees.

Original entry on oeis.org

7, 13, 19, 31, 37, 43, 49, 61, 67, 73, 79, 91, 91, 97, 103, 109, 127, 133, 133, 139, 151, 157, 163, 169, 181, 193, 199, 211, 217, 217, 223, 229, 241, 247, 247, 259, 259, 271, 277, 283, 301, 301, 307, 313, 331, 337, 343, 349, 361, 367, 373, 379, 397, 403, 403, 409, 421, 427, 427, 433, 439, 457

Offset: 1

Bernard Schott, Oct 01 2022

nonn

For the corresponding primitive triples and miscellaneous properties and references, see A357274.
Solutions c of the Diophantine equation c^2 = a^2 + a*b + b^2 with gcd(a,b) = 1 and a < b.
Also, side c can be generated with integers u, v such that gcd(u,v) = 1 and 0 < v < u, c = u^2 + u*v + v^2.
Some properties:
-> Terms are primes of the form 6k+1, or products of primes of the form 6k+1.
-> The lengths c are in A004611 \ {1} without repetition, in increasing order.
-> Every term appears 2^(k-1) (k>=1) times consecutively.
-> The smallest term that appears 2^(k-1) times is precisely A121940(k): see examples.
-> The terms that appear only once in this sequence are in A133290.
-> The terms are the same as in A335895 but frequency is not the same: when a term appears m times consecutively here, it appears 2m times consecutively in A335895. This is because if (a,b,c) is a primitive 120-triple, then both (a,a+b,c) and (a+b,b,c) are 60-triples in A335893 (see Emrys Read link, lemma 2 p. 302).
Differs from A088513, the first 20 terms are the same then a(21) = 151 while A088513(21) = 157.
A050931 gives all the possible values of the largest side c, in increasing order without repetition, for all triangles with an angle of 120 degrees, but not necessarily primitive.

			c = 7 appears once because A121940(1) = 7 with triple (3,5,7) and 7^2 = 3^2 + 3*5 + 5^2.
c = 91 is the smallest term to appear twice because A121940(2) = 91 with primitive 120-triples (11, 85, 91) and (19, 80, 91).
c = 1729 is the smallest term to appear four times because A121940(3) = 1729 with triples (96, 1679, 1729), (249, 1591, 1729), (656, 1305, 1729), (799, 1185, 1729).

Emrys Read, On integer-sided triangles containing angles of 120° or 60°, Mathematical Gazette 90, July 2006, 299-305.

Cf. A357274 (triples), A357275(smallest side), A357276 (middle side), A357278 (perimeter).

Cf. Also A004611, A050931, A088513, A121940, A133290, A335895.

for c from 5 to 500 by 2 do
for a from 3 to c-2 do
b := (-a + sqrt(4*c^2-3*a^2))/2;
if b=floor(b) and gcd(a, b)=1 and a

Formula

a(n) = A357274(n, 3).

A353618 Three-column array giving list of primitive triples for integer-sided triangles whose angle B = 3*C.

Original entry on oeis.org

3, 10, 8, 35, 48, 27, 119, 132, 64, 112, 195, 125, 279, 280, 125, 20, 357, 343, 253, 504, 343, 539, 510, 216, 552, 665, 343, 91, 792, 729, 923, 840, 343, 533, 840, 512, 476, 1035, 729, 1455, 1288, 512, 224, 1485, 1331, 1504, 1575, 729, 17, 1740, 1728, 799

Offset: 1

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Author

Bernard Schott, Apr 30 2022

Keywords

nonn
tabf

Comments

This sequence is inspired by the 1st problem proposed during the 46th Czech and Slovak Mathematical Olympiad 1997 (see link).

The triples (a, b, c) are displayed in increasing order of side b, and if sides b coincide then in increasing order of the side c.

If in triangle ABC, B = 3*C, then the corresponding metric relations between sides are c*a^2= (b-c)^2 * (b+c). <===> a/(b-c) = sqrt(1+b/c).

This metric relation is equivalent to a = m(m^2-2k^2), b = k(m^2-k^2), c = k^3, gcd(k,m) = 1 and sqrt(2) * k < m < 2*k; hence every c is a cube number and always c < b.

When A <> 3*Pi/7 and A <> Pi/5, table below shows there exist these 3 possible configurations: c < b < a; c < a < b and a < c < b:

----------------------------------------------------------------------------

| A | Pi | decr. | 3*Pi/7 | decr. | Pi/5 | decr. | 0 |

---------------------------------------------------------------------------

| B | 0 | incr. | 3*Pi/7 | incr. | 3*Pi/5 | incr. | 3*Pi/4 |

----------------------------------------------------------------------------

| C | 0 | incr. | Pi/7 | incr. | Pi/5 | incr. | Pi/4 |

----------------------------------------------------------------------------

| < | No | c < b < a | c < b=a | c < a < b | c=a < b | a < c < b | No |

----------------------------------------------------------------------------

where 'No' means there is no such corresponding triangle.

If (A,B,C) = (3*Pi/7,3*Pi/7,Pi/7) then a = b with c = 2*a*cos(Pi/7), so isosceles ABC is not an integer-sided triangle.

If (A,B,C) = (Pi/5,3*Pi/5,Pi/5) then a = c with b = a*(1+sqrt(5))/2, so ABC is not an integer-sided triangle.

Examples

The table begins: 3, 10, 8; 35, 48, 27; 119, 132, 64; 112, 195, 125; 279, 280, 125; 20, 357, 343; 253, 504, 343, 539, 510, 216; ................ The smallest such triangle is (3,10,8), it is of type a < c < b with 3/(10-8) = sqrt(1+10/8) = 3/2. The 2nd triple (35, 48, 27) is of type c < a < b with 35/(48-27) = sqrt(1+48/27) = 5/3. The 8th triple (539, 510, 216) is the first of type c < b < a with 539/(510-216) = sqrt(1+510/216) = 11/6.

Links

The IMO Compendium, Problem 1, 46th Czech and Slovak Mathematical Olympiad 1997.

Mathematical Reflections, Solution to problem O467, Issue 5, 2018, p. 26.

Index to sequences related to Olympiads.

Crossrefs

Cf. A335893 (A < B < C are in arithmetic progression), A343063 (B = 2*C).

Programs

Maple

for b from 1 to 2500 do for q from 2 to floor((b-1)^(1/3)) do a := (b-q^3) * sqrt(1+b/q^3); if a= floor(a) and q^3 < b and igcd(a,b,q)=1 and (b-q^3) < a and a < b+q^3 then print(a,b,q^3); end if; end do; end do;

PARI

lista(nn) = {for (b = 1, nn, for (q = 2, sqrtnint(b-1, 3), if (issquare(z=1+b/q^3), a = (b-q^3) * sqrtint(numerator(z))/sqrtint(denominator(z)); if ((q^3 < b) && (gcd([a, b, q]) == 1) && ((b-q^3) < a) && (a < b+q^3), print1([a, b, q^3], ", ")););););} \\ Michel Marcus, May 11 2022

Python

from math import gcd from itertools import count, islice from sympy import integer_nthroot def A353618_gen(): # generator of terms for b in count(1): q, c = 2, 8 while c < b: d = (b-c)**2*(b+c) s, t = divmod(d,c) if t == 0: a, r = integer_nthroot(s,2) if r and b-c < a < b+c and gcd(a,b,q) == 1: yield from (a, b, c) c += q*(3*q+3)+1 q += 1 A353618_list = list(islice(A353618_gen(),30)) # Chai Wah Wu, May 14 2022

A357274 List of primitive triples for integer-sided triangles with angles A < B < C and C = 2*Pi/3 = 120 degrees.

Original entry on oeis.org

3, 5, 7, 7, 8, 13, 5, 16, 19, 11, 24, 31, 7, 33, 37, 13, 35, 43, 16, 39, 49, 9, 56, 61, 32, 45, 67, 17, 63, 73, 40, 51, 79, 11, 85, 91, 19, 80, 91, 55, 57, 97, 40, 77, 103, 24, 95, 109, 13, 120, 127, 23, 120, 133, 65, 88, 133, 69, 91, 139, 56, 115, 151, 25, 143, 157, 75, 112, 163, 15, 161, 169, 104, 105, 181

Offset: 1

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Author

Bernard Schott, Sep 22 2022

Keywords

nonn
tabf

Comments

The only triangles with integer sides that have an angle equal to a whole number of degrees are triangles which have an angle of 60° (A335893), or an angle of 90° (A263728) or an angle of 120° as here (see Keith Selkirk link, p. 251).

The triples are displayed in nondecreasing order of largest side c, and if largest sides coincide then by increasing order of the smallest side a, hence, each triple (a, b, c) is in increasing order.

The corresponding metric relation between sides is c^2 = a^2 + a*b + b^2.

The triples (a, b, c) can be generated with integers u, v such that gcd(u,v) = 1 and 0 < v < u:

-> a = u^2 - v^2

-> b = 2*u*v + v^2

-> c = u^2 + u*v + v^2.

Note that side c cannot be even when the triple is primitive as here.

The (3, 5, 7) triangle is the only primitive triangle with a 120-degree angle and with its integer sides in arithmetic progression (A336750). This smallest triple is obtained for u = 2 and v = 1.

The Fermat point of these triangles is vertex C, then distance FA+FB+FC = CA+CB = b+a is an integer.

If (a,b,c) is a primitive 120-triple, then both (a,a+b,c) and (a+b,b,c) are 60-triples in A335893, see Emrys Read link, lemma 2 p. 302.

Examples

Table of triples begins: 3, 5, 7; 7, 8, 13; 5, 16, 19; 11, 24, 31; 7, 33, 37; ............ (7, 8, 13) is a triple for this sequence because from the law of cosines (see link), cos(C) = (7^2 + 8^2 - 13^2)/(2*7*8) = -1/2.

Links

G. Julia, Triangles dont un angle mesure 120 degrés, Problème Capes (in French).

Emrys Read, On integer-sided triangles containing angles of 120° or 60°, Mathematical Gazette 90, July 2006, 299-305.

Keith Selkirk, Integer-Sided Triangles with an Angle of 120°, Mathematical Gazette, Vol. 67, No. 442 (Dec., 1983), pp. 251-255.

Eric Weisstein's World of Mathematics, Law of Cosines.

Wikipedia, Triangles with an angle of 120°, Eisenstein triple.

Crossrefs

Cf. A357275, A357276, A357277, A357278.

Cf. also A263728, A336750, A335893 (similar with an angle of Pi/3).

Programs

Maple

for c from 5 to 181 by 2 do for a from 3 to c-2 do b := (-a + sqrt(4*c^2-3*a^2))/2; if b=floor(b) and gcd(a,b)=1 and a

Extensions

a(31..33) = 40,51,79 inserted by Georg Fischer, Dec 04 2022

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A336750 Triples for integer-sided triangles whose sides a < b < c are in arithmetic progression.

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A335895 Middle side of primitive triples, in nondecreasing order, for integer-sided triangles whose angles A < B < C are in arithmetic progression.

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A335894 Smallest side of integer-sided primitive triangles whose angles A < B < C are in arithmetic order.

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A335896 Largest side of integer-sided primitive triangles whose angles A < B < C are in arithmetic order.

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A335897 Perimeters of primitive integer-sided triangles whose angles A < B < C are in arithmetic order.

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A343063 Primitive triples (a, b, c) for integer-sided triangles whose angle B = 2*C.

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A339856 Primitive triples for integer-sided triangles whose sides a < b < c form a geometric progression.

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