A335895 -id:A335895 - cp's OEIS frontend

A004611 Divisible only by primes congruent to 1 mod 3.

1, 7, 13, 19, 31, 37, 43, 49, 61, 67, 73, 79, 91, 97, 103, 109, 127, 133, 139, 151, 157, 163, 169, 181, 193, 199, 211, 217, 223, 229, 241, 247, 259, 271, 277, 283, 301, 307, 313, 331, 337, 343, 349, 361, 367, 373, 379, 397, 403, 409, 421, 427, 433, 439, 457

Offset: 1

N. J. A. Sloane

In other words, if a prime p divides n, then p == 1 mod 3.
Equivalently, products of primes == 1 (mod 6), products of elements of A002476.
Positive integers n such that n+d+1 is divisible by 3 for all divisors d of n. For example, a(13)=91 since 91=7*13, 91+1+1=93=3*31, 91+7+1=99=9*11, 91+13+1=105=3*7*5, 91+91+1=183=3*61. The only prime p such that x+d+1 is divisible by p for all divisors d of x is p=3. The sequence consists of 1 and all integers whose prime divisors are of the form 6k+1. - Walter Kehowski, Aug 09 2006
Also z such that z^2 = x^2 + x*y + y^2 and gcd(x,y,z) = 1. - Frank M Jackson, Jul 30 2013
From Jean-Christophe Hervé, Nov 24 2013: (Start)
Apart from the first term (for all in this comment), this sequence is the analog of A008846 (hypotenuses of primitive Pythagorean triangles) for triangles with integer sides and a 120-degree angle: a(n), n>1, is the sequence of lengths of the longest side of the primitive triangles.
Not only the square of these numbers is equal to x^2 + xy + y^2 with x and y > 0, but the numbers themselves also are; the sequence starting at n=2 is then a subsequence of A024606.
(End)
Numbers n such that 3/n cannot be written as the sum of 2 unit fractions. - Carl Schildkraut, Jul 19 2016
a(n), n>1, is the sequence of lengths of the middle side b of the primitive triangles such that A < B < C with an angle B = 60 degrees (A335895). Compare with comment of Nov 24 2013 where a(n), n>1, is the sequence of lengths of the longest side of the primitive triangles that have an angle = 120 degrees. - Bernard Schott, Mar 29 2021

T. D. Noe, Table of n, a(n) for n = 1..10000
J. H. Conway, E. M. Rains and N. J. A. Sloane, On the existence of similar sublattices, Canad. J. Math. 51 (1999), 1300-1306 (Abstract, pdf, ps).
Walter Kehowski, D Numbers.
Index entries for sequences related to A2 = hexagonal = triangular lattice

Cf. A004612, A034017, A120806, A024606, A008846, A335895.

Multiplicative closure of A002476.

[n: n in [1..500] | forall{d: d in PrimeDivisors(n) | d mod 3 eq 1}]; // Vincenzo Librandi, Aug 21 2012

with(numtheory): for n from 1 to 1801 by 6 do it1 := ifactors(n)[2]: it2 := 1: for i from 1 to nops(it1) do if it1[i][1] mod 6 > 1 then it2 := 0; break fi: od: if it2=1 then printf(`%d,`,n) fi: od:
with(numtheory): cnt:=0: L:=[]: for w to 1 do for n from 1 while cnt<100 do dn:=divisors(n); Q:=map(z-> n+z+1, dn); if andmap(z-> z mod 3 = 0, Q) then cnt:=cnt+1; L:=[op(L),[cnt,n]]; fi; od od; L; # Walter Kehowski, Aug 09 2006

ok[1]=True;ok[n_]:=And@@(Mod[#,3]==1&)/@FactorInteger[n][[All,1]];Select[Range[500],ok] (* Vincenzo Librandi, Aug 21 2012 *)
lst={}; maxLen=331; Do[If[Reduce[m^2+m*n+n^2==k^2&&m>=n>=0&&GCD[k, m, n]==1, {m, n}, Integers]===False, Null[], AppendTo[lst, k]], {k, maxLen}]; lst (* Frank M Jackson, Jul 04 2013 from A034017 *)

is(n)=my(f=factor(n)[,1]);for(i=1,#f,if(f[i]%3!=1,return(0)));1 \\ Charles R Greathouse IV, Feb 06 2013

list(lim)=my(v=List([1]), mn, mx, t); forprime(p=7, lim\=1, if(p%6==1, listput(v, p))); if(lim<49, return(Vec(v))); forprime(p=7, sqrtint(lim), if(p%6>1, next); mx=1; while(v[mx+1]*p<=lim, for(i=mn=mx+1, mx=#v, t=p*v[i]; if(t>lim, break); listput(v, t)))); Set(v) \\ Charles R Greathouse IV, Jan 11 2018

More terms from James Sellers, Oct 30 2000

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, May 31 2007

A121940 Product of the first n primes of the form 6k+1.

Original entry on oeis.org

7, 91, 1729, 53599, 1983163, 85276009, 5201836549, 348523048783, 25442182561159, 2009932422331561, 194963444966161417, 20081234831514625951, 2188854596635094228659, 277984533772656967039693, 38639850194399318418517327, 5834617379354297081196116377

Offset: 1

Jonathan Vos Post, Sep 03 2006

For n>1, a(n) is the least positive integer that can be primitively represented as m^2+mn+n^2 with 0<=m<=n and gcd(m,n)=1 in exactly 2^(n-1) ways. - Ray Chandler, Oct 01 2007
From Bernard Schott, Mar 30 2021: (Start)
Also, for n >= 1, a(n) is the smallest positive integer m such that m^2 can be primitively represented as k^2-k*q+q^2 with 1 <= k < q and gcd(k,q)= 1 in exactly 2^n ways. For example (a(1))^2 = 7^2 = 3^2 - 3*8 + 8^2 = 5^2 - 5*8 + 8^2.
It follows that a(n) is the smallest middle side b that appears exactly 2^n times consecutively in the data of A335895, for integer-sided triangles whose angles A < B < C are in arithmetic progression. (End)
Also, a(n) is the smallest largest side c that appears exactly 2^(n-1) times consecutively in the data of A357277 for integer-sided triangles with angles A < B < C = 2*Pi/3 = 120 degrees. - Bernard Schott, Oct 21 2022

			a(4) = 53599 = 7 * 13 * 19 * 31.

Ray Chandler, Table of n, a(n) for n=1..100

Cf. A002476, A002110, A006278, A078586, A335895.

Greedy inverse of A005088.

Rest@FoldList[Times, 1, Select[6 Range[100] + 1, PrimeQ]] (* Ray Chandler, Oct 01 2007 *)

lista(nn) = {my(pr=1, list=List()); forprime(p=1, nn, if ((p%3) == 1, listput(list,pr *= p));); Vec(list);} \\ Michel Marcus, Jul 17 2020

a(n) = Product_{i=1..n} A002476(i).

Extended by Ray Chandler, Oct 01 2007

A335893 Primitive triples for integer-sided triangles whose angles A < B < C are in arithmetic progression.

Original entry on oeis.org

3, 7, 8, 5, 7, 8, 7, 13, 15, 8, 13, 15, 5, 19, 21, 16, 19, 21, 11, 31, 35, 24, 31, 35, 7, 37, 40, 33, 37, 40, 13, 43, 48, 35, 43, 48, 16, 49, 55, 39, 49, 55, 9, 61, 65, 56, 61, 65, 32, 67, 77, 45, 67, 77, 17, 73, 80, 63, 73, 80, 40, 79, 91, 51, 79, 91, 11, 91, 96

Offset: 1

Bernard Schott, Jun 29 2020

The triples are displayed in nondecreasing order of middle side, and if middle sides coincide then by increasing order of the largest side, hence, each triple (a, b, c) is in increasing order.
These three properties below are equivalent:
-> integer-sided triangles whose angles A < B < C are in arithmetic progression,
-> integer-sided triangles such that B = (A+C)/2 with A < C,
-> integer-sided triangles such that A < B < C with B = Pi/3.
When A < B < C are in arithmetic progression with B = A + phi and C = B + phi, then 0 < phi < Pi/3.
The corresponding metric relation between sides is b^2 = a^2 - a*c + c^2.
There exists such primitive triangle iff b^2 is an odd square term of A024612. Hence, the first few middle sides b are 7, 13, 19, 31, 37, 43, 49, 61, 67, ... and b is a term of A004611 \ {1}. Indeed, b cannot be even if the triple is primitive.
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 sines (see link): b/sin(B) = c/sin(C), and c < b/sin(Pi/3) = b * 2/sqrt(3) < 6*b/5. This bound is used in the PARI and Maple programs below.
When triple (a, b, c) is solution, then triple (c-a, b, c) is another solution. Hence, for each b odd solution, there exist 2 triples with same middle side b and same largest side c.
The common tangent to the nine-point circle and the incircle of a triangle ABC is parallel to the Euler line iff angles A < B < C are in arithmetic progression (see Crux Mathematicorum for Indian team selection). - Bernard Schott, Apr 14 2022
These triples are called (primitive) Eisenstein triples (Wikipedia). - Bernard Schott, Sep 21 2022

			(3, 7, 8) is a triple for this sequence because from law of cosines (see link), cos(A) = (7^2 + 8^2 - 3^2)/(2*7*8) = 13/14, cos(B) = (8^2 + 3^2 - 7^2)/(2*8*3) = 1/2 and cos(C) = (3^2 + 7^2 - 8^2)/(2*3*7) = -1/7; then, (A+C)/2 = ( arccos(13/14) + arccos(-1/7) )/2 = Pi/3 = B.
Also, arccos(13/14) ~ 21.787 degrees, arccos(1/2) = 60 degrees, arccos(-1/7) ~ 98.213 degrees, so B-A = C-B ~ 38.213 degrees, hence (A, B, C) are in arithmetic progression.
5^2 - 5*8 + 8^2 = 7^2, hence (5, 7, 8) is another triple for triangle whose angles A < B < C are in arithmetic progression.

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

Bill Sands, Indian team selection test 2007, question 5, Crux Mathematicorum, Vol. 36, No. 5 (2010), p. 278.
Eric Weisstein's World of Mathematics, Law of Cosines.
Eric Weisstein's World of Mathematics, Law of Sines.
Wikipedia, Eisenstein triple.
Index to sequences related to Olympiads.

Cf. A004611, A024612.
Cf. A335894 (smallest side), A335895 (middle side), A335896 (largest side), A335897 (perimeter).
Cf. A103606 (primitive Pythagorean triples), A335034 (primitive triples for triangles with two perpendicular medians).

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,c-a,b,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(a, ", ", b, ", ", c, ", "); print(c-a, ", ", b, ", ", c, ", ");););););} \\ Michel Marcus, Jul 15 2020

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).

A343065 Side b of integer-sided primitive triangles (a, b, c) whose angle B = 2*C.

Original entry on oeis.org

6, 12, 20, 30, 42, 56, 15, 72, 90, 110, 132, 35, 156, 182, 210, 240, 63, 28, 272, 306, 342, 40, 380, 99, 420, 462, 506, 552, 143, 600, 70, 650, 702, 756, 45, 195, 88, 812, 870, 930, 992, 255, 1056, 1122, 130, 1190, 1260, 77, 323, 1332, 154, 1406, 1482, 1560, 399, 1640, 1722, 66, 1806, 208, 1892, 117, 483, 1980, 2070, 238

Offset: 1

Bernard Schott, Apr 11 2021

nonn

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.
This sequence is not increasing because a(7) = 15 < a(6) = 56 (A106430).
If in triangle ABC, B = 2*C, then the corresponding metric relation between sides is a*c + c^2 = c * (a + c) = b^2.
Equivalently, length of side opposite to the greater of the two angles, one being the double of the other.
For the corresponding primitive triples and miscellaneous properties and references, see A343063.

			According to inequalities between a, b, c, there exist 3 types of such triangles:
c < a < b for the first triple (5, 6, 4) with b = 6.
c < b < a for the second triple (16, 15, 9) with b = 15.
a < c < b for the seventh triple (7, 12, 9) with b = 12.

APMEP, Baccalauréat Mathématiques, Lyon, Septembre 1937.

Cf. A335895 (similar for A < B < C in arithmetic progression).
Cf. A343063 (triples), A343064 (side a), A343066 (side c), A343067 (perimeter).
Cf. A106420 (sides b sorted on perimeter), A106430 (sides b in increasing order).

for a from 2 to 100 do
for c from 3 to floor(a^2/2) do
d := c*(a+c);
if issqr(d) and igcd(a,sqrt(d),c)=1 and abs(a-c)

a(n) = A343063(n, 2).

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).

cp's OEIS Frontend

A004611 Divisible only by primes congruent to 1 mod 3.

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A121940 Product of the first n primes of the form 6k+1.

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A335893 Primitive triples 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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A343065 Side b of integer-sided primitive triangles (a, b, c) whose angle B = 2*C.

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