A264826 -id:A264826 - cp's OEIS frontend

A350038 Numbers that are the perimeter of a primitive 60-degree integer triangle.

18, 20, 35, 36, 45, 56, 77, 84, 90, 104, 110, 120, 126, 135, 143, 170, 176, 182, 189, 198, 209, 210, 216, 221, 252, 260, 264, 266, 270, 272, 273, 297, 299, 323, 350, 351, 360, 368, 374, 378, 380, 390, 396, 425, 432, 437, 459, 462, 464, 468, 476, 494, 495, 506, 527, 551, 561, 570, 575, 585, 594, 608, 612

Offset: 1

Seiichi Manyama, Dec 10 2021

nonn

			b(n) = Sum_{k=1..3} A264826(3*n+k-3).
c(n) = Sum_{k=1..3} A201223(3*n+k-3).
b(1) = c(1) = 3+7+8 = 18 = a(1).
b(2) = c(2) = 5+7+8 = 20 = a(2).
b(3) = c(5) = 5+19+21 = 45 = a(5).
b(4) = c(3) = 7+13+15 = 35 = a(3).
b(5) = c(9) = 7+37+40 = 84 = a(8).
b(6) = c(4) = 8+13+15 = 36 = a(4).

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A024364, A201223, A264826, A350013, A350039.

def A(n)
  ary = []
  (1..n).each{|i|
    (i + 1..n).each{|j|
      if i.gcd(j) == 1 && (i - j) % 3 > 0
        x, y, z = j * j, i * j, i * i
        ary << 2 * x + 5 * y + 2 * z
        ary << 3 * x + 3 * y
      end
    }
  }
  ary
end
p A(20).uniq.sort[0..100]

A350013 Number of integer-sided triangles with one side having length n and an adjacent angle of 60 degrees.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 4, 3, 3, 3, 2, 3, 3, 7, 6, 3, 3, 3, 3, 7, 3, 3, 7, 5, 3, 4, 3, 3, 7, 3, 8, 6, 3, 10, 3, 3, 3, 6, 11, 3, 7, 3, 3, 10, 3, 3, 12, 5, 5, 6, 3, 3, 4, 10, 10, 6, 3, 3, 7, 3, 3, 10, 10, 10, 6, 3, 3, 6, 10, 3, 10, 3, 3, 11, 3, 10, 6, 3, 18, 5, 3, 3, 7, 9, 3, 6, 10, 3, 10, 10

Offset: 1

Joseph C. Y. Wong, Dec 08 2021

All terms are greater than or equal to 1 because a triangle with side lengths {n, n, n} is equilateral and has an adjacent angle of 60 degrees.
Number of possible integer solutions to the equation n^2 + x^2 - nx = y^2.
x <= n^2 and y <= n^2. - Seiichi Manyama, Dec 09 2021
From David A. Corneth, Dec 10 2021: (Start)
Solving n^2 + x^2 - nx = y^2 for x using the quadratic formula gives x = (n +- sqrt(4*y^2 - 3*n^2)) / 2.
So we need sqrt(4*y^2 - 3*n^2) to be an integer, say k, i.e., sqrt(4*y^2 - 3*n^2) = k.
Squaring gives 4*y^2 - 3*n^2 = k^2, i.e., (2y - k)*(2y + k) = 4*y^2 - k^2 = 3*n^2
Checking divisors d of 3*n^2 gives all candidates for y = (d + 3*n^2/d)/4 and x = (n +- sqrt(4*y^2 - 3*n^2)) / 2 which must be positive. (End)

			For n = 8, there are 4 possible integer triangles with side length 8 and adjacent angle 60 degrees. Their side lengths are {8, 3, 7}, {8, 5, 7}, {8, 8, 8}, {8, 15, 13}.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A121992, A201223, A264826.

a(n) = sum(x=1, n^2, issquare(x^2 - n * x + n^2)); \\ David A. Corneth, Dec 09 2021

a(n) = { my(n23 = 3*n^2, d = divisors(n23), res = 0); for(i = 1, (#d + 1)\2, y = (d[i] + n23/d[i])/4; if(denominator(y) == 1, x = (n + sqrtint(4*y^2 - n23))/2; if(denominator(x) == 1, res++ ); x = (n - sqrtint(4*y^2 - n23))/2; if(x > 0 && denominator(x) == 1, res++ ); ) ); res } \\ faster than above \\ David A. Corneth, Dec 10 2021

More terms from David A. Corneth, Dec 09 2021

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A264827 (a,b,c) in lexicographic order such that a^2 + b^2 + a*b - c^2 = 0 with a < b < c and gcd(a, b) = 1.

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A350038 Numbers that are the perimeter of a primitive 60-degree integer triangle.

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A350013 Number of integer-sided triangles with one side having length n and an adjacent angle of 60 degrees.

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