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

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

Author

Joseph C. Y. Wong, Dec 08 2021

Keywords

• nonn
• easy

Comments

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)

Examples

			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}.
		

Links

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

Crossrefs

Cf. A121992, A201223, A264826.

Programs

PARI

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

PARI

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

Extensions

More terms from David A. Corneth, Dec 09 2021