OFFSET
1,3
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)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
EXAMPLE
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}.
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Joseph C. Y. Wong, Dec 08 2021
EXTENSIONS
More terms from David A. Corneth, Dec 09 2021
STATUS
approved