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A350013
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Number of integer-sided triangles with one side having length n and an adjacent angle of 60 degrees.
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2
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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
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OFFSET
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1,3
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COMMENTS
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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.
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)
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LINKS
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EXAMPLE
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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}.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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