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A286637
Perimeter of primitive integer automedian triangles (without repetitions).
0
3, 37, 47, 73, 107, 121, 181, 191, 239, 249, 253, 299, 337, 359, 429, 431, 433, 503, 529, 537, 541, 579, 587, 649, 659, 661, 671, 759, 767, 781, 789, 793, 851, 863, 933, 937, 971, 1067, 1069, 1079, 1081, 1093, 1191, 1199, 1237, 1249, 1257, 1261, 1307, 1319
OFFSET
1,1
COMMENTS
An automedian triangle has the square of its sides in an arithmetic progression. Hence its sides (a, b, c) with a <= b <= c are related by a^2+c^2 = 2b^2. Also a+b > c in order to satisfy the triangle inequality. It is so named because its sides and medians are proportional to each other but in a different order. Apart from the unit equilateral triangle all other primitive integer automedian triangles can be generated from primitive Pythagorean triangles (PPT's). If x^2+y^2 = z^2 is a PPT then a = |x-y|, b = z and c = x+y gives an integer automedian triangle (a, b, c) provided that Min(2x, 2y) < z. Note that the PPT (3, 4, 5) will not generate an integer automedian triangle.
FORMULA
Using the Euclidean parameters m, n to generate a PPT where (x, y, z) = (m^2-n^2, 2m*n, m^2+n^2) with m > n, GCD(m, n) = 1 and m+n odd, we can also generate an integer automedian triangle (a, b, c) = (|m^2-n^2-2m*n|, m^2+n^2, m^2-n^2+2m*n) with additionally n < m < n*Sqrt(3) or m > n*(2+Sqrt(3)) to satisfy the triangle inequality.
EXAMPLE
Substituting m=3 and n=2 into the Euclidean parameters in the above formula it gives PPT (5, 12, 13) and it will generate an integer automedian triangle (7, 13, 17) with perimeter 37. It is the second occurrence of a primitive integer automedian triangle.
MATHEMATICA
lst={}; Do[{a, b, c}={Abs[m^2-n^2-2m*n], m^2+n^2, m^2-n^2+2m*n}; If[GCD[m, n]==1&&OddQ[m+n]&&a+b>c, AppendTo[lst, a+b+c]], {m, 1, 100}, {n, 0, m}]; Take[Union@lst, 50]
CROSSREFS
Sequence in context: A042333 A106103 A224471 * A106996 A338751 A300347
KEYWORD
nonn
AUTHOR
Frank M Jackson, May 11 2017
STATUS
approved