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For a triangle with integer sides a,b,c (none over 2000) evaluate the triplet of its medians m_{a }, m_{b }, m_{c }.
Let those three become sides of a new triangle i.e. (a,b,c) =(m_{a }, m_{b }, m_{c }). <end>

It is up to you to find a triplet (a,b,c) such that the above procedure can be executed a maximal number of times, creating sets of “medians“ with integer values only.

The answer should include: (a,b,c) and all interim sets of medians.

For a maximal number of times, it would be where each formed triangle had each side of length 2^{n}, i.e., the triangle would be an equilateral triangle with the inner-most triangle having sides of length 2^{0}, with vertices at the medial points of an equilateral triangle with sides of length 2^{1}, and so on until the vertices of the triangle are the medial points of a triangles with sides of length 2^{10}. (2^{11} = 2024 and exceeds the maximum limit of 2000.)

Where (a,b,c) = (a_{n},b_{n},c_{n}) and n = 0, and each smaller triplet (m_{a }, m_{b }, m_{c}) being designated (a_{n},b_{n},c_{n}) and n++: