FYI, there is no known algorithm for optimal Egyptian fractions. Therefore this problem should allow the trivial solution of p/q as 1/q times p, which is not currently (this could be the worst score, a penalty can be included for repeated denominators). The solution for this problem is a greedy one (which may produce huge denominators), or the combination of non-greedy techniques that breaks the problem into several smaller pieces and which may create a huge sequence. I hope that the author has tested for upper and lower bounds on the test suite numbers since he is requesting an unknown solution and random numbers.
And my advice is if you do find a solution for this which attends the general case, don't publish it here, write a scientific paper.
Sum the entries of each column of a matrix which satisfy a logical condition.
Length of the hypotenuse
Crunch that matrix!
Open coded lock.
Packing oranges - one layer
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