 MEDIUM-HARD

# DRAFT EXPLANATION:

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There are two possible solutions to the problem. Either one comes up with a precise characterization of N-tuples with \mu equal to 0 or one comes up with a clever way to compute \mu. The characterization is very involved and has many cases, so we will describe the other solution. Let f_y(x):= \sum_{d\le x} \mu(yd). Notice that \mu(y) = f_y(0). It is not hard to show f_y(kx)-f_y((k-1)x) = f_{yk}(x) for any k\ge 1. One can now prove by induction on y that:

• f_y(x)=0 if x is not special.

• if x is special, f_y(x) depends only on the first element of x. Therefore, we can write down the values of f_{yk}(x) when x is a nonnegative integer: $$f_{y0}(x) = f_y(x) ,$$ $$f_{y1}(x) = f_y(1) - f_y(x),$$ and if k\ge 2 the values of f_{yk} are $$f_y(k)-f_y(k-1), 0, 0, \dots, 0, -f_y(k-1), f_y(k)-f_y(k-1), f_y(k)-f_y(k-1), \dots,$$ where the zeroes are repeated k-2 times. One can show by induction on y that f_y takes only the values {-1,0,1} and, fixed y, f_y cannot take both -1 and 1. Thus, to compute \mu(y) = f_y(0), one can compute the functions f_{y'}:\N\to\{-1,0,1\} for each prefix y' of y. Notice that the function f_{y'} is encoded by a positive integer in the range [2,M] (denoted with k above) and 4 boolean values. So, we can compute how many N-uples have \mu equal to 0 with a dynamic programming solution which has O(NM) states and O(NM^2) transitions. So, the overall complexity is O(NM^2).