Was thinking about a #maths problem on the plane: prime numbers are those with two factors (1 and n) and that's the minimum attainable by arbitrarily large numbers.   I wondered what the minimum total factor count for n-1 and n+1 was for n prime, and then for those cases how do you minimise factors of n-2 and n+2 and so on.
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It turns out the best you can do is one prime neighbour being a multiple of 4 and the other being a multiple of 6, giving 6+8 total factors respectively.