For any population p of positive integers between pmin and pmax, we can calculate histogram count n for a list of evenly dividing divisors d. Here's how it works.
Consider the population p = [10, 12, 15, 11, 20, 11]. For the list of divisors d = [2, 3, 5] we get a histogram count n of [3, 2, 3]. That is, three of the numbers in the population are evenly divisible by 2, two are evenly divisible by 3, and three are evenly divisible by 5.
You will face the inverse problem. Given a histogram count corresponding to a list of divisors, produce a population of positive integers that fits the distribution. Your answer will not be unique and can be provided in any order. It just has to match the distribution. The elements of p can be no less than pmin and no greater than pmax.
pmin = 5 pmax = 10 d = [1 2 3 4 5] n = [8 3 4 2 2]
One answer (of many possible): p = [5 5 6 8 8 9 9 9]
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