@Absinthe @freemo I forked the repo and added my solution here: https://git.qoto.org/zingbretsen/coconuts/blob/master/coco.py
It should solve the 5 person problem when run as-is, but it could be run for any number of people. I should probably have added that as a command line argument.
It works backwards from possible ending amounts, but is somewhat selective in the numbers it tries. e.g., for the 5 person case, we know that the ending pile must be divisible by both 4 and 5, so the final pile will be a multiple of 20. In addition to that, we know that 5/4 + 1 of the ending pile must be divisible by 5, so 60 is the lowest number that would work for the first round of stepping backwards.
There is a regular pattern that seems to work for all of the numbers of people I tested, not just 5, so we can reduce the search space pretty significantly. For the 5 person problem, I only have to test 13 numbers before finding the answer.
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