The obvious first step is to obtain the list of factors with k f. This list will always be in sorted order, so if there are duplicates they will be right next to each other. A suitable method is to compare the list with a copy of itself shifted over by one.
1: [3, 7, 7, 7, 19] 2: [3, 7, 7, 7, 19] 2: [3, 7, 7, 7, 19, 0]
. 1: [3, 7, 7, 7, 19, 0] 1: [0, 3, 7, 7, 7, 19]
. .
19551 k f RET 0 | TAB 0 TAB |
1: [0, 0, 1, 1, 0, 0] 1: 2 1: 0
. . .
V M a = V R + 0 a =
Note that we have to arrange for both vectors to have the same length so that the mapping operation works; no prime factor will ever be zero, so adding zeros on the left and right is safe. From then on the job is pretty straightforward.
Incidentally, Calc provides the @c{\dfn{M\"obius} $\mu$} Moebius mu function which is zero if and only if its argument is square-free. It would be a much more convenient way to do the above test in practice.