This thing about square-free and distinct factorizations is extremely reminiscent of a result that was discussed on this forum about a decade ago. (Insert obligatory query about archives here.) I remember serving primarily as a listener. If my memory is correct, Gosper told me a similar result: The number of partitions of an integer n into ODD integers equals the number of partitions of the same n into DISTINCT integers. Partitions into odds are analogous (if my intuition is correct) to square-free factorizations; while partitions into distincts are like distinct factorizations in the problem presently under discussion. I will, therefore, digress and discuss the partition problem, returning to factorizations at the end. Consider the case N = 7. Leaving out the plus signs for brevity, partitions of 7 into odd integers are: 7, 511, 331, 31111, 1111111 Partitions into distinct integers are: 7, 61, 52, 43, 421 Note that there are five of each. Somebody noticed, I don't know how, that there are always the same number of both kinds of partitions, regardless of N. Gosper showed me an absurdly intricate _analytic_ proof of this, and bewailed the lack of a _combinatorial_ proof; the latter was discovered later, either by Gosper or by Schroeppel, if memory serves. The analytic proof constructed generating functions from the two sequences, and then showed analytically that the generating functions were equal. Presumably Gosper can reconstruct this proof with less trouble than I, so I will leave it to him. The combinatorial proof shows a direct 1-1 correspondence between partitions of the two types. To convert a partition of the "odd" type to one of the "distinct" type, find a pair of identical piles and merge them. Repeat until there are no identical piles; by construction the resulting partition is of the "distinct" type. The inverse operation is to find an even pile and break it into two equal ones. This operation is repeated until there are no even piles, so the result is a partition of "odd" type. Now that I have typed this out, it is fairly obvious that an analogous argument can be made for distinct and square-free factorizations. A distinct factorization can be converted to a square-free one by repeatedly splitting square factors until none remain; the inverse operation combines identical factors until none remain. QED, sez me. -A