Hi muffin masticators, I now have a proof of an asymptotic result I'd conjectured a while ago. Recall that I write multiplicities after parts and that I'm working with U(m, n), the size of the smallest part in a common refinement of m * n and n * m. Theorem: Given m, the limit, as n goes to infinity over nonmultiples of m, of U(m, n) is m/2. Note that for m not dividing n, we always have U(m, n) <= m/2, since some m must be split, so we just have to prove that the lim inf of U(m, n) over all n is at least m/2. Lemma 1: Given m>=3, let e be 1 if m is odd and 2 if m is even. For k>=1 we have U(m, (km-1)(m-e)/2) >= (km-1)/2k = m/2 - 1/2k. Proof of Lemma 1: We use the multiset (m/2 - 1/2k) * ek(m-e), m/2 * (km-2ek-1)(m-e), (m/2 + 1/2k) * ek(m-e). It is clear that the smallest part has the desired size and that the parts can be paired up to give (km-1)(m-e)/2 copies of m. Also, all the smallest parts can be added together to give e copies of (km-1)(m-e)/2. Finally, we have m/2 * (km-2ek-1) + (m/2 + 1/2k) * ek = km^2/2 - ekm/2 - m/2 + e/2 = (km-1)(m-e)/2, so the rest of the pieces can be combined to make m-e more copies of (km-1)(m-e)/2, and this is a valid common refinement. QED In fact, it's not too hard to show that if we combine as many pairs of m/2's as possible into m's, then we get the optimum partition in my sense of lexically maximizing the increasing list of pieces sizes. In particular, we get equality on U, although this is not needed for the proof of the theorem. Lemma 2: For all positive m, n_1, n_2, ..., n_k we have U(m, n_1 + ... + n_k) >= min_i { U(m, n_i) }. Proof of Lemma 2: Take the union of the optimal partitions for each pair (m, n_i). Together they give m * (n_1 + ... + n_k), and we can combine the n_i * m 's to give (n_1 + ... + n_k) * m. Thus we have a corefinement of m * (n_1 + ... + n_k) and (n_1 + ... + n_k) * m with smallest part equal to min_i { U(m, n_i) }. QED Proof of Theorem: The result is vacuous for m=1 and trivial for m=2, so take m>=3 and define e as in Lemma 1. First, we have U(m, m/e) >= m/2. Now fix k and note that m/e and (km-1)(m-e)/2 are relatively prime. Then for every sufficiently large integer n we can write n = m/e * a + (km-1)(m-e)/2 * b with a,b>=0, and we have U(m, n) = U(m, m/e * a + (km-1)(m-e)/2 * b) >= min( U(m, m/e), U(m, (km-1)(m-e)/2) ) by Lemma 2 >= m/2 - 1/2k. Letting k go to infinity now proves the theorem. QED David P. Moulton