It's easy to show S(k m, k p) >= S(m, p). Proof: I prefer the symmetrical quantity U(m, p) = p S(m, p). A "partition" is an m by p array of numbers x_ij that satisfies sum_{i=1}^{m} x_ij = m sum_{j=1}^{p} x_ij = p. An optimal partition is one which maximizes the minimum non-zero x_ij. The minimum non-zero x_ij is the number U(m, p). Let x_ij be an optimal partition for the pair m,p. Replace each x_ij by k x_ij times the k by k identity matrix. The resulting k m by k p array (of diagonal blocks) satisfies sum_{i=1}^{k m} x_ij = k m sum_{j=1}^{k p} x_ij = k p and has minimal non-zero element k U(m, p). Hence U(k m, k p) >= k U(m, p). This translates to S(k m, k p) >= S(m, p). Veit On Jan 27, 2009, at 11:21 AM, James Propp wrote:
Scott wrote:
Conjecture 6: S(km,kp) = S(m,p)
Has anyone settled this? I don't even see how to prove that S(km,kp) stabilizes as k goes to infinity with m,n fixed.
Jim Propp
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