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T(1, n) and T(2, n) are very easy to determine in general. it looks like there is probably also pattern in T(3, n) .

apparently U(3,3n) = 3 and U(3,3n +/- 1) = (3n-1)/2n.

this translates to T(3,3n) = 1/3n, T(3,3n-1) = 1/6n, and T(3,3n+1) = (3n-1)/6n(n+1).

there's a typo in your formula for T(3, 3n+1) , but other than that, i have the same formula, and now, a proof. it was just a matter of finding the right simplifying device. T(3, 1) = T(1, 3) = 1/3 , which is trivial. T(3, 3n) = 1/(3n) , which is also trivial. T(3, 3n+1) = (3n-1)/(6n(3n+1)) , for n >= 1 , with partition 2 * [n * (3n-1)/(6n(3n+1)) + (n+1) * 1/(6n+2)] + [2n * 1/(6n)] <---> 2n * [(3n-1)/(6n(3n+1)) + 1/(6n)] + (n+1) * [1/(6n+2) + 1/(6n+2)] T(3, 3n+2) = 1/(6n+6) , with partition [(2n+2) * 1/(6n+6)] + 2 * [n * 1/(6n+4) + (n+1) * (3n+4)/(6(n+1)(3n+2))] <---> (2n+2) * [1/(6n+6) + (3n+4)/(6(n+1)(3n+2))] + n * [1/(6n+4) + 1/(6n+4)] T(3, 3n+1): the case when n = 1 is easily to handle separately, so here i'll assume that n > 1 . if some 1/(3n+1) is split into 3 or more parts, then the smallest of these parts is <= 1/(9n+3) , which is too small. here we use that n > 1 . at least one 1/(3n+1) must be split, because 1/3 is not an integer multiple of 1/(3n+1) . therefore we have some part <= 1/(6n+2) . if any 1/(3n+1) is not split, then there is no loss in generality in halving it, because it does not decrease the smallest part size. it might be the case that the two parts go to the same 1/3 , but that's not an issue for us. therefore we may assume that every 1/(3n+1) is split into exactly 2 parts, and thus there are exactly 6n+2 parts in total. therefore, some 1/3 is split into at most 2n parts, so the largest of these is >= 1/(6n) , and its "complement" in a 1/(3n+1) has size <= 1/(3n+1) - 1/(6n) = (3n-1)/(6n(3n+1)) . T(3, 3n+2): the case when n = 0 is easily to handle separately, so here i'll assume that n >= 1 . if some 1/(3n+2) is split into 3 or more parts, then the smallest of these parts is <= 1/(9n+6) , which is too small (for n >= 1 ). at least one 1/(3n+2) must be split, because 1/3 is not an integer multiple of 1/(3n+2) . therefore we have some part <= 1/(6n+4) . as before, if any 1/(3n+2) is not split, then there is no loss in generality in halving it. therefore we may assume that every 1/(3n+2) is split into exactly 2 parts, and thus there are exactly 6n+4 parts in total. now one of the 1/3 's must be split into at least 2n+2 parts, so the smallest of these is <= 1/(6n+6) . mike