A243840 Pair deficit of the most nearly equal in size partition of n into two parts using floor rounding of the expectations for n, floor(n/2) and n- floor(n/2), assuming equal likelihood of states defined by the number of two-cycles.
0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2
Offset: 0
Keywords
Examples
Trivially, for n = 0,1 no pairs are possible so a(0) and a(1) are 0. For n = 2, the expectation, E(n), equals 0.5. So a(2) = floor(E(2)) - floor(E(1)) - floor(E(1)) = 0. For n = 5 = 2 + 3, E(5) = 20/13, E(2) = 0.5 and E(3) = 0.75 and a(5) = floor(E(5)) - floor(E(2)) - floor(E(3)) = 1. Interestingly, for n = 8, E(8) = 532/191 and E(4) = 6/5, so a(n) = 2 - 1 - 1 = 0.