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A243840 Pair deficit of the most 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. 1

%I

%S 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,

%T 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,

%U 2,2,3,2,1,2,2,2,1,2,2,2

%N Pair deficit of the most 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.

%F a(n) = floor(a162970(n)/a000085(n)) - floor(a162970(floor(n/2))/a000085(floor(n/2))) - floor(a162970(n-floor(n/2))/a000085(n-floor(n/2))).

%e 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.

%e Interestingly, for n = 8, E(8) = 532/191 and E(4) = 6/5, so a(n) = 2 - 1 -1 = 0.

%Y A162970 provides the numerator for calculating the expected value.

%Y A000085 provides the denominator for calculating the expected value.

%K nonn

%O 0,12

%A _Rajan Murthy_, Jun 12 2014

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