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