%I
%S 0,1,1,1,1,3,3,4,5,7,8,10,12,16,18,22,26,33,38,45,53,65,75,89,103,124,
%T 143,168,195,230,265,309,357,418,479,556,639,742,850,979,1122,1294,
%U 1478,1696,1935,2220,2528,2889,3287,3752,4261,4850,5502,6257,7084
%N Number of (3,0)separable partitions of n; see Comments.
%C Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)separable if there is an ordering x, h, x, h, ... , x, h, where the number of h's on the ends is 1; next, p is (h,2)separable if there is an ordering h, x, h, ... , x, h. Finally, p is hseparable if it is (h,i)separable for i = 0,1,2.
%e The (3,0)separable partitions of 11 are 731, 632, 434, 23231, so that a(11) = 4.
%t z = 65; 1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]  1], {n, 2, z}] (* A165652 *)
%t 1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]  1], {n, 3, z}] (* A239482 *)
%t 1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]  1], {n, 4, z}] (* A239483 *)
%t 1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]  1], {n, 5, z}] (* A239484 *)
%t 1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]  1], {n, 6, z}] (* A239485 *)
%Y Cf. A239467, A165652, A239482, A239484, A239485.
%K nonn,easy
%O 4,6
%A _Clark Kimberling_, Mar 20 2014
