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%I #10 Jan 28 2022 01:07:47
%S 0,0,1,0,1,2,1,2,3,3,4,6,6,8,11,12,15,20,22,28,35,40,49,61,70,85,103,
%T 120,143,173,200,238,283,329,388,459,531,624,731,848,988,1154,1332,
%U 1548,1797,2072,2395,2772,3184,3672,4228,4850,5569,6396,7314,8378
%N Number of (2,1)-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 h-separable if it is (h,i)-separable for i = 0,1,2.
%e The (2,1)-separable partitions of 11 are 92, 6212, 4232, 321212, so that a(11) = 4.
%t z = 70; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]], {n, 1, z}] (* A008483 *)
%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]], {n, 1, z}] (* A239493 *)
%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]], {n, 1, z}] (* A239494 *)
%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]], {n, 1, z}] (* A239495 *)
%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]], {n, 1, z}] (* A239496 *)
%Y Cf. A230467, A008483, A239494, A239495, A239496.
%K nonn,easy
%O 1,6
%A _Clark Kimberling_, Mar 20 2014
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