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%I #8 Jan 28 2022 01:03:24
%S 0,0,1,2,4,5,9,12,16,22,29,37,48,61,76,95,118,146,179,219,265,323,390,
%T 471,564,677,809,967,1148,1365,1616,1915,2259,2665,3135,3686,4320,
%U 5065,5923,6923,8070,9408,10942,12721,14762,17117,19819,22933,26490,30583
%N Number of partitions of n that are separable by the least part; 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, or 2.
%e Let h represent least part. The (h,0)-separable partitions of 8 are 512, 413, 323, 21212; the (h,1)-separable partitions are 71, 62, 53, 4211, 3311; the (h,2)-separable partitions are 161, 242, 13121. So, there are 5 + 5 + 3 = 12 h-separable partitions of 8.
%t z = 75; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Min[p]] <= Length[p] + 1], {n, 1, z}] (* A239515 *)
%t t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2*Min[p]] <= Length[p] + 1], {n, 1, z}] (* A239516 *)
%t t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Max[p]] <= Length[p] + 1], {n, 1, z}] (* A239517 *)
%t t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Length[p]] <= Length[p] + 1], {n, 1, z}] (* A239518 *)
%Y Cf. A239516, A239517, A239518, A239482.
%K nonn,easy
%O 1,4
%A _Clark Kimberling_, Mar 26 2014
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