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Number of partitions of n that are separable by the greatest part; see Comments.
4

%I #8 Jan 28 2022 00:57:20

%S 0,0,1,2,4,5,8,10,13,16,20,24,29,33,39,46,53,59,67,77,87,97,107,120,

%T 134,147,163,180,196,216,236,259,281,305,332,363,393,423,456,496,534,

%U 577,619,667,718,770,823,887,949,1016,1087,1165,1240,1325,1414,1512

%N Number of partitions of n that are separable by the greatest 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 = max(p). The (h,0)-separable partition of 8 are 161, 251, 341, 242; the (h,1)-separable partitions are 71, 62, 323, 1313; the (h,2)-separable partitions are 323, 21212. So, there are 4 + 4 + 2 = 10 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. A239515, A239516, A239518, A239482.

%K nonn,easy

%O 1,4

%A _Clark Kimberling_, Mar 26 2014