A239499 Number of partitions p of n such that if h = (number of parts of p), then h is an (h,2)-separator of p; see Comments.

0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 9, 9, 11, 12, 14, 15, 17, 18, 20, 22, 24, 26, 29, 31, 34, 37, 40, 43, 48, 51, 56, 61, 67, 72, 80, 86, 94, 102, 111, 119, 131, 140, 152, 164, 178, 190, 207, 221, 239

Offset: 1

Author

Clark Kimberling, Mar 25 2014

Keywords

• nonn
• easy

Comments

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.

Examples

			a(19) counts these 3 partitions: [3,13,3], [5,3,5,1,5], [5,2,5,2,5].
		

Crossrefs

Cf. A239729, A239481, A239456, A239689.

Programs

Mathematica

z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] + 1], {n, 1, z}]  (* A239729 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] + 1], {n, 1, z}] (* A239481 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p]] == Length[p] + 1], {n, 1, z}] (* A239456 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] + 1], {n, 1, z}] (* A239499 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] + 1], {n, 1, z}]  (*A239689 *)