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

1,17

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.

Links

Table of n, a(n) for n=1..69.

Example

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

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 *)

Crossrefs

Cf. A239729, A239481, A239456, A239689.

Sequence in context: A262869 A342678 A108356 * A241139 A294242 A241092

Adjacent sequences: A239496 A239497 A239498 * A239500 A239501 A239502

Keyword

nonn,easy

Author

Clark Kimberling, Mar 25 2014

Status

approved