A239513 Number of partitions p of n such that if h = (number of parts of p), then h is an (h,0)-separator of p; see Comments.

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 5, 5, 7, 8, 9, 10, 12, 13, 15, 17, 19, 21, 25, 27, 31, 35, 40, 44, 50, 55, 62, 68, 76, 83, 93, 101, 112, 122, 136, 147, 163, 177, 196, 213, 235, 255, 281, 305, 335, 363, 398, 431, 471, 510, 556, 601, 654, 706, 768, 828

Offset

1,9

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..62.

Example

a(13) counts these 5 partitions: 931, 832, 634, 535, 15151.

Mathematica

z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239510 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239511 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p] - 1], {n, 1, z}]  (* A237828 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] - 1], {n, 1, z}]  (* A239513 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] - 1], {n, 1, z}] (* A239514 *)

Crossrefs

Cf. A239510, A239511, A238928, A239514, A239482.

Sequence in context: A119620 A240870 A265771 * A029018 A238217 A185226

Adjacent sequences: A239510 A239511 A239512 * A239514 A239515 A239516

Keyword

nonn,easy

Author

Clark Kimberling, Mar 24 2014

Status

approved