A239515 Number of partitions of n that are separable by the least part; see Comments.

0, 0, 1, 2, 4, 5, 9, 12, 16, 22, 29, 37, 48, 61, 76, 95, 118, 146, 179, 219, 265, 323, 390, 471, 564, 677, 809, 967, 1148, 1365, 1616, 1915, 2259, 2665, 3135, 3686, 4320, 5065, 5923, 6923, 8070, 9408, 10942, 12721, 14762, 17117, 19819, 22933, 26490, 30583

Offset

1,4

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, or 2.

Links

Table of n, a(n) for n=1..50.

Example

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.

Mathematica

z = 75; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239515 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2*Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239516 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Max[p]] <= Length[p] + 1], {n, 1, z}]  (* A239517 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Length[p]] <= Length[p] + 1], {n, 1, z}]  (* A239518 *)

Crossrefs

Cf. A239516, A239517, A239518, A239482.

Sequence in context: A039021 A295564 A236336 * A087667 A241444 A082592

Adjacent sequences: A239512 A239513 A239514 * A239516 A239517 A239518

Keyword

nonn,easy

Author

Clark Kimberling, Mar 26 2014

Status

approved