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

Author

Clark Kimberling, Mar 26 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, or 2.

Examples

			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.
		

Crossrefs

Cf. A239516, A239517, A239518, A239482.

Programs

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 *)