A239518 Number of partitions p of n that are separable by the number of parts of p; see Comments.

0, 0, 1, 0, 2, 2, 3, 3, 3, 5, 6, 7, 8, 9, 11, 13, 15, 17, 20, 22, 25, 28, 32, 36, 42, 45, 52, 57, 65, 71, 81, 88, 100, 109, 122, 134, 149, 162, 180, 197, 218, 238, 262, 286, 315, 343, 376, 410, 449, 488, 534, 580, 633, 687, 749, 812, 883, 956, 1038, 1123

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 = number of parts of p. The (h,0)-separable partition of 11 are 92, 731, 632, 434; the (h,1)-separable partition is 2414; the (h,2)-partition is 353. So, there are 4 + 1 + 1 = 6 h-separable partitions of 11.
		

Crossrefs

Cf. A239515, A239516, A239517, 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 *)