A239517 Number of partitions of n that are separable by the greatest part; see Comments.

0, 0, 1, 2, 4, 5, 8, 10, 13, 16, 20, 24, 29, 33, 39, 46, 53, 59, 67, 77, 87, 97, 107, 120, 134, 147, 163, 180, 196, 216, 236, 259, 281, 305, 332, 363, 393, 423, 456, 496, 534, 577, 619, 667, 718, 770, 823, 887, 949, 1016, 1087, 1165, 1240, 1325, 1414, 1512

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..56.

Example

Let h = max(p). The (h,0)-separable partition of 8 are 161, 251, 341, 242; the (h,1)-separable partitions are 71, 62, 323, 1313; the (h,2)-separable partitions are 323, 21212. So, there are 4 + 4 + 2 = 10 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. A239515, A239516, A239518, A239482.

Sequence in context: A227800 A144876 A337483 * A231056 A325363 A000549

Adjacent sequences: A239514 A239515 A239516 * A239518 A239519 A239520

Keyword

nonn,easy

Author

Clark Kimberling, Mar 26 2014

Status

approved