A239482 Number of (2,0)-separable partitions of n; see Comments.

0, 1, 0, 1, 2, 2, 3, 5, 5, 7, 10, 11, 14, 19, 21, 27, 34, 39, 48, 60, 69, 84, 102, 119, 142, 172, 199, 237, 282, 328, 387, 458, 530, 623, 730, 847, 987, 1153, 1331, 1547, 1796, 2071, 2394, 2771, 3183, 3671, 4227, 4849, 5568, 6395, 7313, 8377, 9584, 10940

Offset

3,5

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=3..56.

Example

The (2,0)-separable partitions of 10 are 721, 523, 424, 42121, 1212121, so that a(10) = 5.

Mathematica

z = 65; -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p] - 1], {n, 2, z}]  (* A165652 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p] - 1], {n, 3, z}]  (* A239482 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p] - 1], {n, 4, z}]  (* A239483 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p] - 1], {n, 5, z}]  (* A239484 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p] - 1], {n, 6, z}] (* A239485 *)

Crossrefs

Cf. A239467, A165652, A239483, A239484, A239485.

Sequence in context: A008507 A318683 A028364 * A280470 A011971 A060048

Adjacent sequences: A239479 A239480 A239481 * A239483 A239484 A239485

Keyword

nonn,easy

Author

Clark Kimberling, Mar 20 2014

Status

approved