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

Author

Clark Kimberling, Mar 20 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 = 0,1,2.

Examples

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

Crossrefs

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

Programs

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 *)