A239493 Number of (2,1)-separable partitions of n; see Comments.

0, 0, 1, 0, 1, 2, 1, 2, 3, 3, 4, 6, 6, 8, 11, 12, 15, 20, 22, 28, 35, 40, 49, 61, 70, 85, 103, 120, 143, 173, 200, 238, 283, 329, 388, 459, 531, 624, 731, 848, 988, 1154, 1332, 1548, 1797, 2072, 2395, 2772, 3184, 3672, 4228, 4850, 5569, 6396, 7314, 8378

Offset

1,6

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=1..56.

Example

The (2,1)-separable partitions of 11 are 92, 6212, 4232, 321212, so that a(11) = 4.

Mathematica

z = 70; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]], {n, 1, z}] (* A008483 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]], {n, 1, z}] (* A239493 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]], {n, 1, z}] (* A239494 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]], {n, 1, z}] (* A239495 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]], {n, 1, z}] (* A239496 *)

Crossrefs

Cf. A230467, A008483, A239494, A239495, A239496.

Sequence in context: A116498 A143472 A180235 * A331849 A015739 A015746

Adjacent sequences: A239490 A239491 A239492 * A239494 A239495 A239496

Keyword

nonn,easy

Author

Clark Kimberling, Mar 20 2014

Status

approved