A239496 Number of (5,1)-separable partitions of n; see Comments.

0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 2, 2, 3, 3, 3, 3, 5, 5, 7, 8, 9, 10, 13, 14, 17, 20, 23, 26, 32, 35, 42, 48, 55, 63, 75, 83, 97, 111, 127, 144, 168, 188, 217, 246, 280, 317, 365, 409, 467, 528, 598, 674, 768, 861, 977, 1099, 1239, 1392, 1575, 1762, 1987

Offset

1,12

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..62.

Example

The (5,1)-separable partitions of 14 are 95, 3515, 2525, so that a(14) = 3.

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, A239493, A239494, A239495.

Sequence in context: A355250 A055377 A157524 * A348164 A299962 A340010

Adjacent sequences: A239493 A239494 A239495 * A239497 A239498 A239499

Keyword

nonn,easy

Author

Clark Kimberling, Mar 20 2014

Status

approved