A239485 Number of (5,0)-separable partitions of n; see Comments.

0, 1, 1, 2, 2, 2, 2, 4, 4, 6, 7, 8, 9, 12, 13, 16, 19, 22, 25, 31, 34, 41, 47, 54, 62, 74, 82, 96, 110, 126, 143, 167, 187, 216, 245, 279, 316, 364, 408, 466, 527, 597, 673, 767, 860, 976, 1098, 1238, 1391, 1574, 1761, 1986, 2228, 2502, 2801, 3150, 3518

Offset

6,4

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 = 0,1,2.

Links

Table of n, a(n) for n=6..62.

Example

The (5,0)-separable partitions of 13 are 751, 652, 454, 15151, so that a(13) = 4.

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, A239481, A165652, A239483, A239484.

Sequence in context: A214927 A326115 A355746 * A237756 A010334 A010578

Adjacent sequences: A239482 A239483 A239484 * A239486 A239487 A239488

Keyword

nonn,easy

Author

Clark Kimberling, Mar 20 2014

Status

approved