A239471 Number of 5-separable partitions of n; see Comments.

0, 0, 0, 0, 0, 1, 2, 2, 3, 2, 4, 5, 7, 8, 9, 11, 13, 16, 20, 23, 27, 31, 37, 43, 52, 59, 70, 80, 93, 108, 126, 144, 167, 191, 221, 253, 292, 332, 382, 435, 498, 567, 649, 736, 839, 951, 1082, 1226, 1393, 1573, 1784, 2013, 2277, 2568, 2902, 3266, 3683, 4141

Offset

1,7

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..58.

Example

(5,0)-separable partitions of 7: 151
(5,1)-separable partitions of 7: 52
(5,2)-separable partitions of 7: (none)
5-separable partitions of 7: 151, 52, so that a(7) = 2.

Mathematica

z = 55; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 1] <= Length[p] + 1], {n, 1, z}] (* A239467 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2] <= Length[p] + 1], {n, 1, z}] (* A239468 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 3] <= Length[p] + 1], {n, 1, z}] (* A239469 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 4] <= Length[p] + 1], {n, 1, z}] (* A239470 *)
t5 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 5] <= Length[p] + 1], {n, 1, z}] (* A239472 *)

Crossrefs

Cf. A239467, A239468, A239469, A239470.

Sequence in context: A319706 A305894 A305811 * A241509 A268327 A053023

Adjacent sequences: A239468 A239469 A239470 * A239472 A239473 A239474

Keyword

nonn,easy

Author

Clark Kimberling, Mar 20 2014

Status

approved