A239500 Number of partitions p of n such that if h = (number of parts of p), then h is an (h,1)-separator of p; see Comments.

0, 0, 1, 0, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 3, 4, 4, 5, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 29, 32, 36, 39, 43, 48, 53, 58, 65, 70, 78, 85, 93, 101, 112, 120, 132, 143, 156, 168, 184, 198, 216, 233, 253, 273, 298, 320, 348, 376, 407, 439

Offset

1,10

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..65.

Example

a(12) counts these partitions: 84, 4431, 4422.

Mathematica

z = 35; t1 = Table[Count[IntegerPartitions[n],  p_ /; 2 Count[p, Min[p]] == Length[p]], {n, 1, z}]  (* A239497 *)
t2 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p]], {n, 1, z}] (* A239498 *)
t3 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p]], {n, 1, z}] (* A118096 *)
t4 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Length[p]] == Length[p]], {n, 1, z}] (* A239500 *)
t5 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p]], {n, 1, z}]  (* A239501 *)

Crossrefs

Cf. A239497, A239498, A118096, A239501, A239482.

Sequence in context: A103469 A337932 A029326 * A029204 A241085 A057334

Adjacent sequences: A239497 A239498 A239499 * A239501 A239502 A239503

Keyword

nonn,easy

Author

Clark Kimberling, Mar 24 2014

Status

approved