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A239499
Number of partitions p of n such that if h = (number of parts of p), then h is an (h,2)-separator of p; see Comments.
6
0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 9, 9, 11, 12, 14, 15, 17, 18, 20, 22, 24, 26, 29, 31, 34, 37, 40, 43, 48, 51, 56, 61, 67, 72, 80, 86, 94, 102, 111, 119, 131, 140, 152, 164, 178, 190, 207, 221, 239
OFFSET
1,17
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.
EXAMPLE
a(19) counts these 3 partitions: [3,13,3], [5,3,5,1,5], [5,2,5,2,5].
MATHEMATICA
z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] + 1], {n, 1, z}] (* A239729 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] + 1], {n, 1, z}] (* A239481 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p]] == Length[p] + 1], {n, 1, z}] (* A239456 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] + 1], {n, 1, z}] (* A239499 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] + 1], {n, 1, z}] (*A239689 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 25 2014
STATUS
approved