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A239514
Number of partitions p of n such that if h = max(p) - min(p), then h is an (h,0)-separator of p; see Comments.
4
0, 0, 0, 0, 1, 1, 0, 2, 2, 3, 2, 4, 2, 7, 6, 7, 6, 10, 7, 14, 12, 18, 12, 22, 18, 23, 23, 31, 29, 42, 33, 45, 42, 54, 49, 68, 62, 78, 76, 95, 87, 110, 102, 124, 128, 150, 141, 178, 174, 203, 203, 237, 228, 272, 269, 308, 318, 360, 356, 422, 420, 472, 482
OFFSET
1,8
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.
FORMULA
a(12) counts these partitions: 615, 642, 43131, 3121212.
MATHEMATICA
z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] - 1], {n, 1, z}] (* A239510 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] - 1], {n, 1, z}] (* A239511 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p] - 1], {n, 1, z}] (* A237828 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] - 1], {n, 1, z}] (* A239513 *)
Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] - 1], {n, 1, z}] (* A239514 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 24 2014
STATUS
approved