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A239456
Number of partitions p of n such that if h = max(p), then h is an (h,2)-separator of p; see Comments.
5
0, 0, 0, 0, 1, 0, 1, 2, 1, 1, 4, 2, 3, 4, 4, 5, 7, 5, 7, 9, 9, 9, 14, 11, 13, 16, 17, 19, 22, 20, 25, 28, 29, 30, 38, 37, 41, 45, 48, 51, 60, 59, 67, 73, 76, 82, 93, 94, 103, 111, 121, 127, 142, 143, 158, 171, 180, 191, 211, 218, 236, 252, 270, 284, 309, 320
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.
EXAMPLE
a(11) counts these 4 partitions: 515, 434, 31313, 2121212.
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