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A239513
Number of partitions p of n such that if h = (number of parts of p), then h is an (h,0)-separator of p; see Comments.
4
0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 5, 5, 7, 8, 9, 10, 12, 13, 15, 17, 19, 21, 25, 27, 31, 35, 40, 44, 50, 55, 62, 68, 76, 83, 93, 101, 112, 122, 136, 147, 163, 177, 196, 213, 235, 255, 281, 305, 335, 363, 398, 431, 471, 510, 556, 601, 654, 706, 768, 828
OFFSET
1,9
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(13) counts these 5 partitions: 931, 832, 634, 535, 15151.
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