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A239515
Number of partitions of n that are separable by the least part; see Comments.
4
0, 0, 1, 2, 4, 5, 9, 12, 16, 22, 29, 37, 48, 61, 76, 95, 118, 146, 179, 219, 265, 323, 390, 471, 564, 677, 809, 967, 1148, 1365, 1616, 1915, 2259, 2665, 3135, 3686, 4320, 5065, 5923, 6923, 8070, 9408, 10942, 12721, 14762, 17117, 19819, 22933, 26490, 30583
OFFSET
1,4
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, or 2.
EXAMPLE
Let h represent least part. The (h,0)-separable partitions of 8 are 512, 413, 323, 21212; the (h,1)-separable partitions are 71, 62, 53, 4211, 3311; the (h,2)-separable partitions are 161, 242, 13121. So, there are 5 + 5 + 3 = 12 h-separable partitions of 8.
MATHEMATICA
z = 75; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Min[p]] <= Length[p] + 1], {n, 1, z}] (* A239515 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2*Min[p]] <= Length[p] + 1], {n, 1, z}] (* A239516 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Max[p]] <= Length[p] + 1], {n, 1, z}] (* A239517 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Length[p]] <= Length[p] + 1], {n, 1, z}] (* A239518 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 26 2014
STATUS
approved