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A239516
Number of partitions p of n that are separable by the 2*min(p); see Comments.
4
0, 0, 1, 1, 1, 3, 2, 4, 6, 6, 8, 12, 14, 16, 22, 27, 32, 41, 49, 60, 73, 88, 106, 130, 154, 184, 220, 262, 313, 373, 440, 520, 616, 723, 849, 1002, 1173, 1373, 1606, 1873, 2182, 2543, 2955, 3431, 3979, 4608, 5327, 6160, 7105, 8190, 9435, 10851, 12469, 14317
OFFSET
1,6
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 = 2*min(p). The (h,0)-separable partition of 8 is 521; the (h,1)-separable partition is 3212; the (h,2)-separable partitions are 242, 21212. So, there are 1 + 1 + 2 = 4 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