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A239482
Number of (2,0)-separable partitions of n; see Comments.
18
0, 1, 0, 1, 2, 2, 3, 5, 5, 7, 10, 11, 14, 19, 21, 27, 34, 39, 48, 60, 69, 84, 102, 119, 142, 172, 199, 237, 282, 328, 387, 458, 530, 623, 730, 847, 987, 1153, 1331, 1547, 1796, 2071, 2394, 2771, 3183, 3671, 4227, 4849, 5568, 6395, 7313, 8377, 9584, 10940
OFFSET
3,5
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
The (2,0)-separable partitions of 10 are 721, 523, 424, 42121, 1212121, so that a(10) = 5.
MATHEMATICA
z = 65; -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p] - 1], {n, 2, z}] (* A165652 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p] - 1], {n, 3, z}] (* A239482 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p] - 1], {n, 4, z}] (* A239483 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p] - 1], {n, 5, z}] (* A239484 *)
-1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p] - 1], {n, 6, z}] (* A239485 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 20 2014
STATUS
approved