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A239468
Number of 2-separable partitions of n; see Comments.
5
0, 0, 1, 1, 2, 3, 4, 6, 7, 10, 12, 16, 20, 25, 31, 39, 47, 59, 71, 87, 105, 128, 153, 185, 221, 265, 315, 377, 445, 530, 625, 739, 870, 1025, 1201, 1411, 1649, 1930, 2249, 2625, 3050, 3549, 4116, 4773, 5523, 6391, 7375, 8515, 9806, 11293, 12980, 14917, 17110
OFFSET
1,5
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
(2,0)-separable partitions of 7: 421, 12121;
(2,1)-separable partitions of 7: 52;
(2,2)-separable partitions of 7: 232;
2-separable partitions of 7: 421, 12121, 52, 232, so that a(7) = 4.
MATHEMATICA
z = 55; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 1] <= Length[p] + 1], {n, 1, z}] (* A239467 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2] <= Length[p] + 1], {n, 1, z}] (* A239468 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 3] <= Length[p] + 1], {n, 1, z}] (* A239469 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 4] <= Length[p] + 1], {n, 1, z}] (* A239470 *)
t5 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 5] <= Length[p] + 1], {n, 1, z}] (* A239472 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 20 2014
STATUS
approved