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A239494
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Number of (3,1)-separable partitions of n; see Comments.
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4
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0, 0, 0, 1, 1, 0, 1, 2, 2, 2, 2, 4, 4, 5, 6, 8, 9, 11, 13, 17, 19, 23, 27, 34, 39, 46, 54, 66, 76, 90, 104, 125, 144, 169, 196, 231, 266, 310, 358, 419, 480, 557, 640, 743, 851, 980, 1123, 1295, 1479, 1697, 1936, 2221, 2529, 2890, 3288, 3753, 4262, 4851
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OFFSET
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1,8
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COMMENTS
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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.
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LINKS
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EXAMPLE
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The (3,1)-separable partitions of 14 are [11,3], [7,3,1,3], [6,3,2,3], [4,3,4,3], [2,3,2,3,1,3], so that a(14) = 5.
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MATHEMATICA
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z = 70; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]], {n, 1, z}] (* A008483 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]], {n, 1, z}] (* A239493 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]], {n, 1, z}] (* A239494 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]], {n, 1, z}] (* A239495 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]], {n, 1, z}] (* A239496 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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