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A239495
Number of (4,1)-separable partitions of n; see Comments.
4
0, 0, 0, 0, 1, 1, 1, 0, 1, 2, 2, 3, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 20, 23, 27, 32, 37, 43, 52, 59, 69, 80, 93, 108, 126, 144, 166, 192, 222, 254, 294, 334, 384, 441, 504, 575, 658, 748, 854, 972, 1106, 1254, 1428, 1617, 1834, 2077, 2350, 2656, 3007
OFFSET
1,10
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 (4,1)-separable partitions of 12 are 84, 3414, 2424, so that a(12) = 3.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 20 2014
STATUS
approved