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A239471
Number of 5-separable partitions of n; see Comments.
4
0, 0, 0, 0, 0, 1, 2, 2, 3, 2, 4, 5, 7, 8, 9, 11, 13, 16, 20, 23, 27, 31, 37, 43, 52, 59, 70, 80, 93, 108, 126, 144, 167, 191, 221, 253, 292, 332, 382, 435, 498, 567, 649, 736, 839, 951, 1082, 1226, 1393, 1573, 1784, 2013, 2277, 2568, 2902, 3266, 3683, 4141
OFFSET
1,7
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
(5,0)-separable partitions of 7: 151
(5,1)-separable partitions of 7: 52
(5,2)-separable partitions of 7: (none)
5-separable partitions of 7: 151, 52, so that a(7) = 2.
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