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A239469
Number of 3-separable partitions of n; see Comments.
5
0, 0, 0, 1, 2, 1, 3, 4, 5, 6, 8, 11, 13, 15, 20, 24, 30, 35, 43, 52, 63, 74, 89, 106, 127, 148, 177, 208, 246, 287, 338, 396, 464, 538, 630, 732, 853, 985, 1145, 1324, 1532, 1765, 2038, 2345, 2702, 3098, 3562, 4081, 4679, 5348, 6120, 6987, 7978, 9087, 10359
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
(3,0)-separable partitions of 7: 232;
(3,1)-separable partitions of 7: 43;
(3,2)-separable partitions of 7: 3231;
3-separable partitions of 7: 232, 43, 3231, so that a(7) = 3.
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