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A239500
Number of partitions p of n such that if h = (number of parts of p), then h is an (h,1)-separator of p; see Comments.
4
0, 0, 1, 0, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 3, 4, 4, 5, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 29, 32, 36, 39, 43, 48, 53, 58, 65, 70, 78, 85, 93, 101, 112, 120, 132, 143, 156, 168, 184, 198, 216, 233, 253, 273, 298, 320, 348, 376, 407, 439
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
a(12) counts these partitions: 84, 4431, 4422.
MATHEMATICA
z = 35; t1 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p]], {n, 1, z}] (* A239497 *)
t2 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p]], {n, 1, z}] (* A239498 *)
t3 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p]], {n, 1, z}] (* A118096 *)
t4 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Length[p]] == Length[p]], {n, 1, z}] (* A239500 *)
t5 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p]], {n, 1, z}] (* A239501 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 24 2014
STATUS
approved