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A239498
Number of partitions p of n such that if h = 2*min(p), then h is an (h,1)-separator of p; see Comments.
4
0, 0, 1, 0, 0, 2, 0, 1, 3, 1, 2, 5, 4, 4, 8, 7, 9, 15, 15, 18, 23, 26, 32, 43, 47, 57, 72, 80, 98, 120, 138, 163, 198, 227, 267, 323, 372, 438, 517, 596, 696, 818, 944, 1098, 1282, 1477, 1711, 1989, 2285, 2637, 3049, 3496, 4023, 4633, 5303, 6080, 6976, 7968
OFFSET
1,6
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(9) counts these partitions: 63, 4212, 212121.
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