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A239497
Number of partitions p of n such that if h = min(p), then h is an (h,1)-separator of p; see Comments.
4
0, 0, 1, 1, 2, 3, 4, 5, 7, 9, 11, 15, 17, 22, 28, 34, 40, 52, 60, 75, 90, 109, 129, 160, 186, 225, 268, 321, 376, 455, 530, 632, 743, 878, 1028, 1219, 1416, 1667, 1947, 2281, 2648, 3103, 3593, 4189, 4857, 5638, 6516, 7564, 8715, 10080, 11614, 13394, 15392
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
a(7) counts these partitions: 61, 52, 43, 3211.
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