

A239501


Number of partitions p of n such that if h = max(p)  min(p), then h is an (h,1)separator of p; see Comments.


4



0, 0, 1, 0, 0, 2, 0, 1, 2, 0, 3, 3, 2, 2, 3, 5, 4, 8, 4, 5, 9, 6, 13, 10, 11, 15, 14, 17, 16, 20, 21, 26, 29, 30, 33, 36, 35, 41, 47, 47, 61, 61, 66, 71, 73, 85, 88, 98, 102, 114, 122, 131, 148, 154, 163, 182, 188, 205, 220, 231, 249, 271, 293, 306, 338, 359
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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 hseparable if it is (h,i)separable for i = 0,1,2.


LINKS

Table of n, a(n) for n=1..66.


EXAMPLE

a(11) counts these partitions: 4313, 4232, 321212.


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

Cf. A239497, A239498, A239499, A239500, A239482.
Sequence in context: A025679 A071491 A156537 * A145316 A137298 A136487
Adjacent sequences: A239498 A239499 A239500 * A239502 A239503 A239504


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 24 2014


STATUS

approved



