

A239493


Number of (2,1)separable partitions of n; see Comments.


4



0, 0, 1, 0, 1, 2, 1, 2, 3, 3, 4, 6, 6, 8, 11, 12, 15, 20, 22, 28, 35, 40, 49, 61, 70, 85, 103, 120, 143, 173, 200, 238, 283, 329, 388, 459, 531, 624, 731, 848, 988, 1154, 1332, 1548, 1797, 2072, 2395, 2772, 3184, 3672, 4228, 4850, 5569, 6396, 7314, 8378
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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..56.


EXAMPLE

The (2,1)separable partitions of 11 are 92, 6212, 4232, 321212, so that a(11) = 4.


MATHEMATICA

z = 70; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]], {n, 1, z}] (* A008483 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]], {n, 1, z}] (* A239493 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]], {n, 1, z}] (* A239494 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]], {n, 1, z}] (* A239495 *)
Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]], {n, 1, z}] (* A239496 *)


CROSSREFS

Cf. A230467, A008483, A239494, A239495, A239496.
Sequence in context: A116498 A143472 A180235 * A331849 A015739 A015746
Adjacent sequences: A239490 A239491 A239492 * A239494 A239495 A239496


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 20 2014


STATUS

approved



