

A239485


Number of (5,0)separable partitions of n; see Comments.


4



0, 1, 1, 2, 2, 2, 2, 4, 4, 6, 7, 8, 9, 12, 13, 16, 19, 22, 25, 31, 34, 41, 47, 54, 62, 74, 82, 96, 110, 126, 143, 167, 187, 216, 245, 279, 316, 364, 408, 466, 527, 597, 673, 767, 860, 976, 1098, 1238, 1391, 1574, 1761, 1986, 2228, 2502, 2801, 3150, 3518
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OFFSET

6,4


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=6..62.


EXAMPLE

The (5,0)separable partitions of 13 are 751, 652, 454, 15151, so that a(13) = 4.


MATHEMATICA

z = 65; 1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]  1], {n, 2, z}] (* A165652 *)
1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]  1], {n, 3, z}] (* A239482 *)
1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]  1], {n, 4, z}] (* A239483 *)
1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]  1], {n, 5, z}] (* A239484 *)
1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]  1], {n, 6, z}] (* A239485 *)


CROSSREFS

Cf. A239467, A239481, A165652, A239483, A239484.
Sequence in context: A325710 A214927 A326115 * A237756 A010334 A010578
Adjacent sequences: A239482 A239483 A239484 * A239486 A239487 A239488


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 20 2014


STATUS

approved



