

A239468


Number of 2separable partitions of n; see Comments.


5



0, 0, 1, 1, 2, 3, 4, 6, 7, 10, 12, 16, 20, 25, 31, 39, 47, 59, 71, 87, 105, 128, 153, 185, 221, 265, 315, 377, 445, 530, 625, 739, 870, 1025, 1201, 1411, 1649, 1930, 2249, 2625, 3050, 3549, 4116, 4773, 5523, 6391, 7375, 8515, 9806, 11293, 12980, 14917, 17110
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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 hseparable if it is (h,i)separable for i = 0,1,2.


LINKS

Table of n, a(n) for n=1..53.


EXAMPLE

(2,0)separable partitions of 7: 421, 12121
(2,1)separable partitions of 7: 52
(2,2)separable partitions of 7: 232
2separable partitions of 7: 421, 12121, 52, 232, so that a(7) = 4.


MATHEMATICA

z = 55; t1 = 1 + Table[Count[IntegerPartitions[n], p_ /; Length[p]  1 <= 2 Count[p, 1] <= Length[p] + 1], {n, 1, z}] (* A239467 *)
t2 = 1 + Table[Count[IntegerPartitions[n], p_ /; Length[p]  1 <= 2 Count[p, 2] <= Length[p] + 1], {n, 1, z}] (* A239468 *)
t3 = 1 + Table[Count[IntegerPartitions[n], p_ /; Length[p]  1 <= 2 Count[p, 3] <= Length[p] + 1], {n, 1, z}] (* A239469 *)
t4 = 1 + Table[Count[IntegerPartitions[n], p_ /; Length[p]  1 <= 2 Count[p, 4] <= Length[p] + 1], {n, 1, z}] (* A239470 *)
t5 = 1 + Table[Count[IntegerPartitions[n], p_ /; Length[p]  1 <= 2 Count[p, 5] <= Length[p] + 1], {n, 1, z}] (* A239472 *)


CROSSREFS

Cf. A239467, A239469, A239470, A239471.
Sequence in context: A137606 A320224 A328172 * A119793 A181436 A199118
Adjacent sequences: A239465 A239466 A239467 * A239469 A239470 A239471


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 20 2014


STATUS

approved



