

A239471


Number of 5separable partitions of n; see Comments.


4



0, 0, 0, 0, 0, 1, 2, 2, 3, 2, 4, 5, 7, 8, 9, 11, 13, 16, 20, 23, 27, 31, 37, 43, 52, 59, 70, 80, 93, 108, 126, 144, 167, 191, 221, 253, 292, 332, 382, 435, 498, 567, 649, 736, 839, 951, 1082, 1226, 1393, 1573, 1784, 2013, 2277, 2568, 2902, 3266, 3683, 4141
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OFFSET

1,7


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..58.


EXAMPLE

(5,0)separable partitions of 7: 151
(5,1)separable partitions of 7: 52
(5,2)separable partitions of 7: (none)
5separable partitions of 7: 151, 52, so that a(7) = 2.


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, A239468, A239469, A239470.
Sequence in context: A319706 A305894 A305811 * A241509 A268327 A053023
Adjacent sequences: A239468 A239469 A239470 * A239472 A239473 A239474


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 20 2014


STATUS

approved



