

A239470


Number of 4separable partitions of n; see Comments.


5



0, 0, 0, 0, 1, 2, 2, 2, 3, 5, 6, 7, 8, 11, 13, 17, 19, 23, 27, 34, 40, 47, 55, 66, 77, 92, 106, 125, 145, 171, 198, 231, 266, 310, 358, 416, 477, 552, 633, 731, 838, 963, 1100, 1263, 1442, 1651, 1880, 2147, 2442, 2785, 3163, 3597, 4078, 4631, 5244, 5946
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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

(4,0)separable partitions of 7: 241
(4,1)separable partitions of 7: 43
(4,2)separable partitions of 7: (none)
4separable partitions of 7: 241, 43, 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, A239471.
Sequence in context: A077018 A007918 A278167 * A320786 A126111 A296103
Adjacent sequences: A239467 A239468 A239469 * A239471 A239472 A239473


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 20 2014


STATUS

approved



