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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 h-separable if it is (h,i)-separable for i = 0,1,2.

(2,0)-separable partitions of 7: 421, 12121;

(2,1)-separable partitions of 7: 52;

(2,2)-separable partitions of 7: 232;

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

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z = 55; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 1] <= Length[p] + 1], {n, 1, z}] (* A239467 *)

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allocated for Clark KimberlingNumber of 2-separable partitions of n; see Comments.

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

1,5

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 h-separable if it is (h,i)-separable for i = 0,1,2.

(2,0)-separable partitions of 7: 421, 12121

(2,1)-separable partitions of 7: 52

(2,2)-separable partitions of 7: 232

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

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 *)

Cf. A239467, A239469, A239470, A239471.

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nonn,easy

Clark Kimberling, Mar 20 2014

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allocated for Clark Kimberling

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