%I #8 Jan 28 2022 01:12:00

%S 0,0,1,1,2,3,4,6,7,10,12,16,20,25,31,39,47,59,71,87,105,128,153,185,

%T 221,265,315,377,445,530,625,739,870,1025,1201,1411,1649,1930,2249,

%U 2625,3050,3549,4116,4773,5523,6391,7375,8515,9806,11293,12980,14917,17110

%N Number of 2-separable partitions of n; see Comments.

%C 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.

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

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

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

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

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

%t t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2] <= Length[p] + 1], {n, 1, z}] (* A239468 *)

%t t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 3] <= Length[p] + 1], {n, 1, z}] (* A239469 *)

%t t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 4] <= Length[p] + 1], {n, 1, z}] (* A239470 *)

%t t5 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 5] <= Length[p] + 1], {n, 1, z}] (* A239472 *)

%Y Cf. A239467, A239469, A239470, A239471.

%K nonn,easy

%O 1,5

%A _Clark Kimberling_, Mar 20 2014