%I #7 Jan 28 2022 01:17:17
%S 0,0,1,2,3,4,6,8,11,14,19,24,31,39,50,62,78,96,120,147,181,220,270,
%T 327,397,478,578,693,833,994,1189,1414,1683,1994,2365,2792,3297,3880,
%U 4568,5359,6287,7354,8602,10036,11704,13618,15841,18387,21332,24702,28591
%N Number of 1-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 (1,0)-separable partitions of 7: 421, 313;
%e (1,1)-separable partitions of 7: 61, 3121;
%e (1,2)-separable partitions of 7: 151, 12121;
%e 1-separable partitions of 7: 421, 313, 61, 3121, 151, 12121, so that a(7) = 6.
%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. A239468, A239469, A239470, A239471.
%K nonn,easy
%O 1,4
%A _Clark Kimberling_, Mar 20 2014