%I #8 Jan 28 2022 00:58:42

%S 0,0,0,0,0,1,1,1,1,0,1,2,2,3,3,3,3,5,5,7,8,9,10,13,14,17,20,23,26,32,

%T 35,42,48,55,63,75,83,97,111,127,144,168,188,217,246,280,317,365,409,

%U 467,528,598,674,768,861,977,1099,1239,1392,1575,1762,1987

%N Number of (5,1)-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 The (5,1)-separable partitions of 14 are 95, 3515, 2525, so that a(14) = 3.

%t z = 70; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p]], {n, 1, z}] (* A008483 *)

%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p]], {n, 1, z}] (* A239493 *)

%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p]], {n, 1, z}] (* A239494 *)

%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p]], {n, 1, z}] (* A239495 *)

%t Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p]], {n, 1, z}] (* A239496 *)

%Y Cf. A230467, A008483, A239493, A239494, A239495.

%K nonn,easy

%O 1,12

%A _Clark Kimberling_, Mar 20 2014