%I
%S 0,0,0,0,0,1,2,2,3,2,4,5,7,8,9,11,13,16,20,23,27,31,37,43,52,59,70,80,
%T 93,108,126,144,167,191,221,253,292,332,382,435,498,567,649,736,839,
%U 951,1082,1226,1393,1573,1784,2013,2277,2568,2902,3266,3683,4141
%N Number of 5separable 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 hseparable if it is (h,i)separable for i = 0,1,2.
%e (5,0)separable partitions of 7: 151
%e (5,1)separable partitions of 7: 52
%e (5,2)separable partitions of 7: (none)
%e 5separable partitions of 7: 151, 52, so that a(7) = 2.
%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, A239468, A239469, A239470.
%K nonn,easy
%O 1,7
%A _Clark Kimberling_, Mar 20 2014
