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A239470 Number of 4-separable partitions of n; see Comments.

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%I A239470 #5 Jan 28 2022 01:13:03
%S A239470 0,0,0,0,1,2,2,2,3,5,6,7,8,11,13,17,19,23,27,34,40,47,55,66,77,92,106,
%T A239470 125,145,171,198,231,266,310,358,416,477,552,633,731,838,963,1100,
%U A239470 1263,1442,1651,1880,2147,2442,2785,3163,3597,4078,4631,5244,5946
%N A239470 Number of 4-separable partitions of n; see Comments.
%C A239470 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 A239470 (4,0)-separable partitions of 7: 241;
%e A239470 (4,1)-separable partitions of 7: 43;
%e A239470 (4,2)-separable partitions of 7: (none);
%e A239470 4-separable partitions of 7: 241, 43, so that a(7) = 2.
%t A239470 z = 55; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 1] <= Length[p] + 1], {n, 1, z}] (* A239467 *)
%t A239470 t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2] <= Length[p] + 1], {n, 1, z}] (* A239468 *)
%t A239470 t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 3] <= Length[p] + 1], {n, 1, z}] (* A239469 *)
%t A239470 t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 4] <= Length[p] + 1], {n, 1, z}] (* A239470 *)
%t A239470 t5 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 5] <= Length[p] + 1], {n, 1, z}] (* A239472 *)
%Y A239470 Cf. A239467, A239468, A239469, A239471.
%K A239470 nonn,easy
%O A239470 1,6
%A A239470 _Clark Kimberling_, Mar 20 2014