cp's OEIS Frontend

A239499 Number of partitions p of n such that if h = (number of parts of p), then h is an (h,2)-separator of p; see Comments.

%I A239499 #13 Jan 28 2022 01:15:01
%S A239499 0,0,0,0,0,0,1,1,0,1,1,1,1,1,1,1,2,2,3,3,3,3,4,4,5,5,6,6,7,7,9,9,11,
%T A239499 12,14,15,17,18,20,22,24,26,29,31,34,37,40,43,48,51,56,61,67,72,80,86,
%U A239499 94,102,111,119,131,140,152,164,178,190,207,221,239
%N A239499 Number of partitions p of n such that if h = (number of parts of p), then h is an (h,2)-separator of p; see Comments.
%C A239499 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 A239499 a(19) counts these 3 partitions: [3,13,3], [5,3,5,1,5], [5,2,5,2,5].
%t A239499 z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] + 1], {n, 1, z}]  (* A239729 *)
%t A239499 Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] + 1], {n, 1, z}] (* A239481 *)
%t A239499 Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p]] == Length[p] + 1], {n, 1, z}] (* A239456 *)
%t A239499 Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] + 1], {n, 1, z}] (* A239499 *)
%t A239499 Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] + 1], {n, 1, z}]  (*A239689 *)
%Y A239499 Cf. A239729, A239481, A239456, A239689.
%K A239499 nonn,easy
%O A239499 1,17
%A A239499 _Clark Kimberling_, Mar 25 2014