cp's OEIS Frontend

A239482 Number of (2,0)-separable partitions of n; see Comments.

%I A239482 #9 Jan 28 2022 01:19:45
%S A239482 0,1,0,1,2,2,3,5,5,7,10,11,14,19,21,27,34,39,48,60,69,84,102,119,142,
%T A239482 172,199,237,282,328,387,458,530,623,730,847,987,1153,1331,1547,1796,
%U A239482 2071,2394,2771,3183,3671,4227,4849,5568,6395,7313,8377,9584,10940
%N A239482 Number of (2,0)-separable partitions of n; see Comments.
%C A239482 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 A239482 The (2,0)-separable partitions of 10 are 721, 523, 424, 42121, 1212121, so that a(10) = 5.
%t A239482 z = 65; -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p] - 1], {n, 2, z}]  (* A165652 *)
%t A239482 -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p] - 1], {n, 3, z}]  (* A239482 *)
%t A239482 -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p] - 1], {n, 4, z}]  (* A239483 *)
%t A239482 -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p] - 1], {n, 5, z}]  (* A239484 *)
%t A239482 -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p] - 1], {n, 6, z}] (* A239485 *)
%Y A239482 Cf. A239467, A165652, A239483, A239484, A239485.
%K A239482 nonn,easy
%O A239482 3,5
%A A239482 _Clark Kimberling_, Mar 20 2014