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Number of partitions p of n such that if h = min(p), then h is an (h,1)-separator of p; see Comments.
4

%I #12 Jan 28 2022 00:59:13

%S 0,0,1,1,2,3,4,5,7,9,11,15,17,22,28,34,40,52,60,75,90,109,129,160,186,

%T 225,268,321,376,455,530,632,743,878,1028,1219,1416,1667,1947,2281,

%U 2648,3103,3593,4189,4857,5638,6516,7564,8715,10080,11614,13394,15392

%N Number of partitions p of n such that if h = min(p), then h is an (h,1)-separator of p; 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 a(7) counts these partitions: 61, 52, 43, 3211.

%t z = 35; t1 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p]], {n, 1, z}] (* A239497 *)

%t t2 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p]], {n, 1, z}] (* A239498 *)

%t t3 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p]], {n, 1, z}] (* A118096 *)

%t t4 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Length[p]] == Length[p]], {n, 1, z}] (* A239500 *)

%t t5 = Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p]], {n, 1, z}] (* A239501 *)

%Y Cf. A239498, A118096, A239500, A239501, A239482.

%K nonn,easy

%O 1,5

%A _Clark Kimberling_, Mar 24 2014