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%I #9 Jan 28 2022 01:07:21
%S 0,1,1,2,2,2,2,4,4,6,7,8,9,12,13,16,19,22,25,31,34,41,47,54,62,74,82,
%T 96,110,126,143,167,187,216,245,279,316,364,408,466,527,597,673,767,
%U 860,976,1098,1238,1391,1574,1761,1986,2228,2502,2801,3150,3518
%N Number of (5,0)-separable 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 h-separable if it is (h,i)-separable for i = 0,1,2.
%e The (5,0)-separable partitions of 13 are 751, 652, 454, 15151, so that a(13) = 4.
%t z = 65; -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 1] == Length[p] - 1], {n, 2, z}] (* A165652 *)
%t -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2] == Length[p] - 1], {n, 3, z}] (* A239482 *)
%t -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 3] == Length[p] - 1], {n, 4, z}] (* A239483 *)
%t -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 4] == Length[p] - 1], {n, 5, z}] (* A239484 *)
%t -1 + Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 5] == Length[p] - 1], {n, 6, z}] (* A239485 *)
%Y Cf. A239467, A239481, A165652, A239483, A239484.
%K nonn,easy
%O 6,4
%A _Clark Kimberling_, Mar 20 2014
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