A239471 - OEIS

%I #5 Jan 28 2022 01:05:57

%S 0,0,0,0,0,1,2,2,3,2,4,5,7,8,9,11,13,16,20,23,27,31,37,43,52,59,70,80,

%T 93,108,126,144,167,191,221,253,292,332,382,435,498,567,649,736,839,

%U 951,1082,1226,1393,1573,1784,2013,2277,2568,2902,3266,3683,4141

%N Number of 5-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 (5,0)-separable partitions of 7: 151

%e (5,1)-separable partitions of 7: 52

%e (5,2)-separable partitions of 7: (none)

%e 5-separable partitions of 7: 151, 52, so that a(7) = 2.

%t z = 55; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 1] <= Length[p] + 1], {n, 1, z}] (* A239467 *)

%t t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2] <= Length[p] + 1], {n, 1, z}] (* A239468 *)

%t t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 3] <= Length[p] + 1], {n, 1, z}] (* A239469 *)

%t t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 4] <= Length[p] + 1], {n, 1, z}] (* A239470 *)

%t t5 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 5] <= Length[p] + 1], {n, 1, z}] (* A239472 *)

%Y Cf. A239467, A239468, A239469, A239470.

%K nonn,easy

%O 1,7

%A _Clark Kimberling_, Mar 20 2014