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A330147 Number of partitions p of n such that (number of numbers in p that have multiplicity 1) != (number of numbers in p having multiplicity > 1). 0

%I #5 Feb 04 2020 14:57:19

%S 0,1,2,3,4,4,8,8,15,20,30,40,63,78,110,143,190,238,313,389,501,621,

%T 786,975,1231,1522,1901,2344,2930,3595,4451,5448,6700,8147,9974,12087,

%U 14651,17672,21326,25558,30709,36657,43770,52069,61902,73357,86921,102697

%N Number of partitions p of n such that (number of numbers in p that have multiplicity 1) != (number of numbers in p having multiplicity > 1).

%C For each partition of n, let

%C d = number of terms that are not repeated;

%C r = number of terms that are repeated.

%C a(n) is the number of partitions such that d != r.

%F a(n) + A241274(n) = A000041(n) for all n >= 0.

%e The partitions of 6 are 6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111.

%e These have d > r: 6, 51, 42, 321

%e These have d = r: 411, 3222, 21111

%e These have d < r: 33, 222, 2211, 111111

%e Thus, a(6) = 8

%t z = 30; d[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] == 1 &]]];

%t r[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] > 1 &]]]; Table[ Count[IntegerPartitions[n], p_ /; d[p] != r[p]], {n, 0, z}]

%Y Cf. A000041, A241274, A329976.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Feb 03 2020

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