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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).
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%I #12 Apr 07 2023 08:57:30

%S 0,1,1,2,2,3,4,6,9,14,18,27,38,50,66,89,113,145,186,234,297,374,468,

%T 585,737,912,1140,1407,1758,2153,2668,3254,4007,4855,5946,7170,8705,

%U 10451,12626,15068,18125,21551,25766,30546,36365,42958,50976,60062,70987

%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.

%C Also the number of integer partitions of n with median multiplicity 1. - _Gus Wiseman_, Mar 20 2023

%F a(n) + A241274(n) + A330001(n) = A000041(n) for 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) = 4.

%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 For parts instead of multiplicities we have A027336

%Y The complement is counted by A330001.

%Y A000041 counts integer partitions, strict A000009.

%Y A116608 counts partitions by number of distinct parts.

%Y A237363 counts partitions with median difference 0.

%Y Cf. A000975, A027193, A067538, A240219, A241274, A325347, A359893, A360005.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Feb 03 2020