A329976 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, 1, 1, 2, 2, 3, 4, 6, 9, 14, 18, 27, 38, 50, 66, 89, 113, 145, 186, 234, 297, 374, 468, 585, 737, 912, 1140, 1407, 1758, 2153, 2668, 3254, 4007, 4855, 5946, 7170, 8705, 10451, 12626, 15068, 18125, 21551, 25766, 30546, 36365, 42958, 50976, 60062, 70987

Offset: 0

Author

Clark Kimberling, Feb 03 2020

Keywords

• nonn
• easy

Comments

For each partition of n, let

d = number of terms that are not repeated;

r = number of terms that are repeated.

a(n) is the number of partitions such that d > r.

Also the number of integer partitions of n with median multiplicity 1. - Gus Wiseman, Mar 20 2023

Examples

			The partitions of 6 are 6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111.
These have d > r:  6, 51, 42, 321
These have d = r:  411, 3222, 21111
These have d < r:  33, 222, 2211, 111111
Thus, a(6) = 4.
		

Crossrefs

For parts instead of multiplicities we have A027336

The complement is counted by A330001.

A000041 counts integer partitions, strict A000009.

A116608 counts partitions by number of distinct parts.

A237363 counts partitions with median difference 0.

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

Programs

Mathematica

z = 30; d[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] == 1 &]]];
r[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] > 1 &]]]; Table[Count[IntegerPartitions[n], p_ /; d[p] > r[p]], {n, 0, z}]

Formula

a(n) + A241274(n) + A330001(n) = A000041(n) for n >= 0.