login
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).
6

%I #8 Feb 04 2020 14:56:54

%S 1,0,0,0,1,3,3,7,7,10,12,16,14,23,25,33,41,59,72,101,126,171,216,280,

%T 344,436,535,666,788,970,1153,1394,1649,1996,2336,2796,3326,3965,4689,

%U 5627,6629,7926,9404,11192,13273,15777,18637,22057,26067,30672,36122

%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).

%F a(n) + A329976(n) + A330001(n) = A000041(n) for n >= 0.

%e a(6) counts these 3 partitions: 411, 3111, 21111.

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

%Y Cf. A241090, A241131, A241132.

%K nonn,easy

%O 0,6

%A _Clark Kimberling_, Apr 24 2014