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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 #8 Jul 27 2024 20:08:34

%S 1,0,1,1,3,4,7,9,13,16,24,29,39,51,69,87,118,152,199,256,330,418,534,

%T 670,838,1046,1296,1603,1960,2412,2936,3588,4342,5288,6364,7713,9272,

%U 11186,13389,16117,19213,23032,27408,32715,38810,46176,54582,64692,76286

%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) + A329976(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) = 7

%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 0,5

%A _Clark Kimberling_, Feb 03 2020