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A240200
Number of partitions p of n such that mean(p) < multiplicity(max(p)).
5
0, 0, 1, 1, 1, 2, 3, 3, 4, 5, 7, 9, 11, 14, 17, 21, 24, 33, 37, 47, 56, 67, 79, 100, 109, 137, 161, 189, 217, 272, 297, 365, 416, 485, 560, 685, 726, 891, 1029, 1176, 1314, 1600, 1728, 2085, 2336, 2637, 3020, 3621, 3802, 4554, 5171, 5820, 6461, 7691, 8266
OFFSET
0,6
FORMULA
a(n) = A240201(n) - A116900(n) for n >= 0.
a(n) + A116900(n) + A240202(n) = A000041(n) for n >= 1.
EXAMPLE
a(6) counts these 3 partitions: 222, 2211, 111111.
MATHEMATICA
z = 60; f[n_] := f[n] = IntegerPartitions[n];
t1 = Table[Count[f[n], p_ /; Mean[p] < Count[p, Max[p]]], {n, 0, z}] (* A240200 *)
t2 = Table[Count[f[n], p_ /; Mean[p] <= Count[p, Max[p]]], {n, 0, z}] (* A240201 *)
t3 = Table[Count[f[n], p_ /; Mean[p] == Count[p, Max[p]]], {n, 0, z}] (* A116900 *)
t4 = Table[Count[f[n], p_ /; Mean[p] > Count[p, Max[p]]], {n, 0, z}] (* A240202 *)
t5 = Table[Count[f[n], p_ /; Mean[p] >= Count[p, Max[p]]], {n, 0, z}] (* A116901 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 03 2014
STATUS
approved