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A240305
Number of partitions p of n such that (maximal multiplicity of the parts of p) < (number of distinct parts of p).
4
1, 0, 0, 1, 1, 2, 3, 5, 7, 12, 14, 21, 29, 38, 50, 70, 90, 117, 156, 196, 253, 324, 411, 514, 650, 809, 1015, 1259, 1555, 1917, 2365, 2898, 3536, 4318, 5248, 6365, 7691, 9297, 11180, 13446, 16115, 19296, 23019, 27474, 32653, 38838, 46002, 54511, 64371, 76012
OFFSET
0,6
FORMULA
a(n) = A240306(n) - A239964(n) for n >= 0.
a(n) + A239964(n) + A240309(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 3 partitions: 51, 42, 321.
MATHEMATICA
z = 60; f[n_] := f[n] = IntegerPartitions[n]; m[p_] := Max[Map[Length, Split[p]]] (* maximal multiplicity *); d[p_] := d[p] = Length[DeleteDuplicates[p]] (* number of distinct terms *)
t1 = Table[Count[f[n], p_ /; m[p] < d[p]], {n, 0, z}] (* A240305 *)
t2 = Table[Count[f[n], p_ /; m[p] <= d[p]], {n, 0, z}] (* A240306 *)
t3 = Table[Count[f[n], p_ /; m[p] == d[p]], {n, 0, z}] (* A239964 *)
t4 = Table[Count[f[n], p_ /; m[p] >= d[p]], {n, 0, z}] (* A240308 *)
t5 = Table[Count[f[n], p_ /; m[p] > d[p]], {n, 0, z}] (* A240309 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 05 2014
STATUS
approved