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A240309
Number of partitions p of n such that (maximal multiplicity of the parts of p) > (number of distinct parts of p).
4
0, 0, 1, 1, 2, 2, 5, 6, 9, 13, 17, 23, 33, 42, 59, 76, 100, 128, 173, 212, 275, 350, 447, 552, 704, 870, 1094, 1346, 1672, 2048, 2540, 3084, 3775, 4595, 5592, 6764, 8180, 9853, 11865, 14250, 17075, 20404, 24376, 29024, 34498, 41012, 48550, 57463, 67873
OFFSET
0,5
FORMULA
a(n) = A240308(n) - A239964(n) for n >= 0.
a(n) + A240305(n) + A239964(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 5 partitions: 33, 3111, 222, 21111, 111111.
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