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A240205
Number of partitions p of n such that mean(p) = multiplicity(min(p)).
5
0, 1, 0, 0, 1, 0, 1, 0, 2, 1, 2, 0, 4, 0, 2, 7, 3, 0, 17, 0, 5, 26, 2, 0, 60, 1, 2, 61, 59, 0, 91, 0, 149, 119, 2, 34, 480, 0, 2, 215, 788, 0, 288, 0, 1147, 923, 2, 0, 2528, 1, 1585, 611, 3319, 0, 1150, 3963, 5366, 986, 2, 0, 20317
OFFSET
0,9
COMMENTS
a(n) = 0 if and only if n = 0 or n is a prime.
FORMULA
a(n) = A240204(n) - A240203(n) for n >= 0.
a(n) + A240203(n) + A240206(n) = A000041(n) for n >= 0.
EXAMPLE
a(12) counts these 4 partitions: 9111, 6222, 422211, 332211.
MATHEMATICA
z = 60; f[n_] := f[n] = IntegerPartitions[n];
t1 = Table[Count[f[n], p_ /; Mean[p] < Count[p, Min[p]]], {n, 0, z}] (* A240203 *)
t2 = Table[Count[f[n], p_ /; Mean[p] <= Count[p, Min[p]]], {n, 0, z}] (* A240204 *)
t3 = Table[Count[f[n], p_ /; Mean[p] == Count[p, Min[p]]], {n, 0, z}] (* A240205 *)
t4 = Table[Count[f[n], p_ /; Mean[p] > Count[p, Min[p]]], {n, 0, z}] (* A240206 *)
t5 = Table[Count[f[n], p_ /; Mean[p] >= Count[p, Min[p]]], {n, 0, z}] (* A240079 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 03 2014
STATUS
approved