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A240209
Number of partitions p of n such that median(p) = multiplicity(max(p)).
5
0, 1, 0, 0, 2, 3, 3, 4, 7, 9, 13, 18, 24, 30, 41, 50, 70, 85, 117, 140, 182, 225, 287, 348, 442, 537, 672, 818, 1010, 1225, 1509, 1810, 2208, 2655, 3210, 3834, 4629, 5508, 6605, 7851, 9364, 11086, 13188, 15553, 18422, 21682, 25568, 29999, 35285, 41279, 48378
OFFSET
0,5
FORMULA
a(n) = A240208(n) + A240207(n) for n >= 0.
a(n) + A240207(n) + A240210 = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 3 partitions: 411, 3111, 21111.
MATHEMATICA
z = 60; f[n_] := f[n] = IntegerPartitions[n];
t1 = Table[Count[f[n], p_ /; Median[p] < Count[p, Max[p]]], {n, 0, z}] (* A240207 *)
t2 = Table[Count[f[n], p_ /; Median[p] <= Count[p, Max[p]]], {n, 0, z}] (* A240208 *)
t3 = Table[Count[f[n], p_ /; Median[p] == Count[p, Max[p]]], {n, 0, z}] (* A240209 *)
t4 = Table[Count[f[n], p_ /; Median[p] > Count[p, Max[p]]], {n, 0, z}] (* A240210 *)
t5 = Table[Count[f[n], p_ /; Median[p] >= Count[p, Max[p]]], {n, 0, z}] (* A240211 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 03 2014
STATUS
approved