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A240213
Number of partitions p of n such that median(p) <= multiplicity(min(p)).
5
0, 1, 1, 1, 3, 3, 6, 8, 13, 18, 24, 31, 45, 57, 78, 102, 136, 174, 230, 291, 379, 478, 608, 761, 965, 1196, 1497, 1852, 2299, 2822, 3487, 4258, 5222, 6350, 7737, 9370, 11362, 13687, 16512, 19825, 23806, 28460, 34041, 40552, 48312, 57365, 68081, 80578, 95334
OFFSET
0,5
FORMULA
a(n) = A240212(n) + A240213(n) for n >= 0.
a(n) + A240215(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 13 partitions: 611, 5111, 422, 4211, 41111, 3311, 32111, 311111, 2222, 22211, 221111, 2111111, 11111111.
MATHEMATICA
z = 40; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Median[p] < Count[p, Min[p]]], {n, 0, z}] (* A240212 *)
t2 = Table[Count[f[n], p_ /; Median[p] <= Count[p, Min[p]]], {n, 0, z}] (* A240213 *)
t3 = Table[Count[f[n], p_ /; Median[p] == Count[p, Min[p]]], {n, 0, z}] (* A240214 *)
t4 = Table[Count[f[n], p_ /; Median[p] > Count[p, Min[p]]], {n, 0, z}] (* A240215 *)
t5 = Table[Count[f[n], p_ /; Median[p] >= Count[p, Min[p]]], {n, 0, z}] (* A240216 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 04 2014
STATUS
approved