login
A241090
Number of partitions p of n such that (number of numbers in p having multiplicity > 1) = number of 1s in p.
3
1, 0, 1, 1, 1, 3, 3, 5, 6, 10, 12, 16, 21, 29, 36, 47, 58, 77, 93, 121, 146, 185, 225, 280, 338, 419, 505, 612, 743, 888, 1075, 1283, 1539, 1822, 2190, 2575, 3073, 3612, 4287, 5022, 5936, 6938, 8158, 9527, 11151, 12983, 15156, 17617, 20468, 23770, 27531
OFFSET
0,6
EXAMPLE
a(6) counts these 3 partitions: 6, 42, 2211.
MATHEMATICA
z = 30; f[n_] := f[n] = IntegerPartitions[n]; e[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] > 1 &]]]; Table[Count[f[n], p_ /; e[p] == Count[p, 1]], {n, 0, z}]
CROSSREFS
Sequence in context: A048274 A351012 A059892 * A091607 A276434 A183561
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 24 2014
STATUS
approved