OFFSET
1,6
COMMENTS
Also, the number of partitions of n such that (greatest part) > 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) < 0.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..1000
EXAMPLE
a(6) = 2 counts these partitions: 21111, 111111.
MATHEMATICA
z = 55; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] < Length[p]], {n, z}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 13 2014
STATUS
approved