A108711 Number of partitions of n with floor(2n/3) parts.

0, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5, 7, 7, 7, 11, 11, 11, 15, 15, 15, 22, 22, 22, 30, 30, 30, 42, 42, 42, 56, 56, 56, 77, 77, 77, 101, 101, 101, 135, 135, 135, 176, 176, 176, 231, 231, 231, 297, 297, 297, 385, 385, 385, 490, 490, 490, 627, 627, 627, 792, 792, 792, 1002

Offset: 1

Author

John W. Layman, Jun 20 2005

Keywords

• nonn

Comments

It would be interesting to know whether the sequence continues with runs of length 3 of terms of equal values.

The number of partitions of n with floor(2n/3) = A004523(n) parts equals the number of partitions of n with maximum part floor(2n/3). This leaves n-floor(2n/3) = ceiling(n/3) = A002264(n+2) as the sum of all the other parts, with no further restriction since floor(2n/3) >= ceiling(n/3) remains the largest part for any partition of the remainder, at least for n > 1. Since A002264 triplicates the integers, this sequence here triplicates the entries of A000041. - R. J. Mathar, Jul 31 2010, Feb 22 2012

Examples

			The partitions of 6 are {{6},{5,1},{4,2},{4,1,1},{3,3},{3,2,1},{3,1,1,1},{2,2,2},{2,2,1,1},{2,1,1,1,1},{1,1,1,1,1,1}}, of which 2 have 4 parts. Thus a(6)=2.
		

Crossrefs

Cf. A066639.

Extensions

Sequence extended by R. J. Mathar, Jul 31 2010