1,3

Table of n, a(n) for n=1..50.

a(5)=2 because there are 2 such partition of 5: {5}, {2,3}.

<< DiscreteMath`Combinatorica`; np[n_]:= Length@Select[Mod[ #, 3]& /@ Partitions[n], (Length@# != Length@Union@#)&]; lst = Array[np, 50]

Sequence in context: A050493 A085454 A083403 * A209580 A166008 A194291

Adjacent sequences: A114088 A114089 A114090 * A114092 A114093 A114094

nonn

Giovanni Resta, Feb 06 2006

approved