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Number of distinct prime factors in floor((4/3)^n).
0

%I #8 Nov 21 2013 12:47:35

%S 0,0,1,1,1,1,1,1,1,1,1,1,3,2,2,2,2,2,2,3,4,3,2,3,2,3,2,2,3,2,2,4,2,3,

%T 3,2,3,1,2,2,3,1,2,2,3,2,3,3,3,3,4,3,2,2,4,3,3,3,1,4,4,3,2,2,5,6,6,5,

%U 2,4,2,2,2,3,4,3,2,4,5,5,5,4,2,3,4,4,3,5,5,4,3,4,3,3,3,1,3,3,3,4,3,3,4,2,3

%N Number of distinct prime factors in floor((4/3)^n).

%t PrimeNu/@Floor[(4/3)^Range[110]] (* _Harvey P. Dale_, Nov 02 2011 *)

%K easy,nonn

%O 1,13

%A _Benoit Cloitre_, Feb 23 2002