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a(n) = absolute value of ( a([ n/2 ]) - a([ n/3 ]) + ... + ((-1)^n)a([ n/n ]) ).
3

%I #5 Mar 30 2012 18:55:51

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

%T 3,1,0,0,1,1,2,2,1,0,0,2,1,1,3,3,2,1,0,0,1,1,2,2,1,0,1,1,1,5,4,4,3,2,

%U 1,3,2,4,3,3,3,2,1,1,0,2,2,0,1,2,1,1,0,0,1,1,4,1,2,2,1,1,2,4,4

%N a(n) = absolute value of ( a([ n/2 ]) - a([ n/3 ]) + ... + ((-1)^n)a([ n/n ]) ).

%K nonn

%O 0,13

%A _Clark Kimberling_