%I #8 Mar 30 2012 18:50:19

%S 1,2,5,4,17,18,73,8,65,66,529,68,545,546,4369,16,257,258,4129,260,

%T 4161,4162,66593,264,4225,4226,67617,4228,67649,67650,1082401,32,1025,

%U 1026,32833,1028,32897,32898,1052737,1032,33025,33026,1056833,33028,1056897

%N Reduced binary string self-substitutions: a(n) is obtained by substituting n for each 1-bit in the binary expansion of n, then dividing by n.

%C By convention a(0)=0. a(2^n)=2^n. a(4n+2)=1+a(4n+1). a(n)=A065158(n,n)=A065157(n,n)/n=A065159(n)/n.

%F a(2^n)=2^n. a(4n+2)=1+a(4n+1). a(n)=A065158(n, n)=A065157(n, n)/n=A065159(n)/n.

%F a(n)=z(n, A062383(n)) with z(u, v) = if u=0 then 0 else if u mod 2 = 0 then z(u/2, v)*2 else z([u/2], v)*v+1. - _Reinhard Zumkeller_, Feb 15 2004

%e a(5): 5=101->(101)0(101)=1010101=85; 85/5=17.

%Y Cf. A065157, A065158. Equals A065159(n)/n.

%K base,easy,nonn

%O 1,2

%A _Marc LeBrun_, Oct 18 2001