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A065160
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.
3
1, 2, 5, 4, 17, 18, 73, 8, 65, 66, 529, 68, 545, 546, 4369, 16, 257, 258, 4129, 260, 4161, 4162, 66593, 264, 4225, 4226, 67617, 4228, 67649, 67650, 1082401, 32, 1025, 1026, 32833, 1028, 32897, 32898, 1052737, 1032, 33025, 33026, 1056833, 33028, 1056897
OFFSET
1,2
COMMENTS
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.
FORMULA
a(2^n)=2^n. a(4n+2)=1+a(4n+1). a(n)=A065158(n, n)=A065157(n, n)/n=A065159(n)/n.
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
EXAMPLE
a(5): 5=101->(101)0(101)=1010101=85; 85/5=17.
CROSSREFS
Cf. A065157, A065158. Equals A065159(n)/n.
Sequence in context: A093727 A286146 A281922 * A255544 A019152 A013621
KEYWORD
base,easy,nonn
AUTHOR
Marc LeBrun, Oct 18 2001
STATUS
approved