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A065160
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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.
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3
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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Table of n, a(n) for n=1..45.
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FORMULA
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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
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EXAMPLE
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a(5): 5=101->(101)0(101)=1010101=85; 85/5=17.
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CROSSREFS
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Cf. A065157, A065158. Equals A065159(n)/n.
Sequence in context: A189942 A002518 A093727 * A019152 A013621 A107732
Adjacent sequences: A065157 A065158 A065159 * A065161 A065162 A065163
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KEYWORD
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base,easy,nonn
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AUTHOR
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Marc LeBrun, Oct 18 2001
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STATUS
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approved
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