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Search: id:A065159
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| A065159 |
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Binary string self-substitutions: a(n) is obtained by substituting n for each 1-bit in binary expansion of n. |
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+0
4
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| 0, 1, 4, 15, 16, 85, 108, 511, 64, 585, 660, 5819, 816, 7085, 7644, 65535, 256, 4369, 4644, 78451, 5200, 87381, 91564, 1531639, 6336, 105625, 109876, 1825659, 118384, 1961821, 2029500, 33554431, 1024, 33825, 34884, 1149155, 37008, 1217189
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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a(0)=0. a(2^n)=4^n. a(4n+2)=(4n+2)*(1+a(4n+1)/(4n+1)). a(n)=A065157(n,n)=A065158(n,n)*n=A065160(n)*n.
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FORMULA
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a(n)=z(n, 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)*A062383(v)+v. - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Feb 15 2004
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EXAMPLE
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a(5): 5=101->(101)0(101)=1010101=85.
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CROSSREFS
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Cf. A065157, A065158, A065160.
Sequence in context: A135658 A066862 A103540 this_sequence A051956 A032826 A022133
Adjacent sequences: A065156 A065157 A065158 this_sequence A065160 A065161 A065162
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KEYWORD
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base,easy,nonn
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AUTHOR
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Marc LeBrun (mlb(AT)well.com), Oct 18 2001
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