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.

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.

Links

Table of n, a(n) for n=1..45.

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

Adjacent sequences: A065157 A065158 A065159 * A065161 A065162 A065163

Keyword

base,easy,nonn

Author

Marc LeBrun, Oct 18 2001

Status

approved