The On-Line Encyclopedia of Integer Sequences (OEIS)

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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.
(history; published version)

#8 by Russ Cox at Fri Mar 30 18:50:19 EDT 2012

FORMULA

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 (reinhard.zumkeller(AT)gmail.com), _, Feb 15 2004

Discussion

Fri Mar 30

18:50

OEIS Server: https://oeis.org/edit/global/246

#7 by Russ Cox at Fri Mar 30 17:38:10 EDT 2012

AUTHOR

_Marc LeBrun (mlb(AT)well.com), _, Oct 18 2001

Discussion

Fri Mar 30

17:38

OEIS Server: https://oeis.org/edit/global/183

#6 by N. J. A. Sloane at Fri Jan 09 03:00:00 EST 2009

FORMULA

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 (reinhard.zumkeller(AT)lhsystemsgmail.com), Feb 15 2004

KEYWORD

base,easy,nonn,new

#5 by N. J. A. Sloane at Fri Feb 24 03:00:00 EST 2006

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 (reinhard.zumkeller(AT)lhsystems.com), Feb 15 2004

KEYWORD

base,easy,nonn,new

#4 by N. J. A. Sloane at Wed Sep 22 03:00:00 EDT 2004

FORMULA

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 (reinhard.zumkeller(AT)lhsystems.com), Feb 15 2004

KEYWORD

base,easy,nonn,new

#3 by N. J. A. Sloane at Sat Jun 12 03:00:00 EDT 2004

FORMULA

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 (reinhard.zumkeller(AT)lhsystems.com), Feb 15 2004

KEYWORD

base,easy,nonn,new

#2 by N. J. A. Sloane at Thu Feb 19 03:00:00 EST 2004

NAME

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.

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.

CROSSREFS

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

KEYWORD

base,easy,nonn,new

#1 by N. J. A. Sloane at Fri May 16 03:00:00 EDT 2003

NAME

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

DATA

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.

EXAMPLE

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

CROSSREFS

A065157, A065158, A065159.

KEYWORD

base,easy,nonn

AUTHOR

Marc LeBrun (mlb(AT)well.com), Oct 18 2001

STATUS

approved