A038572 a(n) = n rotated one binary place to the right.

0, 1, 1, 3, 2, 6, 3, 7, 4, 12, 5, 13, 6, 14, 7, 15, 8, 24, 9, 25, 10, 26, 11, 27, 12, 28, 13, 29, 14, 30, 15, 31, 16, 48, 17, 49, 18, 50, 19, 51, 20, 52, 21, 53, 22, 54, 23, 55, 24, 56, 25, 57, 26, 58, 27, 59, 28, 60, 29, 61, 30, 62, 31, 63, 32, 96, 33, 97, 34, 98, 35, 99, 36, 100

Offset: 0

Author

Marc LeBrun

Keywords

• easy
• base
• nonn
• nice

Comments

Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010

Examples

			For n = 35, 35_10 = 100011_2, which after rotating one binary place to the right becomes 110001. Now, 110001_2 = 49_10. So, a(35) = 49. - _Indranil Ghosh_, Jan 21 2017
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..20000 (first 1024 terms from T. D. Noe)

Crossrefs

Cf. A006257, A088146.

Programs

Haskell

a038572 0 = 0
a038572 n = a053645 n * m + n' where (n', m) = divMod n 2
-- Reinhard Zumkeller, Dec 03 2012

Maple

A038572 := proc(n)
    convert(n,base,2) ;
    ListTools[Rotate](%,1) ;
    add( op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, May 20 2016

Mathematica

Table[ FromDigits[ RotateRight[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v *)

PARI

a(n)=if(n<2,return(n)); my(d=binary(n)); fromdigits(concat(d[#d], d[1..#d-1]),2) \\ Charles R Greathouse IV, Sep 02 2015

Python

def A038572(n):
    x = bin(n)[2:]
    return int(x[-1]+x[:-1],2) # Indranil Ghosh, Jan 21 2017

Python

def A038572(n): return (n>>1)+(1<Chai Wah Wu, Jan 22 2023

Formula

a(n) = A053645(n) * A000035(n) + A004526(n) = most significant bit(n) * least significant bit(n) + floor(n/2).

a(0)=0, a(1)=1, a(2n) = n, a(2n+1) = 2a(n) + 2a(n+1) - n. - Ralf Stephan, Oct 24 2003