A036044 BCR(n): write in binary, complement, reverse.

1, 0, 2, 0, 6, 2, 4, 0, 14, 6, 10, 2, 12, 4, 8, 0, 30, 14, 22, 6, 26, 10, 18, 2, 28, 12, 20, 4, 24, 8, 16, 0, 62, 30, 46, 14, 54, 22, 38, 6, 58, 26, 42, 10, 50, 18, 34, 2, 60, 28, 44, 12, 52, 20, 36, 4, 56, 24, 40, 8, 48, 16, 32, 0, 126, 62, 94, 30, 110, 46, 78, 14, 118, 54, 86

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• base
• nice
• look

Comments

a(0) could be considered to be 0 if the binary representation of zero were chosen to be the empty string. - Jason Kimberley, Sep 19 2011

From Bernard Schott, Jun 15 2021: (Start)

Except for a(0) = 1, every term is even.

For each q >= 0, there is one and only one odd number h such that a(n) = 2*q iff n = h*2^m-1 for m >= 1 when q = 0, and for m >= 0 when q >= 1 (see A345401 and some examples below).

a(n) = 0 iff n = 2^m-1 for m >= 1 (Mersenne numbers) (A000225).

a(n) = 2 iff n = 3*2^m-1 for m >= 0 (A153893).

a(n) = 4 iff n = 7*2^m-1 for m >= 0 (A086224).

a(n) = 6 iff n = 5*2^m-1 for m >= 0 (A153894).

a(n) = 8 iff n = 15*2^m-1 for m >= 0 (A196305).

a(n) = 10 iff n = 11*2^m-1 for m >= 0 (A086225).

a(n) = 12 iff n = 13*2^m-1 for m >= 0 (A198274).

For k >= 1, a(n) = 2^k iff n = (2^(k+1)-1)*2^m - 1 for m >= 0.

Explanation for a(n) = 2:

For m >= 0, A153893(m) = 3*2^m-1 -> 1011...11 -> 0100...00 -> 10 -> 2 where 1011...11_2 is 10 followed by m 1's. (End)

Examples

			4 -> 100 -> 011 -> 110 -> 6.
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..10000 (first 1024 terms from T. D. Noe)

Crossrefs

Cf. A030101, A056539, A329369, A345401.

Cf. A035928 (fixed points), A195063, A195064, A195065, A195066.

Indices of terms 0, 2, 4, 6, 8, 10, 12, 14, 18, 22, 26, 30: A000225 \ {0}, A153893, A086224, A153894, A196305, A086225, A198274, A052996\{1,3}, A291557, A198276, A171389, A198275.

Programs

Haskell

import Data.List (unfoldr)
a036044 0 = 1
a036044 n = foldl (\v d -> 2 * v + d) 0 (unfoldr bc n) where
   bc 0 = Nothing
   bc x = Just (1 - m, x') where (x',m) = divMod x 2
-- Reinhard Zumkeller, Sep 16 2011

Magma

A036044:=func; // Jason Kimberley, Sep 19 2011

Maple

A036044 := proc(n)
    local bcr ;
    if n = 0 then
        return 1;
    end if;
    convert(n,base,2) ;
    bcr := [seq(1-i,i=%)] ;
    add(op(-k,bcr)*2^(k-1),k=1..nops(bcr)) ;
end proc:
seq(A036044(n),n=0..200) ; # R. J. Mathar, Nov 06 2017

Mathematica

dtn[ L_ ] := Fold[ 2#1+#2&, 0, L ]; f[ n_ ] := dtn[ Reverse[ 1-IntegerDigits[ n, 2 ] ] ]; Table[ f[ n ], {n, 0, 100} ]
Table[FromDigits[Reverse[IntegerDigits[n,2]/.{1->0,0->1}],2],{n,0,80}] (* Harvey P. Dale, Mar 08 2015 *)

PARI

a(n)=fromdigits(Vecrev(apply(n->1-n,binary(n))),2) \\ Charles R Greathouse IV, Apr 22 2015

Python

def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
print([BCR(n) for n in range(75)]) # Michael S. Branicky, Jun 14 2021

Python

def A036044(n): return -int((s:=bin(n)[-1:1:-1]),2)-1+2**len(s) # Chai Wah Wu, Feb 04 2022

Formula

a(2n) = 2*A059894(n), a(2n+1) = a(2n) - 2^floor(log_2(n)+1). - Ralf Stephan, Aug 21 2003

Conjecture: a(n) = (-1)^A023416(n)*b(n) for n > 0 with a(0) = 1 where b(2^m) = (-1)^m*(2^(m+1) - 2) for m >= 0, b(2n+1) = b(n) for n > 0, b(2n) = b(n) + b(n - 2^f(n)) + b(2n - 2^f(n)) for n > 0 and where f(n) = A007814(n) (see A329369). - Mikhail Kurkov, Dec 13 2024