A063946 Write n in binary and complement second bit (from the left), with a(0)=0 and a(1)=1.

0, 1, 3, 2, 6, 7, 4, 5, 12, 13, 14, 15, 8, 9, 10, 11, 24, 25, 26, 27, 28, 29, 30, 31, 16, 17, 18, 19, 20, 21, 22, 23, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 96, 97, 98, 99, 100, 101, 102

Offset: 0

Author

Henry Bottomley, Sep 03 2001

Keywords

• easy
• nonn
• base

Comments

From Yosu Yurramendi, Mar 21 2017: (Start)

This sequence is a self-inverse permutation of the integers. Except for fixed points 0, 1, it consists completely of 2-cycles: (2^(m+1)+k, 2^(m+1)+2^m+k), m >= 0, 0 <= k < 2^m.

A071766(a(n)) = A229742(n), A229742(a(n)) = A071766(n), n > 0.

A245325(a(n)) = A245326(n), A245326(a(n)) = A245325(n), n > 0.

A065190(a(n)) = a(A065190(n)), n > 0.

A054429(a(n)) = a(A054429(n)) = A117120(n), n > 0.

A258746(a(n)) = a(A258746(n)), n > 0.

A258996(a(n)) = a(A258996(n)), n > 0. (End)

A324337(a(n)) = A324338(n), A324338(a(n)) = A324337(n), n > 0. - Yosu Yurramendi, Nov 04 2019

Examples

			a(11)=15 since 11 is written in binary as 1011, which changes to 1111, i.e., 15; a(12)=8 since 12 is written as 1100 which changes to 1000, i.e., 8.
		

Links

• Yosu Yurramendi, Table of n, a(n) for n = 0..32767

Crossrefs

Cf. A004442, A053645, A054429.

Programs

Maple

a:= proc(n) option remember;
  if n<2 then n
elif n<4 then 5-n
elif `mod`(n,2)=0 then 2*a(n/2)
else 2*a((n-1)/2) + 1
  fi; end proc;
seq(a(n), n = 0..80); # G. C. Greubel, Dec 08 2019

Mathematica

bc[n_]:=Module[{idn2=IntegerDigits[n,2]},If[idn2[[2]]==1,idn2[[2]]=0, idn2[[2]]=1];FromDigits[idn2,2]]; Join[{0,1},Array[bc,80,2]] (* Harvey P. Dale, May 31 2012 *)
a[n_]:= a[n]= If[n<2, n, If[n<4, 5-n, If[EvenQ[n], 2*a[n/2], 2*a[(n-1)/2] +1]]];  Table[a[n], {n,0,80}] (* G. C. Greubel, Dec 08 2019 *)

PARI

a(n)=if(n<2,n>0,3/2*2^floor(log(n)/log(2))-2^floor(log(4/3*n)/log(2))+n) /* Ralf Stephan */

PARI

a(n) = if(n<2,n, bitxor(n, 1<<(logint(n,2)-1))); \\ Kevin Ryde, Apr 09 2020

Python

import math
def a(n): return n if n<2 else 3/2*2**int(math.floor(math.log(n)/math.log(2))) - 2**int(math.floor(math.log(4/3*n)/math.log(2))) + n # Indranil Ghosh, Mar 22 2017

R

maxrow <- 8 # by choice
b01 <- 1
for(m in 0:(maxrow-1)){
  b01 <- c(b01,rep(0,2^(m+1))); b01[2^(m+1):(2^(m+1)+2^m-1)] <- 1
}
a <- c(1,3,2)
for(m in 0:(maxrow-2))
  for(k in 0:(2^m-1)){
    a[2^(m+2) +                 k] <- a[2^(m+1) + 2^m + k] + 2^((m+1) + b01[2^(m+2) +                 k])
    a[2^(m+2) +         + 2^m + k] <- a[2^(m+1) +       k] + 2^((m+1) + b01[2^(m+2) +         + 2^m + k])
    a[2^(m+2) + 2^(m+1) +       k] <- a[2^(m+1) + 2^m + k] + 2^((m+1) + b01[2^(m+2) + 2^(m+1) +       k])
    a[2^(m+2) + 2^(m+1) + 2^m + k] <- a[2^(m+1) +       k] + 2^((m+1) + b01[2^(m+2) + 2^(m+1) + 2^m + k])
}
(a <- c(0,a))  # Yosu Yurramendi, Mar 30 2017

R

a <- c(1,3,2)
maxn <- 63 # by choice
for(n in 2:maxn){ a[2*n  ] <- 2*a[n]
                  a[2*n+1] <- 2*a[n] + 1  }
(a <- c(0,a))  # Yosu Yurramendi, Nov 12 2019

Sage

@CachedFunction
def a(n):
    if (n<2): return n
    elif (n<4): return 5-n
    elif (mod(n,2)==0): return 2*a(n/2)
    else: return 2*a((n-1)/2) + 1
[a(n) for n in (0..80)] # G. C. Greubel, Dec 08 2019

Formula

If 2*2^k <= n < 3*2^k then a(n) = n + 2^k; if 3*2^k <= n < 4*2^k then a(n) = n - 2^k.

a(0)=0, a(1)=1, a(2)=3, a(3) = 2, a(2n) = 2*a(n), a(2n+1) = 2*a(n) + 1. - Ralf Stephan, Aug 23 2003