|
%I #51 Apr 10 2020 07:56:24
%S 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,
%T 18,19,20,21,22,23,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,32,
%U 33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,96,97,98,99,100,101,102
%N Write n in binary and complement second bit (from the left), with a(0)=0 and a(1)=1.
%C From _Yosu Yurramendi_, Mar 21 2017: (Start)
%C 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.
%C A071766(a(n)) = A229742(n), A229742(a(n)) = A071766(n), n > 0.
%C A245325(a(n)) = A245326(n), A245326(a(n)) = A245325(n), n > 0.
%C A065190(a(n)) = a(A065190(n)), n > 0.
%C A054429(a(n)) = a(A054429(n)) = A117120(n), n > 0.
%C A258746(a(n)) = a(A258746(n)), n > 0.
%C A258996(a(n)) = a(A258996(n)), n > 0. (End)
%C A324337(a(n)) = A324338(n), A324338(a(n)) = A324337(n), n > 0. - _Yosu Yurramendi_, Nov 04 2019
%H Yosu Yurramendi, <a href="/A063946/b063946.txt">Table of n, a(n) for n = 0..32767</a>
%F 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.
%F 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
%e 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.
%p a:= proc(n) option remember;
%p if n<2 then n
%p elif n<4 then 5-n
%p elif `mod`(n,2)=0 then 2*a(n/2)
%p else 2*a((n-1)/2) + 1
%p fi; end proc;
%p seq(a(n), n = 0..80); # _G. C. Greubel_, Dec 08 2019
%t 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 *)
%t 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 *)
%o (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_ */
%o (PARI) a(n) = if(n<2,n, bitxor(n, 1<<(logint(n,2)-1))); \\ _Kevin Ryde_, Apr 09 2020
%o (Python)
%o import math
%o 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
%o (R)
%o maxrow <- 8 # by choice
%o b01 <- 1
%o for(m in 0:(maxrow-1)){
%o b01 <- c(b01,rep(0,2^(m+1))); b01[2^(m+1):(2^(m+1)+2^m-1)] <- 1
%o }
%o a <- c(1,3,2)
%o for(m in 0:(maxrow-2))
%o for(k in 0:(2^m-1)){
%o a[2^(m+2) + k] <- a[2^(m+1) + 2^m + k] + 2^((m+1) + b01[2^(m+2) + k])
%o a[2^(m+2) + + 2^m + k] <- a[2^(m+1) + k] + 2^((m+1) + b01[2^(m+2) + + 2^m + k])
%o 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])
%o 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])
%o }
%o (a <- c(0,a)) # _Yosu Yurramendi_, Mar 30 2017
%o (R)
%o a <- c(1,3,2)
%o maxn <- 63 # by choice
%o for(n in 2:maxn){ a[2*n ] <- 2*a[n]
%o a[2*n+1] <- 2*a[n] + 1 }
%o (a <- c(0,a)) # _Yosu Yurramendi_, Nov 12 2019
%o (Sage)
%o @CachedFunction
%o def a(n):
%o if (n<2): return n
%o elif (n<4): return 5-n
%o elif (mod(n,2)==0): return 2*a(n/2)
%o else: return 2*a((n-1)/2) + 1
%o [a(n) for n in (0..80)] # _G. C. Greubel_, Dec 08 2019
%Y Cf. A004442, A053645, A054429.
%K easy,nonn,base
%O 0,3
%A _Henry Bottomley_, Sep 03 2001
|
