A092569 Permutation of integers a(a(n)) = n. In binary representation of n, transformation of inner bits, 1 <-> 0, gives binary representation of a(n).

0, 1, 2, 3, 6, 7, 4, 5, 14, 15, 12, 13, 10, 11, 8, 9, 30, 31, 28, 29, 26, 27, 24, 25, 22, 23, 20, 21, 18, 19, 16, 17, 62, 63, 60, 61, 58, 59, 56, 57, 54, 55, 52, 53, 50, 51, 48, 49, 46, 47, 44, 45, 42, 43, 40, 41, 38, 39, 36, 37, 34, 35, 32, 33, 126, 127, 124, 125, 122, 123, 120

Offset: 0

Author

Zak Seidov, Feb 28 2004

Keywords

• nonn
• base
• easy

Comments

Primes which stay primes under transformation "opposite inner bits", A092570.

This permutation transforms the enumeration system of positive irreducible fractions A020651/A020650 into the enumeration system A245327/A245326, and vice versa. - Yosu Yurramendi, Jun 16 2015

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

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

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

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

a(n) = A054429(A065190(n)) = A065190(A054429(n)), n > 0. - Yosu Yurramendi, Mar 23 2017

Examples

			a(9)=15 because 9_10 = 1001_2, transformation of inner bits gives 1001_2 -> 1111_2 = 15_10.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A092570.

Programs

Mathematica

bb={0, 1, 2, 3};Do[id=IntegerDigits[n, 2];Do[id[[i]]=1-id[[i]], {i, 2, Length[id]-1}];bb=Append[bb, FromDigits[id, 2]], {n, 4, 1000}];fla=Flatten[bb]
(* Second program: *)
Table[If[n < 2, n, Function[b, FromDigits[#, 2] &@ Join[{First@ b}, Most[Rest@ b] /. { 0 -> 1, 1 -> 0}, {Last@ b}]]@ IntegerDigits[n, 2]], {n, 0, 70}] (* Michael De Vlieger, Apr 03 2017 *)

PARI

T(n)={pow2=2;v=binary(n);L=#v-1;forstep(k=L,2,-1,if(v[k],n-=pow2,n+=pow2);pow2*=2);return(n)};
for(n=0,70,print1(T(n),", ")) \\ Washington Bomfim, Jan 18 2011

R

maxrow <- 8 # by choice
a <- 1:3 # If it were c(1, 3, 2), it would be A054429
for(m in 1:maxrow) for(k in 0:(2^m-1)){
a[2^(m+1)+    k] = a[2^m+k] + 2^(m+1)
a[2^(m+1)+2^m+k] = a[2^m+k] + 2^m
}
a
# Yosu Yurramendi, Apr 10 2017

Formula

a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 3, a(2^(m+1) +k) = a(2^m+k) + 2^(m+1),

a(2^(m+1)+2^m+k) = a(2^m+k) + 2^m, m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Apr 02 2017