A139707 -id:A139707 - cp's OEIS frontend

A139706 Take n in binary. Rotate the binary digits to the right until a 1 once again appears as the leftmost digit. Convert back into decimal for a(n).

1, 2, 3, 4, 6, 5, 7, 8, 12, 10, 13, 9, 14, 11, 15, 16, 24, 20, 25, 18, 26, 21, 27, 17, 28, 22, 29, 19, 30, 23, 31, 32, 48, 40, 49, 36, 50, 41, 51, 34, 52, 42, 53, 37, 54, 43, 55, 33, 56, 44, 57, 38, 58, 45, 59, 35, 60, 46, 61, 39, 62, 47, 63, 64, 96, 80, 97, 72, 98, 81, 99, 68

Offset: 1

Leroy Quet, Apr 30 2008

This sequence written in binary is A139707.
This is a permutation of the positive integers. A139708 is the inverse permutation.
Moreover, the first 2^n terms are a permutation of the first 2^n positive integers. Fixed points of the permutation are A272919. - Ivan Neretin, May 10 2016

Cf. A139707 (in binary), A139708 (inverse), A272919 (fixed points).

A139706 := proc(n) local a; a := ListTools[Rotate](convert(n,base,2),1) ; while op(-1,a) = 0 do a := ListTools[Rotate](a,1) ; od: add(op(i,a)*2^(i-1),i=1..nops(a)) : end: seq(A139706(n),n=1..100) ; # R. J. Mathar, May 04 2008

Table[FromDigits[RotateRight[IntegerDigits[n, 2], IntegerExponent[n, 2] + 1], 2], {n, 72}] (* Ivan Neretin, May 10 2016 *)

More terms from R. J. Mathar, May 04 2008

A139709 Take n in binary. Rotate the binary digits to the left until a 1 once again appears as the leftmost digit. a(n) is result written in binary.

Original entry on oeis.org

1, 10, 11, 100, 110, 101, 111, 1000, 1100, 1010, 1110, 1001, 1011, 1101, 1111, 10000, 11000, 10100, 11100, 10010, 10110, 11010, 11110, 10001, 10011, 10101, 10111, 11001, 11011, 11101, 11111, 100000, 110000, 101000, 111000, 100100, 101100

Offset: 1

Leroy Quet, Apr 30 2008

This sequence written in decimal is A139708.

			For n = 11 (in decimal): 11 (in decimal) = 1011 in binary. Rotate once to the left, getting 0111. The leftmost digit is a 0, so rotate again to the left, getting 1110. A 1 is the leftmost digit, so stop here. a(11) therefore is 1110 (which is 14 in decimal).

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A007088, A139707, A139708.

A007088 := proc(L) local i ; add(op(i,L)*10^(i-1),i=1..nops(L)) : end: A139709 := proc(n) local a; a := ListTools[Rotate](convert(n,base,2),-1) ; while op(-1,a) = 0 do a := ListTools[Rotate](a,-1) ; od: A007088(a) ; end: seq(A139709(n),n=1..80) ; # R. J. Mathar, May 04 2008

Array[FromDigits@ NestWhile[RotateLeft, IntegerDigits[#, 2], First@ # == 0 &, {2, 1}] &, 37] (* Michael De Vlieger, Oct 22 2017 *)

rot(n) = if(#Str(n)==1, v=vector(1), v=vector(#n-1)); for(i=2, #n, v[i-1]=n[i]); u=vector(#n); for(i=1, #n, u[i]=n[i]); v=concat(v, u[1]); v
a(n) = my(b=rot(binary(n))); while(b[1]==0, b=rot(b)); subst(Pol(b), x, 10) \\ Felix Fröhlich, Oct 22 2017

More terms from R. J. Mathar, May 04 2008

cp's OEIS Frontend

A139706 Take n in binary. Rotate the binary digits to the right until a 1 once again appears as the leftmost digit. Convert back into decimal for a(n).

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