A281497 Write n in binary reflected Gray code and sum the positions where there is a '1' followed immediately to the left by a '0', counting the rightmost digit as position 1.

0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 2, 1, 0, 0, 1, 2, 2, 0, 0, 1, 0, 3, 4, 3, 3, 2, 2, 1, 0, 0, 1, 2, 2, 3, 3, 4, 3, 0, 1, 0, 0, 2, 2, 1, 0, 4, 5, 6, 6, 4, 4, 5, 4, 3, 4, 3, 3, 2, 2, 1, 0, 0, 1, 2, 2, 3, 3, 4, 3, 4, 5, 4, 4, 6, 6, 5, 4, 0, 1, 2, 2, 0, 0, 1, 0, 3, 4, 3, 3, 2, 2, 1, 0, 5, 6, 7, 7, 8, 8, 9, 8, 5, 6, 5, 5, 7, 7, 6

Offset: 1

Author

Indranil Ghosh, Jan 23 2017

Keywords

• nonn
• base

Examples

			For n = 12, the binary reflected Gray code for 12 is '1010'. In '1010', the position of '1' followed immediately to the left by a '0' counting from right is 2. So, a(12) = 2.
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003188, A014550, A049502, A281388.

Programs

Mathematica

Table[If[Length@ # == 0, 0, Total[#[[All, 1]]]] &@ SequencePosition[ Reverse@ IntegerDigits[#, 2] &@ BitXor[n, Floor[n/2]], {1, 0}], {n, 120}] (* Michael De Vlieger, Jan 23 2017, Version 10.1, after Robert G. Wilson v at A003188 *)

Python

def G(n):
    return bin(n^(n/2))[2:]
def a(n):
    x=G(n)[::-1]
    s=0
    for i in range(1,len(x)):
        if x[i-1]=="1" and x[i]=="0":
            s+=i
    return s

Formula

a(n) = A049502(A003188(n)).