A281388 Write n in binary reflected Gray code and sum the positions where there is a '1' followed immediately to the right by a '0', counting the leftmost digit as position 1.

0, 0, 1, 2, 0, 1, 1, 2, 2, 0, 3, 4, 1, 1, 1, 2, 2, 2, 6, 4, 0, 3, 3, 4, 4, 1, 5, 5, 1, 1, 1, 2, 2, 2, 7, 7, 2, 6, 6, 4, 4, 0, 5, 8, 3, 3, 3, 4, 4, 4, 9, 6, 1, 5, 5, 5, 5, 1, 6, 6, 1, 1, 1, 2, 2, 2, 8, 8, 2, 7, 7, 7, 7, 2, 8, 12, 6, 6, 6, 4, 4, 4, 10, 6, 0, 5, 5, 8, 8, 3, 9, 9, 3, 3, 3

Offset: 1

Author

Indranil Ghosh, Jan 21 2017

Keywords

• nonn
• base

Examples

			For n = 11, the binary reflected Gray code for 11 is '1110'. In '1110', the position of '1' followed immediately to the right by '0' counting from left is 3. So, a(11) = 3.
For n = 12, the binary reflected Gray code for 12 is '1010'. In '1010', the positions of '1' followed immediately to the right by '0' counting from left are 1 and 3. So, a(12) = 1 + 3 = 4.
		

Links

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

Crossrefs

Cf. A003188, A014550, A049501.

Programs

Python

def g(n):
    return bin(n^(n//2))[2:]
def a(n):
    x=g(n)
    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) = A049501(A003188(n)).