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

1,4

Links

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

Formula

a(n) = A049501(A003188(n)).

Example

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.

Prog

(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

Crossrefs

Cf. A003188, A014550, A049501.

Sequence in context: A236532 A077763 A030218 * A127440 A118198 A348691

Adjacent sequences: A281385 A281386 A281387 * A281389 A281390 A281391

Keyword

nonn,base

Author

Indranil Ghosh, Jan 21 2017

Status

approved