

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.


3



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
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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[i1]=="1" and x[i]=="0":
............s+=i
....return s


CROSSREFS

Cf. A003188, A014550, A049501.
Sequence in context: A236532 A077763 A030218 * A127440 A118198 A290885
Adjacent sequences: A281385 A281386 A281387 * A281389 A281390 A281391


KEYWORD

nonn,base


AUTHOR

Indranil Ghosh, Jan 21 2017


STATUS

approved



