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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.
2

%I #9 Jan 23 2017 19:15:22

%S 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,

%T 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,

%U 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

%N 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.

%H Indranil Ghosh, <a href="/A281497/b281497.txt">Table of n, a(n) for n = 1..10000</a>

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

%e 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.

%t 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 *)

%o (Python)

%o def G(n):

%o ....return bin(n^(n/2))[2:]

%o def a(n):

%o ....x=G(n)[::-1]

%o ....s=0

%o ....for i in range(1,len(x)):

%o ........if x[i-1]=="1" and x[i]=="0":

%o ............s+=i

%o ....return s

%Y Cf. A003188, A014550, A049502, A281388.

%K nonn,base

%O 1,12

%A _Indranil Ghosh_, Jan 23 2017