A281388 -id:A281388 - cp's OEIS frontend

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

Indranil Ghosh, Jan 23 2017

			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.

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

Cf. A003188, A014550, A049502, A281388.

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 *)

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

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

A281552 Write n in the Elias gamma 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.

Original entry on oeis.org

0, 1, 1, 2, 2, 6, 2, 3, 3, 9, 3, 8, 8, 9, 3, 4, 4, 12, 4, 11, 11, 12, 4, 10, 10, 18, 10, 11, 11, 12, 4, 5, 5, 15, 5, 14, 14, 15, 5, 13, 13, 23, 13, 14, 14, 15, 5, 12, 12, 22, 12, 21, 21, 22, 12, 13, 13, 23, 13, 14, 14, 15, 5, 6, 6, 18, 6, 17, 17, 18, 6, 16, 16, 28, 16, 17, 17, 18, 6

Offset: 1

Indranil Ghosh, Jan 24 2017

			For n = 6 , the Elias gamma code for n is '11010'. In '11010', the positions of '1' followed immediately to the right by '0' counting from left are 2 and 4. So, a(6) = 2 + 4 = 6.
For n = 10, the Elias gamma code for n is '1110010'. In '1110010', the positions of '1' followed immediately to the right by '0' counting from left are 3 and 6. So, a(10) = 3 + 6 = 9.

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

Cf. A049501, A171885, A281149, A281388, A281497.

def a(n):
    x= A281149(n)
    s=0
    for i in range(1,len(x)):
        if x[i-1]=="1" and x[i]=="0":
            s+=i
    return s

a(n) = A049501(A171885(n)) for n > = 1.

cp's OEIS Frontend

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.

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