A243589 Numbers returned when each binary digit of n is replaced by the sum modulo 2 of the digits to its (wrapped) left and (wrapped) right.

0, 0, 0, 3, 5, 6, 0, 5, 15, 0, 10, 15, 5, 10, 0, 9, 27, 12, 30, 3, 17, 6, 20, 29, 15, 24, 10, 23, 5, 18, 0, 17, 51, 20, 54, 27, 57, 30, 60, 5, 39, 0, 34, 15, 45, 10, 40, 57, 27, 60, 30, 51, 17, 54, 20, 45, 15, 40, 10, 39, 5, 34, 0, 33, 99, 36, 102, 43, 105, 46

Offset

1,4

Comments

a(n) = ror(n) XOR rol(n), where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 24 2016

Numbers returned by the following function: take the t binary digits of n, d(1)..d(t), and replace each with the sum d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) maps to t and (i+1 > t) maps to 1.

Links

Anthony Sand, Table of n, a(n) for n = 1..1000

Formula

for digits d(1)..d(t), d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) -> t, (i+1 > t) -> 1.

Example

For 1, the function returns d(1) = (d(1) + d(1)) mod 2 = (1 + 1) mod 2 = 0.
For 5, the initial digits are (1,0,1).
d(1) = (d(3) + d(2)) mod 2 = (1 + 0) mod 2 = 1; d(2) = (d(1) + d(3)) mod 2 = (1 + 1) mod 2 = 0; d(3) = (d(2) + d(1)) mod 2 = (0 + 1) mod 2 = 1.
The function returns (1,0,1) = 101 = 5 in base 10.

Prog

(Python)
for n in range(1, 100):
    BL = len(bin(n))-2
    x = (n>>1) + ((n&1) << (BL-1))   # A038572(n)
    x^= (n*2) - (1<<BL) + 1   # A006257(n)  for n>0
    print str(x)+', ',

Crossrefs

Cf. A006257, A035327, A038572, A055120.

Sequence in context: A186969 A111950 A274792 * A178255 A154467 A152713

Adjacent sequences: A243586 A243587 A243588 * A243590 A243591 A243592

Keyword

nonn,easy,base,look

Author

Anthony Sand, Jun 07 2014

Status

approved