A281553 Write n in binary reflected Gray code, rotate one binary place to the right and convert the code back to decimal.
0, 1, 3, 1, 3, 7, 6, 2, 6, 14, 15, 7, 5, 13, 12, 4, 12, 28, 29, 13, 15, 31, 30, 14, 10, 26, 27, 11, 9, 25, 24, 8, 24, 56, 57, 25, 27, 59, 58, 26, 30, 62, 63, 31, 29, 61, 60, 28, 20, 52, 53, 21, 23, 55, 54, 22, 18, 50, 51, 19, 17, 49, 48, 16, 48, 112, 113, 49, 51, 115, 114, 50, 54, 118
Offset: 0
Examples
For n = 5, the binary reflected Gray code for n is '111'. Rotating one binary place to the right , '111' gives back '111' again. 111_2 = 7_10. So, a(5) = 7. (For n = 5, A003188(n) = 7). For n = 15, the binary reflected Gray code for n is '1000'. Rotating one binary place to the right, '1000' gives '0100'. 100_2 = 4_10. So, a(15) = 4.
Links
- Indranil Ghosh, Table of n, a(n) for n = 0..10000
Programs
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Python
def rotation(n): x=bin(n^(n/2))[2:] return int(x[-1]+x[:-1],2)
Comments