A317050 a(0) = 0 and for any n >= 0, a(n+1) is obtained by changing the rightmost possible digit in the negabinary representation of a(n) so as to get a value not yet in the sequence.
0, 1, -1, -2, 2, 3, 5, 4, -4, -3, -5, -6, -10, -9, -7, -8, 8, 9, 7, 6, 10, 11, 13, 12, 20, 21, 19, 18, 14, 15, 17, 16, -16, -15, -17, -18, -14, -13, -11, -12, -20, -19, -21, -22, -26, -25, -23, -24, -40, -39, -41, -42, -38, -37, -35, -36, -28, -27, -29, -30
Offset: 0
Examples
The first terms, alongside their negabinary representation, are: n a(n) nega(a(n)) -- ---- ---------- 0 0 0 1 1 1 2 -1 11 3 -2 10 4 2 110 5 3 111 6 5 101 7 4 100 8 -4 1100 9 -3 1101 10 -5 1111 11 -6 1110 12 -10 1010 13 -9 1011 14 -7 1001 15 -8 1000 16 8 11000 17 9 11001 18 7 11011 19 6 11010 20 10 11110 a(8) = -4 because nega(a(7)) = 100. Changing the rightmost digit gives 101 of which the decimal value in the sequence. Similarily, changing to 110 and 000 gives no new term. Changing to 1100 does so a(8) is the decimal value of 1100 which is -4. - _David A. Corneth_, Jul 22 2018
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..16383
- Eric Weisstein's World of Mathematics, Negabinary.
- Wikipedia, Negative base.
Programs
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PARI
a(n) = fromdigits(binary(bitxor(n, n>>1)), -2)
Comments