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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.

%I A317050 #25 Feb 06 2024 19:31:19
%S A317050 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,
%T A317050 19,18,14,15,17,16,-16,-15,-17,-18,-14,-13,-11,-12,-20,-19,-21,-22,
%U A317050 -26,-25,-23,-24,-40,-39,-41,-42,-38,-37,-35,-36,-28,-27,-29,-30
%N 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.
%C A317050 Binary Gray code, interpreted as negabinary number.
%C A317050 This sequence is a bijection from nonnegative integers to signed integers.
%C A317050 This sequence has similarities with A317018; in both sequences, the negabinary representations of consecutive terms differ exactly by one digit.
%H A317050 Rémy Sigrist, <a href="/A317050/b317050.txt">Table of n, a(n) for n = 0..16383</a>
%H A317050 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Negabinary.html">Negabinary</a>.
%H A317050 Wikipedia, <a href="https://en.wikipedia.org/wiki/Negative_base">Negative base</a>.
%F A317050 a(n) = A053985(A003188(n)).
%e A317050 The first terms, alongside their negabinary representation, are:
%e A317050   n   a(n)  nega(a(n))
%e A317050   --  ----  ----------
%e A317050    0     0        0
%e A317050    1     1        1
%e A317050    2    -1       11
%e A317050    3    -2       10
%e A317050    4     2      110
%e A317050    5     3      111
%e A317050    6     5      101
%e A317050    7     4      100
%e A317050    8    -4     1100
%e A317050    9    -3     1101
%e A317050   10    -5     1111
%e A317050   11    -6     1110
%e A317050   12   -10     1010
%e A317050   13    -9     1011
%e A317050   14    -7     1001
%e A317050   15    -8     1000
%e A317050   16     8    11000
%e A317050   17     9    11001
%e A317050   18     7    11011
%e A317050   19     6    11010
%e A317050   20    10    11110
%e A317050 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
%o A317050 (PARI) a(n) = fromdigits(binary(bitxor(n, n>>1)), -2)
%Y A317050 Cf. A003188, A039724, A053985, A212529, A317018.
%K A317050 sign,look,base
%O A317050 0,4
%A A317050 _Rémy Sigrist_, Jul 20 2018