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The larger term of the pair (a(n), a(n+1)) is always odd and the larger digit of any pair of adjacent digits is also odd.
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%I #18 Jun 25 2019 23:01:48

%S 0,1,3,2,5,4,7,6,9,8,91,10,11,13,15,17,19,23,25,27,29,31,30,33,32,35,

%T 37,39,45,47,49,51,50,53,52,55,54,57,59,67,69,71,70,73,72,75,74,77,76,

%U 79,89,93,90,95,92,97,94,99,96,701,98,901,101,103,105,107,109,111,110,113,115,117,119,131,130,133,132,301,135,137,139

%N The larger term of the pair (a(n), a(n+1)) is always odd and the larger digit of any pair of adjacent digits is also odd.

%C The sequence is started with a(1) = 0 and always extended with the smallest integer not yet present and not leading to a contradiction.

%C If two successive digits are equal (e.g., 3,3) we accept that there is a "larger one" (3).

%H Jean-Marc Falcoz, <a href="/A282664/b282664.txt">Table of n, a(n) for n = 1..10001</a>

%e In the 1st pair of integers (0,1) the larger term is (1), which is odd;

%e in the 2nd pair of integers (1,3) the larger term is (3), which is odd;

%e in the 3rd pair of integers (3,2) the larger term is (3), which is odd;

%e in the 4th pair of integers (2,5) the larger term is (5), which is odd;

%e ...

%e in the 9th pair of integers (9,8) the larger term is (9), which is odd;

%e in the 10th pair of integers (8,91) the larger term is (91), which is odd;

%e in the 11th pair of integers (91,10) the larger term is (91), which is odd; etc.

%e In the 1st pair of digits (0,1) the larger digit is (1), which is odd;

%e in the 2nd pair of digits (1,3) the larger digit is (3), which is odd;

%e in the 3rd pair of digits (3,2) the larger digit is (3), which is odd;

%e in the 4th pair of digits (2,5) the larger digit is (5), which is odd;

%e ...

%e in the 9th pair of digits (9,8) the larger digit is (9), which is odd;

%e in the 10th pair of digits (8,9) the larger digit is (9), which is odd;

%e in the 11th pair of digits (9,1) the larger digit is (9), which is odd; etc.

%Y Cf. A282665 (even rather than odd).

%K nonn,base

%O 1,3

%A _Eric Angelini_ and _Jean-Marc Falcoz_, Feb 20 2017