%I
%S 2,3,13,14,22,23,16,15,32,25,12,27,17,33,5,24,42,29,19,102,35,18,34,
%T 26,28,4,43,37,103,52,105,104,45,106,107,53,44,202,203,109,112,47,108,
%U 36,46,49,113,55,114,402,205,116,115,117,204,118,206,62,207,122,39,119,208,123,38,209,124,7,302,125,126,127,128,212,403,133,134,213,136,129,303,64,135,304,405,132,215,137,214,406,217,138,54,6,218,219,305,8
%N Divide a(n) by the last digit of a(n1); the remainder is the first digit of a(n+1). The sequence is started with a(1)=2 and always extended with the smallest integer not yet present in the sequence and not leading to a contradiction.
%C No term can end with the digits 0 or 1 as those would produce a division by 0. This means that no term a(n) in the sequence is divisible by the last digit of a(n1)
%H Eric Angelini, <a href="/A273617/b273617.txt">Table of n, a(n) for n = 1..10000</a>
%e 3 divided by 2 leaves 1 (this "1" starts the next integer)
%e 13 divided by 3 leaves 1 (this "1" starts the next integer)
%e 14 divided by 3 leaves 2 (this "2" starts the next integer)
%e 22 divided by 4 leaves 2 (this "2" starts the next integer)
%e 23 divided by 2 leaves 1 (this "1" starts the next integer)
%e 16 divided by 3 leaves 1 (this "1" starts the next integer)
%e 15 divided by 6 leaves 3 (this "3" starts the next integer)
%e 32 divided by 5 leaves 2 (this "2" starts the next integer)
%e 25 divided by 2 leaves 1 (this "1" starts the next integer)
%e 12 divided by 5 leaves 2 (this "2" starts the next integer)
%e 27 divided by 2 leaves 1 (this "1" starts the next integer)
%K nonn,base
%O 1,1
%A _Eric Angelini_ and _JeanMarc Falcoz_, Jun 16 2016
