%I
%S 0,10,1011,3110,102113,13311210,10411223,1322311410,1041142322,
%T 3213243110,1031331422,2214313310,1031331422,2214313310,1031331422,
%U 2214313310,1031331422,2214313310,1031331422,2214313310,1031331422
%N Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 0.
%C After a while sequence has period 2 > {1031331422,2214313310}
%e To get the term after 102113, we say: one 3's, three 1's, one 2's, one 0's, so 13311210.
%Y Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506A022513, A138485A138493.
%K easy,nonn,base
%O 0,2
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Mar 20 2008
