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%I #3 Aug 24 2012 10:50:01
%S 5,15,1511,3115,152113,13311215,15411223,1322311415,1541142322,
%T 3213243115,1531331422,2214313315,1531331422,2214313315,1531331422,
%U 2214313315,1531331422,2214313315,1531331422,2214313315,1531331422
%N Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 5.
%C After a while sequence has period 2 -> {1531331422,2214313315}
%e To get the term after 152113, we say: one 3's, three 1's, one 2's, one 5's, so 13311215
%Y Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138488, A138490-A138493.
%K easy,nonn,base
%O 0,1
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Mar 20 2008