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A321021 a(0)=0, a(1)=1; thereafter a(n) = a(n-2)+a(n-1), keeping just the digits that appear exactly once.

%I A321021 #25 Nov 19 2018 14:03:22
%S A321021 0,1,1,2,3,5,8,13,21,34,0,34,34,68,102,170,7,1,8,9,17,26,43,69,2,71,
%T A321021 73,1,74,75,149,4,153,157,310,467,0,467,467,934,40,974,4,978,982,1960,
%U A321021 94,2054,2148,40,21,61,82,143,5,148,153,301,5,306,3
%N A321021 a(0)=0, a(1)=1; thereafter a(n) = a(n-2)+a(n-1), keeping just the digits that appear exactly once.
%C A321021 a(n) = A320486(a(n-2)+a(n-1)).
%C A321021 This must eventually enter a cycle, since there are only finitely many pairs of numbers that both have distinct digits. In fact, at step 171, enters a cycle of length 100 (see A321022).
%C A321021 Another entry into this cycle would be to start with 2, 1 and use the same rule, in which case the sequence would begin (2, 1, 3, 4, 7, 0, 7, 7, 14, 21, 35, 56, 91, 147, 238, 385, 623, ..., 40, 80, 120), a cycle of length 100 that repeats (cf. A321022).
%H A321021 N. J. A. Sloane, <a href="/A321021/b321021.txt">Table of n, a(n) for n = 0..1000</a>
%p A321021 f:= proc(n) local F, S;
%p A321021   F:= convert(n, base, 10);
%p A321021   S:= select(t -> numboccur(t, F)>1, [$0..9]);
%p A321021   if S = {} then return n fi;
%p A321021   F:= subs(seq(s=NULL, s=S), F);
%p A321021   add(F[i]*10^(i-1), i=1..nops(F))
%p A321021 end proc: # A320486
%p A321021 x:=0: y:=1: lprint(x); lprint(y);
%p A321021 for n from 2 to 500 do
%p A321021 z:=f(x+y); lprint(z); x:=y; y:=z; od:
%Y A321021 Cf. A000045 (Fibonacci), A320486 (Angelini's contraction), A321022.
%K A321021 nonn,base
%O A321021 0,4
%A A321021 _N. J. A. Sloane_, Nov 19 2018