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

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, 73, 1, 74, 75, 149, 4, 153, 157, 310, 467, 0, 467, 467, 934, 40, 974, 4, 978, 982, 1960, 94, 2054, 2148, 40, 21, 61, 82, 143, 5, 148, 153, 301, 5, 306, 3

Offset: 0

Author

N. J. A. Sloane, Nov 19 2018

Keywords

• nonn
• base

Comments

a(n) = A320486(a(n-2)+a(n-1)).

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).

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).

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000045 (Fibonacci), A320486 (Angelini's contraction), A321022.

Programs

Maple

f:= proc(n) local F, S;
  F:= convert(n, base, 10);
  S:= select(t -> numboccur(t, F)>1, [$0..9]);
  if S = {} then return n fi;
  F:= subs(seq(s=NULL, s=S), F);
  add(F[i]*10^(i-1), i=1..nops(F))
end proc: # A320486
x:=0: y:=1: lprint(x); lprint(y);
for n from 2 to 500 do
z:=f(x+y); lprint(z); x:=y; y:=z; od: