A005678 -id:A005678 - cp's OEIS frontend

1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 3, 1, 2, 1, 3, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 3, 1, 2, 1, 3, 1, 2, 3, 2, 1, 3, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 2, 1, 3, 1, 2, 3, 2, 1, 3, 1, 2, 1, 3, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1

Offset: 0

N. J. A. Sloane, Dec 25 2009

nonn

A concatenation of blocks u_k, k >= 0, where u_k has length 5^k. The sequence is defined recursively - see the Maple code.
From Kevin Ryde, Aug 11 2020: (Start)
Bollobás gives this sequence intending it to be a squarefree ternary word, where squarefree means nowhere a repeat w w for a block w of any length.  However, squares do occur in it, for example a(18) onwards is 3212 3212, or a(19) onwards is 2123 2123.
In Bollobás' proof, the signs sequence is A337004.  For blocks w of length l=4, the second signs subsequence presented (which should stop at length 7), does in fact occur, as does one other.
  - - +  +  - - +    \ two l=4 signs subsequences
  - + +  -  - + +    / in A337004 making squares here
All else in the argument holds, and in particular the "peaks" reduction means the only squares are lengths l = 4*5^k.
Zolotov shows this word is cubefree, and weakly squarefree (no x w w x where x is a single symbol and w is a block, possibly empty).  However uniform cyclic squarefree must wait for Leech's order 13 morphism in A337005.
(End)

B. Bollobas, The Art of Mathematics: Coffee Time in Memphis, Cambridge, 2006, pp. 226-228.

B. Bollobas, The Art of Mathematics: Coffee Time in Memphis, Cambridge 2006, scan of pages 226, 227 annotated by _N. J. A. Sloane_, Jul 31 2020.
Boris Zolotov, Another Solution to the Thue Problem of Non-Repeating Words, arXiv:1505.00019 [math.CO], 2015. (Section 5 morphism 1, then section 6.)
Index entries for sequences that are fixed points of mappings

Cf. A010060, A005678, A005679, A005680, A005681, A006156, A007413.

Cf. A337004 (first differences as +1,-1).

a:=[1,2,3,2,1]; b:=[2,3,1,3,2]; c:=[3,1,2,1,3]; S:=[1];
for m from 1 to 6 do S:=subs({1=a[],2=b[],3=c[]},S); od: S;

my(table=[0,1,2,1,0]); a(n) = my(v=digits(n,5)); sum(i=1,#v,table[v[i]+1]) %3+1; \\ Kevin Ryde, Jul 31 2020

cp's OEIS Frontend

A170823 An infinite word on the alphabet 1, 2, 3 by Bollobas.

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