A170823 - OEIS

%I #28 Dec 19 2021 09:08:29

%S 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,

%T 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,

%U 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

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

%C A concatenation of blocks u_k, k >= 0, where u_k has length 5^k. The sequence is defined recursively - see the Maple code.

%C From _Kevin Ryde_, Aug 11 2020: (Start)

%C 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.

%C 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.

%C   - - +  +  - - +    \ two l=4 signs subsequences

%C   - + +  -  - + +    / in A337004 making squares here

%C All else in the argument holds, and in particular the "peaks" reduction means the only squares are lengths l = 4*5^k.

%C 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.

%C (End)

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

%H B. Bollobas, <a href="/A170823/a170823.pdf">The Art of Mathematics: Coffee Time in Memphis</a>, Cambridge 2006, scan of pages 226, 227 annotated by _N. J. A. Sloane_, Jul 31 2020.

%H Boris Zolotov, <a href="https://arxiv.org/abs/1505.00019">Another Solution to the Thue Problem of Non-Repeating Words</a>, arXiv:1505.00019 [math.CO], 2015.  (Section 5 morphism 1, then section 6.)

%H <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>

%p a:=[1,2,3,2,1]; b:=[2,3,1,3,2]; c:=[3,1,2,1,3]; S:=[1];

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

%o (PARI) 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

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

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

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Dec 25 2009