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%I M0406
%S 1,2,3,1,3,2,1,2,3,2,1,3,1,2,3,1,3,2,1,3,1,2,3,2,1,2,3,1,3,2,1,2,3,2,
%T 1,3,1,2,3,2,1,2,3,1,3,2,1,3,1,2,3,1,3,2,1,2,3,2,1,3,1,2,3,1,3,2,1,3,
%U 1,2,3,2,1,2,3,1,3,2,1,3,1,2,3,1,3,2,1,2,3,2,1,3,1,2,3,2,1,2,3,1,3,2,1,2,3
%N A squarefree (or Thue-Morse) ternary sequence: closed under 1->123, 2->13, 3->2. Start with 1.
%C a(n)=2 if and only if n-1 is in A079523. - Benoit Cloitre, Mar 10 2003.
%C Partial sums modulo 4 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ...- Philippe Deléham, Mar 04 2004
%C To construct the sequence : start with 1 and concatenate 4 -1 = 3 : 1, 3, then change the last term (2 -> 1, 3 ->2 ) gives : 1, 2. Concatenate 1, 2 with 4 -1 = 3, 4 - 2 = 2 : 1, 2, 3, 2 and change the last term : 1, 2, 3, 1. Concatenate 1, 2, 3, 1 with 4 - 1 = 3, 4 - 2 = 2, 4 - 3 = 1, 4 - 1 = 3 : 1, 2, 3, 1, 3, 2, 1, 3 and change the last term : 1, 2, 3, 1, 3, 2, 1, 2 etc.- Philippe Deléham, Mar 04 2004
%C To construct the sequence : start with the Thue-Morse sequence A010060 = 0, 1, 1, 0, 1, 0, 0, 1, ... Then change 0 -> 1, 2, 3, _ and 1 -> 3, 2, 1, _ gives : 1, 2, 3, _, 3, 2, 1, _,3, 2, 1, _, 1, 2, 3, _, 3, 2, 1, _, ...and fill in the successive holes with the successive terms of the sequence itself.- Philippe Deléham, Mar 04 2004
%C To construct the sequence : to insert the number 2 between the A003156(k)-th term and the (1 + A003156(k))-th term of the sequence 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, ...- Philippe Deléham, Mar 04 2004
%C Conjecture. The sequence is formed by the numbers of 1's between every pair of consecutive 2's in A076826. - Vladimir Shevelev, May 31 2009
%D James D. Currie, Palindrome positions in ternary square-free words, Theoretical Computer Science, 396 (2008) 254-257
%D J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 18.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A. Thue. Ueber unendliche Zeichenreihe. Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, 7:1a22, 1906.
%H Roger L. Bagula, <a href="/A007413/a007413.txt">Description of sequence as successive rows of a triangle</a>
%H V. Keranen, <a href="http://south.rotol.ramk.fi/keranen/ias2002/NewAbelianSquare-FreeDT0L-LanguagesOver4Letters.nb">New Abelian Square-Free DT0L-Languages over 4 Letters</a>
%H S. Kitaev and T. Mansour, <a href="http://arXiv.org/abs/math.CO/0210170">Counting the occurrences of generalized patterns...</a>.
%F a(n) modulo 2 = A035263(n). a(A036554(n)) = 2. a(A003159(n)) = 1 if n odd. a(A003159(n)) = 3 if n even. a(n) = A033485(n) mod 4. a(n) = 4 - A036585(n-1).- Philippe Deléham, Mar 04 2004
%F a(n) = 2 - A029883(n) = 3 - A036577(n) . - Philippe Deléham, Mar 20 2004
%F For n>=1, we have: 1) a(A108269(n))=A010684(n-1); 2) a(A079523(n))=A010684(n-1); 3) a(A081706(2n))=A010684(n). - Vladimir Shevelev, Jun 22 2009
%e Here are the first 5 stages in the construction of this sequence, together with Mma code, taken from Keranen's article. His alphabet is a,b,c rather than 1,2,3.
%e productions = {"a" → "abc ", "b" → "ac ", "c" → "b ", " " -> ""};
%e NestList[g, "a", 5] // TableForm
%e a
%e abc
%e abc ac b
%e abc ac b abc b ac
%e abc ac b abc b ac abc ac b ac abc b
%e abc ac b abc b ac abc ac b ac abc b abc ac b abc b ac abc b abc ac b ac
%t Nest[ Flatten[ # /. {1 -> {1, 2, 3}, 2 -> {1, 3}, 3 -> {2}}] &, {1}, 7] (from Robert G. Wilson v, May 07 2005)
%o (PARI) a(n)=if(n<1|valuation(n,2)%2,2,2+(-1)^subst(Pol(binary(n)),x,1))
%Y Cf. A001285, A010060.
%Y First differences of A000069.
%Y Equals A036580(n-1) + 1.
%Y Cf. A115384 A159481 A007413 A000120.
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_.
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