A285961 - OEIS

%I #10 Jul 29 2022 21:27:15

%S 4,7,12,16,20,23,27,31,36,39,44,48,52,57,60,64,68,71,76,80,84,87,91,

%T 95,100,103,107,111,114,119,123,127,132,135,140,144,148,151,155,159,

%U 164,167,172,176,180,185,188,192,196,199,204,208,212,217,220,225,229

%N Positions of 0 in A285960; complement of A285962.

%C Conjecture: lim_{n->oo} a(n)/n = 4.

%C From _Michel Dekking_, Apr 19 2022: (Start)

%C Kimberling's conjecture is equivalent to the property that the frequency of 1's in A285960 is equal to 1/4. This can be computed via the comments of A285960.

%C But more is true. The first difference sequence (d(n)) = 3,5,4,4,3,4,4,5,3,... of (a(n)) is a morphic sequence. Let

%C      A=011, B=010, C=0110, and D=01.

%C From the representation of A285960 as fixed point of the derived morphism

%C      tau: A -> AB, B -> CD, C -> ABD, D -> C,

%C decorated by the morphism

%C        delta: A-> 11, B-> 10, C-> 110, D-> 1,

%C we do not see how to obtain the differences between occurrences of 0's.  However, A285960 is also represented as fixed point of

%C      tau^2: A->ABCD, B->ABDC, C->ABCDC, D->ABD, and delta, where the four images turn into

%C       delta(tau^2(A))=11101101, delta(tau^2(B))=11101110,

%C     delta(tau^2(C))=11101101110, delta(tau^2(D))=11101.

%C The corresponding consecutive distances between 0's are 35, 44, 344, and 5.

%C The "natural" algorithm to obtain (d(n)) as a letter-to-letter image of a morphic sequence from this decoration yields (for example) a morphism mu on a six-letter alphabet {a,b,c,d,e,f} given by

%C       mu: a->ab, b->cd, c->aed, d->f, e->cd, f->aed,

%C with the letter-to-letter map

%C     lambda: a->3, b->5, c->4, d->4, e->4, f->5.

%C We have (d(n)) = lambda(z), where z is the fixed point z = abcdaed... of mu.

%C Note that it turns out that mu is the same morphism as the one used to obtain A286047 and A286048 as morphic sequences.

%C Another way to prove Kimberling's conjecture: a(n)/n = (1/n)*Sum_{k=1..n} d(k), the average value M of the differences.

%C To obtain the limit, you need the frequencies M(a),...,M(f) of a,b,...,f in the fixed point of mu. These are computed with the Perron-Frobenius theorem.

%C We have M(a)=M(d)=1/4, M(b)=M(c)=M(d)=M(f)=1/8. So M = (2/8)*3 + (4/8)*4 + (2/8)*5 = (1/8)*(6+16+10) = 32/8 = 4.

%C (End)

%H Clark Kimberling, <a href="/A285961/b285961.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A285960 = 111011011110111011..., in which 0 is in positions 4,7,12,16,...

%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 8] (* Thue-Morse, A010060 *)

%t w = StringJoin[Map[ToString, s]]

%t w1 = StringReplace[w, {"01" -> "1"}]

%t st = ToCharacterCode[w1] - 48 (* A285960  *)

%t Flatten[Position[st, 0]]  (* A285961 *)

%t Flatten[Position[st, 1]]  (* A285962 *)

%Y Cf. A010060, A285960, A285962, A286047.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 05 2017