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Positions of 0 in A284620.
2

%I #10 Mar 17 2020 12:14:48

%S 1,5,11,15,21,27,31,37,41,47,53,57,63,69,73,79,83,89,95,99,105,109,

%T 115,121,125,131,137,141,147,151,157,163,167,173,179,183,189,193,199,

%U 205,209,215,219,225,231,235,241,247,251,257,261,267,273,277,283,287

%N Positions of 0 in A284620.

%C This sequence and A005843 and A130568 partition the positive integers into sequences with slopes t = 1+sqrt(5), u = 3+sqrt(5), v = 2, where 1/t + 1/u + 1/v = 1. The positions of 1 in A284620 are given by A005843, and of 2, by A130568.

%C From _Michel Dekking_, Mar 17 2020: (Start)

%C This sequence is a generalized Beatty sequence.

%C It was shown in the Comments of A284620 that A284620 is the letter-to-letter image of the fixed point x = ABCDABCDCD... of the morphism

%C mu: A->AB, B->CD, C->ABCD, D->CD,

%C with the letter-to-letter map lambda defined by

%C lambda: A->0, B->1, C->2, D->1.

%C Note that A284620(n)=0 iff x(n) = A, where x = ABCDABCDCD... is the fixed point of mu. The return words of A in x are ABCD and ABCDCD. Coding these two return words by their lengths, mu induces a morphism rho on the coded return words given by

%C rho(4) = 46, rho(4) = 466.

%C The difference sequence (a(n+1)-a(n)) equals the unique fixed point r = 4646646466... of rho.

%C The morphism g on the alphabet {a,b} given by

%C g(a) = ab, g(b) =abb

%C was introduced in A284620. We see that rho is just an alphabet change of the morphism g.

%C Let f given by f(b) = ba, f(a) = b be the Fibonacci morphism on the alphabet {b,a} with fixed point x_F = babbababba....

%C Let x_G = ababbababb... be the fixed point of g. It is well-known (see, e.g., Lemma 12 in "Morphic words..."), that x_G = a x_F.

%C In general the partial sums of x_F are equal to the generalized Beatty sequence V given by V(n) = p*floor(n*phi) +q*n+r, where p = a-b and q = 2*b-a. See Lemma 8 in the Allouche and Dekking paper. Here we obtain p = 2, q = 2. So a(n) = 2*floor((n-1)*phi) + 2*n - 1, for n>0.

%C (End)

%H Clark Kimberling, <a href="/A284621/b284621.txt">Table of n, a(n) for n = 1..10000</a>

%H J.-P. Allouche, F. M. Dekking, <a href="https://doi.org/10.2140/moscow.2019.8.325">Generalized Beatty sequences and complementary triples</a>, Moscow Journal of Combinatorics and Number Theory 8, 325-341 (2019).

%H M. Dekking, <a href="https://doi.org/10.1016/j.tcs.2019.12.036">Morphic words, Beatty sequences and integer images of the Fibonacci language</a>, Theoretical Computer Science 809, 407-417 (2020).

%F a(n+1) = 2*A001950(n) + 1, n>0. - _Michel Dekking_, Mar 17 2020

%e As a word, A284620 = 012101212101210121..., in which 0 is in positions 1,5,11,15,...

%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13] (* A003849 *)

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

%t w1 = StringReplace[w, {"00" -> "2"}]

%t st = ToCharacterCode[w1] - 48 (* A284620 *)

%t Flatten[Position[st, 0]] (* A284621 *)

%t Flatten[Position[st, 1]] (* A005843 *)

%t Flatten[Position[st, 2]] (* A130568 *)

%Y Cf. A005843, A130568, A284620, A001950.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, May 02 2017

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