

A118006


Define a sequence of binary words by w(1) = 01 and w(n+1) = w(n)w(n)Reverse[w(n)]. Sequence gives the limiting word w(infinity).


1



0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1
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OFFSET

1,1


COMMENTS

The article by W. Hebisch and M. Rubey gives a conjectured functional equation for the g.f. for this sequence.  N. J. A. Sloane, Sep 08 2010
The formula in Hebisch and Rubey has an extra left parenthesis before f(x).  Michael Somos, Jan 03 2011
From Michel Dekking, Sep 10 2020: (Start)
This sequence is an automatic sequence, i.e., the lettertoletter image of the fixed point of a uniform morphism mu.
In fact, one can take the alphabet {1,2,3,4} with the morphism
mu: 1>121, 2>234, 3>123, 4>434,
and the lettertoletter map g defined by
g: 1>0, 2>1, 3>1, 4>0.
Then (a(n)) = g(x), where x = 121234121... is the fixed point of the morphism mu starting with 1.
This is obtained by translating the recursion relation for w to the morphism a>aab, b>abb, and then decorating the fixed point aabaababb.... of this morphism with a>01, b>10, since the recursion starts at w(1) = 01.
It is wellknown that decorated fixed points of morphisms are morphic sequences, and the 'natural' algorithm to achieve this (see my paper "Morphic words, Beatty sequences and integer images of the Fibonacci language") yields a morphism on an alphabet of 2+2 = 4 symbols. In general there are several choices for mu. Here we have chosen mu such that it has constant length, i.e., the morphism is uniform.
(End)
Morphism a>aab, b>abb is Stewart's choral sequence A116178 (ternary lowest non1 digit, halved). Decoration 01 and 10 is then an interleaving of that sequence and its complement so a(2n+1) = A116178(n) and a(2n+2) = 1A116178(n) = A136442(n).  Kevin Ryde, Sep 29 2020


LINKS

Table of n, a(n) for n=1..105.
M. Dekking, Morphic words, Beatty sequences and integer images of the Fibonacci language, Theoretical Computer Science 809, 407417 (2020).
W. Hebisch and M. Rubey, Extended Rate, more GFUN, arXiv:math/0702086 [math.CO], 20072010.
Eric Weisstein's World of Mathematics, Reverend Back's Abbey Floor


FORMULA

G.f. f(x) satisfies: (1+x+x^2) * f(x)  x^2 * f(x^3) = x * (1x^4)^2 / ((1x) * (1x^2) * (1x^6)) = A096285(x)/x^2.  Michael Somos, Jan 03 2011


EXAMPLE

01, 010110, 010110010110011010, ...


MATHEMATICA

m = maxExponent = 105;
f[_] = 0;
Do[f[x_] = (x^2 f[x^3] + ((1x^4)^2 x)/((1x)(1x^2)(1x^6)))/(1+x+x^2) + O[x]^m // Normal, {m}];
CoefficientList[f[x], x] (* JeanFrançois Alcover, Aug 17 2018, after Michael Somos *)


PROG

(PARI) a(n) = my(b=n%2, d); n=(n1)>>1; while([n, d]=divrem(n, 3); d==1, ); d==2*b; \\ Kevin Ryde, Sep 29 2020


CROSSREFS

Cf. A116178 (odd bisection), A136442 (even bisection).
Sequence in context: A244992 A286685 A284878 * A285196 A189673 A189014
Adjacent sequences: A118003 A118004 A118005 * A118007 A118008 A118009


KEYWORD

nonn


AUTHOR

Eric W. Weisstein, Apr 09 2006


EXTENSIONS

Edited by N. J. A. Sloane, Feb 13 2008


STATUS

approved



