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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 (list; graph; refs; listen; history; text; internal format)
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 letter-to-letter 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 letter-to-letter 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 well-known 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 non-1 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) = 1-A116178(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,  407-417 (2020).

W. Hebisch and M. Rubey, Extended Rate, more GFUN, arXiv:math/0702086 [math.CO], 2007-2010.

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 * (1-x^4)^2 / ((1-x) * (1-x^2) * (1-x^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] + ((1-x^4)^2 x)/((1-x)(1-x^2)(1-x^6)))/(1+x+x^2) + O[x]^m // Normal, {m}];

CoefficientList[f[x], x] (* Jean-Fran├žois Alcover, Aug 17 2018, after Michael Somos *)

PROG

(PARI) a(n) = my(b=n%2, d); n=(n-1)>>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

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Last modified September 17 06:16 EDT 2021. Contains 347478 sequences. (Running on oeis4.)