OFFSET
1,1
COMMENTS
This is a cube-free sequence [Allouche and Shallit]. - N. J. A. Sloane, Aug 29 2023
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
REFERENCES
J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 28, #49.
LINKS
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, ...
MAPLE
with(ListTools);
f:=proc(S) global f2;
[op(S), op(S), op(Reverse(S))]; end;
S:=[0, 1];
for n from 1 to 6 do S:=f(S): od:
S; # N. J. A. Sloane, Aug 29 2023
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-François 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
KEYWORD
nonn
AUTHOR
Eric W. Weisstein, Apr 09 2006
EXTENSIONS
Edited by N. J. A. Sloane, Feb 13 2008
STATUS
approved