

A221150


The generalized Fibonacci word f^[3].


10



0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0


COMMENTS

(a(n)) is the [0>01, 1>0]transform of the Fibonacci word A005614, or alternatively (see Ramirez et al.) the [0>0, 1>01]transform of the Fibonacci word A003849. (a(n)) is the homogeneous Sturmian word with slope r = (5sqrt(5))/10. Since the algebraic conjugate (5+sqrt(5)/10 of r is also in (0,1), (a(n)) is NOT a fixed point of a morphism (by Allauzen's criterion).  Michel Dekking, Oct 14 2017
From Michel Dekking, Oct 04 2018: (Start)
Let psi_3 be the elementary Sturmian morphism given by
psi_3(0)=0, psi_3(1)=01,
and let x = A003849 be the Fibonacci word. Then, see previous comment, (a(n)) = psi_3(x). We show that (a(n)) is a fixed point of an automorphism sigma of the free group generated by 0 and 1.
To see this, let gamma be the Fibonacci morphism given by gamma(0)=01, gamma(1)=0. Then gamma(x) = x, and so
psi_3(gamma(x)) = psi_3(x) = a,
implying that a = (a(n)) is fixed by
sigma := psi_3 gamma psi_3^{1}.
One easily computes psi_3^{1}: 0>0, 1>0^{1}1, which gives sigma:
sigma(0) = 001, sigma(1) = 1^{1}0^{1}.
(End)
From Michel Dekking, Sep 17 2020: (Start)
Although not a fixed point of a morphism, this sequence is a morphic sequence, i.e., the lettertoletter image of the fixed point of a morphism mu. In fact, one can take the alphabet {1,2,3} with the morphism
mu: 1>1231, 2>12, 3>3,
and the lettertoletter map g defined by
g: 1>0, 2>0, 3>1.
Then (a(n)) = g(x), where x = 1231123123... is the fixed point of the morphism mu starting with 1.
This is obtained by noting that (a(n)) is a decoration of the Fibonacci word abaababaab... by a>0, b>01, fixed point of a>aba, b>ab.
It is wellknown that decorated fixed points of morphisms are morphic sequences, and the 'natural' algorithm to achieve this (see my paper on morphic words) yields a morphism on an alphabet of 1+2=3 symbols.
(End)


REFERENCES

Dale Gerdemann, Problem 5.1, Problem Proposals, Ed. Clark Kimberling, Sixteenth International Conference on Fibonacci Numbers and Their Applications, Rochester Institute of Technology, Rochester, New York, July 24, 2014. Fibonacci Quarterly, to appear. [Mentions a sequence that appears to match this entry]


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194198.
P. G. Anderson, T. C. Brown, P. J.S. Shiue, A simple proof of a remarkable continued fraction identity Proc. Amer. Math. Soc. 123 (1995), 20052009.
Eunice Y. S. Chan, Robert M. Corless, Laureano GonzalezVega, J. Rafael Sendra, Juana Sendra, Steven E. Thornton, Bohemian Upper Hessenberg Toeplitz Matrices, arXiv:1809.10664 [cs.SC], 2018.
M.Dekking, Morphic words, Beatty sequences and integer images of the Fibonacci language, Theoretical Computer Science 809, 407417 (2020).
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 20122014.


FORMULA

Set S_0=0, S_1=001; thereafter S_n = S_{n1}S_{n2}; sequence is S_{oo}.
From Peter Bala, Nov 19 2013: (Start)
a(n) = floor((n + 2)/(phi + 2))  floor((n + 1)/(phi + 2)) where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio.
If we read the present sequence as the digits of a decimal constant c = 0.00100 01001 00010 00100 .... then we have the series representation c = Sum_{n >= 1} 1/10^floor(n*(phi + 2)). An alternative representation is c = 9*Sum_{n >= 1} floor(n*(5  sqrt(5))/10) /10^n.
The constant 9*c has the simple continued fraction representation [0; 111, 10, 10^3, 10^4, 10^7, ..., 10^Lucas(n), ...] (see Adams and Davison). Compare with A230900.
Using this result we can find the alternating series representation c = 9*Sum_{n >= 1} (1)^(n+1)*(1 + 10^Lucas(3*n))/( (10^Lucas(3*n  2)  1)*(10^Lucas(3*n + 1)  1) ). The series converges very rapidly: for example, the first 10 terms of the series give a value for c accurate to more than 7.8 million decimal places. Cf. A005614. (End)


EXAMPLE

(a(n)) can be obtained by iteration of sigma starting with 0.
sigma(0) = 001,
sigma^2(0) = 0010011^{1}0^{1} = 0010,
sigma^3(0) = 0010011^{1}0^{1}001 = 0010001.
sigma^4(0) = 0010011^{1}0^{1}0010010011^{1}0^{1} = 00100010010.


MAPLE

fibi := proc(n, i)
option remember;
local j;
if n = 0 then
[0] ;
elif n = 1 then
[seq(0, j=1..i1), 1] ;
else
[op(procname(n1, i)), op(procname(n2, i))] ;
end if;
end proc:
fibonni := proc(n, i)
local fn;
for fn from 0 do
Fn := fibi(fn, i) ;
if nops( Fn) >= n+1 and nops(Fn) > i+3 then
return op(n+1, Fn) ;
end if;
end do:
end proc:
A221150 := proc(n)
fibonni(n, 3) ;
end proc: # R. J. Mathar, Jul 09 2013


MATHEMATICA

Table[Floor[(n + 2)/(GoldenRatio + 2)]  Floor[(n + 1)/(GoldenRatio + 2)], {n, 0, 120}] (* Michael De Vlieger, Apr 03 2016 *)
fibi[n_, i_] := fibi[n, i] = Which[n == 0, {0}, n == 1, Append[Table[0, {j, 1, i  1}], 1], True, Join[fibi[n  1, i], fibi[n  2, i]]];
fibonni[n_, i_] := fibonni[n, i] = Module[{fn, Fn}, For[fn = 0, True, fn++, Fn = fibi[fn, i]; If[Length[ Fn] >= n + 1 && Length[Fn] > i + 3, Return[ Fn[[n + 1]]]]]];
a[n_] := fibonni[n, 3]; Table[a[n], {n, 0, 100}] (* JeanFrançois Alcover, Nov 21 2017, after R. J. Mathar *)


PROG

(Magma) [Floor((n+2)/(1/2*(1+Sqrt(5))+2))Floor((n+1)/(1/2*(1 +Sqrt(5))+2)): n in [0..100]]; // Vincenzo Librandi, Oct 15 2017


CROSSREFS

Cf. A003849, A005614. A000204, A221151, A221152, A230900.
Sequence in context: A285495 A073059 A288173 * A288997 A276794 A289025
Adjacent sequences: A221147 A221148 A221149 * A221151 A221152 A221153


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Jan 03 2013


STATUS

approved



