OFFSET
0
COMMENTS
More generally, for k = 0, 1, 2, ..., we can define a sequence of words S_k(n) by S_k(0) = 0, S_k(1) = 0...01 (k 0's) and for n >= 1, S_k(n+1) = S_k(n)S_k(n)S_k(n)S_k(n-1). Then the limit word S_k(infinity) is a Sturmian word whose terms are given by the formula a(n) = floor((n + 2)/(k + alpha)) - floor((n + 1)/(k + alpha)), where alpha = 1/2*(sqrt(13) - 1). This sequence corresponds to the case k = 0. Compare with A080764.
(a(n)) is the unique fixed point of the substitution 0 -> 1, 1 -> 1110. - Michel Dekking, Feb 02 2017
LINKS
Robert Israel, Table of n, a(n) for n = 0..10000
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
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), 2005-2009.
Wikipedia, Sturmian word
FORMULA
Let alpha = 1/2*(sqrt(13) - 1). Then a(n) = floor((n + 2)/alpha) - floor((n + 1)/alpha).
If we read the sequence as the decimal constant C = 0.11101 11011 10111 10111 01110 11110 ... then C = sum {n >= 1} 1/10^floor(n*alpha).
EXAMPLE
The sequence of words S(n) begins
S(0) = 0
S(1) = 1
S(2) = 1110
S(3) = 1110 1110 1110 1
S(4) = 1110111011101 1110111011101 1110111011101 1110
MAPLE
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 22 2013
STATUS
approved