0

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

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

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).

The real number 9*C has the simple continued fraction expansion [0; 1, 1110, 1000100010, 1000000000000100000000000010000, ...] = [0; 1, 1110, 10^9 + 10^5 + 10^1, 10^30 + 10^17 + 10^4, 10^99 + 10^56 + 10^13, ..., 10^(3*A006190(n+1)) + 10^A180148(n) + A003688(n), ...]. See Adams and Davison.

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

Digits := 50: u := evalf((sqrt(13) - 1)/2): A230901 := n -> floor((n+2)/u) - floor((n+1)/u): seq(A230901(n), n = 0..80);

Cf. A003688, A006190, A080764, A180148.

Sequence in context: A353457 A112299 A358839 * A285671 A267773 A325321

Adjacent sequences: A230898 A230899 A230900 * A230902 A230903 A230904

nonn,easy

Peter Bala, Nov 22 2013

approved