A258703 a(n) = floor(n/sqrt(2) - 1/2).

0, 0, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 17, 18, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 29, 30, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47, 48, 48, 49, 50, 51, 51, 52, 53, 53

Offset

1,4

Comments

From Michel Dekking, Aug 11 2022: (Start)

By definition, (a(n)) is an inhomogeneous Beatty sequence. The associated Sturmian word is s(alpha, rho) = (floor((n + 1)*alpha + rho) - floor(n*alpha + rho), n= 0, 1, 2,...) = 1,0,1,1,1,0,1,1,0,1,1,..., with slope alpha = sqrt(2)/2, and intercept rho = -1/2.

Also, s(alpha, rho) = s(alpha,rho+1) - 1. Since 0 < alpha < 1 and 0 < rho +1 < 1, with algebraic conjugates

alpha* = -sqrt(2)/2, and (rho +1)* = 1/2,

Yasutomi's criterion gives that s(alpha, rho) is fixed point of a morphism.

The morphism can be found following the ideas of Chapter 2 in Lothaire's book and Section 4 of my paper "Substitution invariant Sturmian words and binary trees" (cf. A006340).

For a better fit with the literature we will determine the morphism that fixes the binary complement s(1-alpha, 1-(1+rho) ) = 0,1,0,0,0,1,0,0,1,0,0....

Let psi_1 and psi_8 be the elementary Sturmian morphisms given by

psi_1(0)=01 , psi_1(1)=0; psi_8(0)=01, psi(8)1=1.

Let psi := psi_1 psi_8. Then psi is given by

psi(0)=010 , psi(1)=0.

We see that psi fixes the Octanacci sequence A324772.

That psi is the right morphism can be proved by checking that (x,y) = (1-alpha, -rho) is fixed point of the composition T_1 T_8 of the fractional linear maps

T_1(x,y) = ((1-x)/(2-x), (1-y)/(2-x)),

T_8(x,y) = ((1/(2-x), y/(2-x))).

Conclusion, taking the binary complement of psi: the Sturmian word equals A104521. (End)

Links

Michel Dekking, Table of n, a(n) for n = 1..10000

Michel Dekking, Substitution invariant Sturmian words and binary trees, arXiv:1705.08607 [math.CO], (2017).

Michel Dekking, Substitution invariant Sturmian words and binary trees, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.

M. Lothaire, Algebraic combinatorics on words, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.

K. O'Bryant, The sequence of fractional parts of roots, arXiv preprint, arXiv:1410.2927 [math.NT], 2014-2015.

Formula

a(n) = floor(1/(exp(sqrt(2)/n)-1)) for all positive integers n [O'Bryant].

a(n) = floor((n*sqrt(2) - 1) / 2). - Reinhard Zumkeller, Jun 09 2015

Mathematica

Table[Floor[n/Sqrt[2] - 1/2], {n, 1, 100}] (* Vincenzo Librandi, Jun 09 2015 *)

Prog

(Magma) [Floor(n/Sqrt(2) - 1/2): n in [1..80]]; // Vincenzo Librandi, Jun 09 2015
(Haskell)
a258703 = floor . (/ 2) . subtract 1 . (* sqrt 2) . fromIntegral
-- Reinhard Zumkeller, Jun 09 2015
(PARI) vector(100, n, n--; floor(n/sqrt(2) - 1/2)) \\ G. C. Greubel, Sep 30 2018

Crossrefs

Cf. A078608, A129935, A104521.

Cf. A049472, A001951.

Sequence in context: A175406 A210435 A181534 * A067022 A113818 A136746

Adjacent sequences: A258700 A258701 A258702 * A258704 A258705 A258706

Keyword

nonn

Author

N. J. A. Sloane, Jun 09 2015

Extensions

Offset changed from 0 to 1 Michel Dekking, Aug 11 2022

Status

approved