cp's OEIS Frontend

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

%I A258703 #30 Sep 08 2022 08:46:12
%S A258703 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,
%T A258703 19,20,20,21,22,22,23,24,24,25,26,27,27,28,29,29,30,31,32,32,33,34,34,
%U A258703 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
%N A258703 a(n) = floor(n/sqrt(2) - 1/2).
%C A258703 From _Michel Dekking_, Aug 11 2022: (Start)
%C A258703 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.
%C A258703 Also, s(alpha, rho) = s(alpha,rho+1) - 1. Since 0 < alpha < 1 and 0 < rho +1 < 1, with algebraic conjugates
%C A258703 alpha* = -sqrt(2)/2, and (rho +1)* = 1/2,
%C A258703 Yasutomi's criterion gives that s(alpha, rho) is fixed point of a morphism.
%C A258703 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).
%C A258703 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....
%C A258703 Let psi_1 and psi_8 be the elementary Sturmian morphisms given by
%C A258703     psi_1(0)=01 , psi_1(1)=0;  psi_8(0)=01, psi(8)1=1.
%C A258703 Let psi := psi_1 psi_8. Then psi is given by
%C A258703     psi(0)=010 , psi(1)=0.
%C A258703 We see that psi  fixes the Octanacci sequence A324772.
%C A258703 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
%C A258703     T_1(x,y) = ((1-x)/(2-x), (1-y)/(2-x)),
%C A258703     T_8(x,y) = ((1/(2-x), y/(2-x))).
%C A258703 Conclusion, taking the binary complement of psi: the Sturmian word equals A104521. (End)
%H A258703 Michel Dekking, <a href="/A258703/b258703.txt">Table of n, a(n) for n = 1..10000</a>
%H A258703 Michel Dekking, <a href="http://arxiv.org/abs/1705.08607">Substitution invariant Sturmian words and binary trees</a>, arXiv:1705.08607 [math.CO], (2017).
%H A258703 Michel Dekking, <a href="http://math.colgate.edu/~integers/sjs7/sjs7.Abstract.html">Substitution invariant Sturmian words and binary trees</a>, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.
%H A258703 M. Lothaire, <a href="http://tomlr.free.fr/Math%E9matiques/Fichiers%20Claude/Auteurs/aaaDivers/Lothaire%20-%20Algebraic%20Combinatorics%20On%20Words.pdf">Algebraic combinatorics on words</a>, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.
%H A258703 K. O'Bryant, <a href="http://arxiv.org/abs/1410.2927">The sequence of fractional parts of roots</a>, arXiv preprint, arXiv:1410.2927 [math.NT], 2014-2015.
%F A258703 a(n) = floor(1/(exp(sqrt(2)/n)-1)) for all positive integers n [O'Bryant].
%F A258703 a(n) = floor((n*sqrt(2) - 1) / 2). - _Reinhard Zumkeller_, Jun 09 2015
%t A258703 Table[Floor[n/Sqrt[2] - 1/2], {n, 1, 100}] (* _Vincenzo Librandi_, Jun 09 2015 *)
%o A258703 (Magma) [Floor(n/Sqrt(2) - 1/2): n in [1..80]]; // _Vincenzo Librandi_, Jun 09 2015
%o A258703 (Haskell)
%o A258703 a258703 = floor . (/ 2) . subtract 1 . (* sqrt 2) . fromIntegral
%o A258703 -- _Reinhard Zumkeller_, Jun 09 2015
%o A258703 (PARI) vector(100, n, n--; floor(n/sqrt(2) - 1/2)) \\ _G. C. Greubel_, Sep 30 2018
%Y A258703 Cf. A078608, A129935, A104521.
%Y A258703 Cf. A049472, A001951.
%K A258703 nonn
%O A258703 1,4
%A A258703 _N. J. A. Sloane_, Jun 09 2015
%E A258703 Offset changed from 0 to 1 _Michel Dekking_, Aug 11 2022