cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

N. J. A. Sloane, Jun 09 2015

Keywords

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)

Crossrefs

Programs

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

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

Extensions

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