A188011 a(n) = floor(n*r) - floor(n*r-k*r) - floor(k*r), where r=(1+sqrt(5))/2, k=3.

1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0

Offset

1

Comments

Essentially the same as A187944.

From Michel Dekking, Sep 28 2017: (Start)

a(n) = floor(n*r) - floor((n-3)*r) - floor(3*r) = b(n) + b(n-1) + b(n-2) - 4, where b(n) = floor(n*r) - floor((n-1)*r) equals A014675(n-2) for n >= 2, the infinite Fibonacci word on {1,2}.

The words w1w2w3 = 212, 122, 221, and 121 of length 3 occurring in (b(n)) yield the sums w1+w2+w3 = 5, 5, 5, and 4, respectively. This implies that the 0's in (a(n)) occur exactly at the 4's in A138967, shifted by 3. So a(n) = 0 <=> n = BB(k)+3 for some k <=> n = 3A(k)+2k+3, where (A(k)) = ([k.r]), and (B(k)) = ([k.r]+k) are the lower and upper Wythoff sequences. (End)

Links

Muniru A Asiru, Table of n, a(n) for n = 1..5000

Formula

a(n+3) = A187944(n) for n >= 1. - Michel Dekking, Sep 17 2016

a(n) = floor(n*r) - floor(n*r-3*r) - floor(3*r), where r=(1+sqrt(5))/2.

a(n) = 2 + floor(n*phi) - floor((n+R)*phi), where phi=(sqrt(5)+1)/2 and R=0.70980344286... is the rabbit constant. - Federico Provvedi, Nov 21 2018

Maple

r:=(1+sqrt(5))/2: a:=n->floor(n*r)-floor(n*r-3*r)-floor(3*r): [a(n)$n=1..140]; # Muniru A Asiru, Nov 22 2018

Mathematica

r=(1+5^(1/2))/2; k=3;
t=Table[Floor[n*r]-Floor[(n-k)*r]-Floor[k*r], {n, 1, 220}]   (* A188011 *)
Flatten[Position[t, 0]]  (* A188012 *)
Flatten[Position[t, 1]]  (* A188013 *)

Prog

(PARI) vector(200, n, floor(n*(1+sqrt(5))/2) - floor((n-3)*(1+sqrt(5))/2) - floor(3*(1+sqrt(5))/2)) \\ G. C. Greubel, Nov 22 2018
(Magma) [Floor(n*(1+Sqrt(5))/2) - Floor((n-3)*(1+Sqrt(5))/2) - Floor(3*(1+Sqrt(5))/2): n in [1..200]]; // G. C. Greubel, Nov 22 2018
(Sage) [floor(n*(1+sqrt(5))/2) - floor((n-3)*(1+sqrt(5))/2) - floor(3*(1+sqrt(5))/2) for n in (1..200)] # G. C. Greubel, Nov 22 2018

Crossrefs

Cf. A014565 (rabbit constant), A188014, A188012, A188013, A187950.

Sequence in context: A179850 A267919 A097343 * A040051 A108788 A103583

Adjacent sequences: A188008 A188009 A188010 * A188012 A188013 A188014

Keyword

nonn

Author

Clark Kimberling, Mar 19 2011

Status

approved