A123564 The infinite Fibonacci word reencoded by writing successive non-overlapping pairs of bits as decimal numbers.

2, 3, 1, 1, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 3, 1, 2, 2, 3, 1, 1, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 3, 1, 2, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 3, 1

Offset: 1

Author

Alexandre Losev, Nov 12 2006

Keywords

• easy
• nonn

Comments

The algorithm used here suggests multiple variations such as using more than 2 bits, allowing overlap of successive subwords, using other numbers for the encoding of subwords or using other binary sequences. (E.g. overlapping: a(n) = 2*A005614(n) + A005614(n+1) )

Essentially equal to A143667. - Michel Dekking, Sep 26 2017

Examples

			a(1) = 2*1+0 = 2;
a(2) = 2*1+1 = 3;
a(3) = 2*0+1 = 1.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Michel Dekking and Mike Keane, Two-block substitutions and morphic words, arXiv:2202.13548 [math.CO], 2022.

Crossrefs

Cf. A005614, A143667

Programs

Mathematica

f := 1/GoldenRatio; T[n_] := Floor[2*n*f] - 2*Floor[(2*n - 1)*f] + Floor[(2*n + 1)*f]; Table[T[n], {n, 100}] (* G. C. Greubel, Oct 16 2017 *)

PARI

f=(sqrt(5)-1)/2; a(n)= my(m=2*n); floor(m*f)-2*floor((m-1)*f)+floor((m+1)*f); \\ Michel Marcus, Sep 26 2017

Formula

f = (sqrt(5)-1)/2; m = 2*n; a(n) = floor(m*f) - 2*floor((m-1)*f) + floor((m+1)*f);

a(n) = 2*A005614(2n-1) + A005614(2n), using the infinite Fibonacci word A005614.