OFFSET
1,1
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
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.
FORMULA
EXAMPLE
a(1) = 2*1+0 = 2;
a(2) = 2*1+1 = 3;
a(3) = 2*0+1 = 1.
MATHEMATICA
f := 1/GoldenRatio; T[n_] := Floor[2*n*f] - 2*Floor[(2*n - 1)*f] + Floor[(2*n + 1)*f]; Transpose[{Range[1, 50], Table[T[n], {n, 1, 50}] (* G. C. Greubel, Oct 16 2017 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Alexandre Losev, Nov 12 2006
STATUS
approved