A119346 Sequence of nim-values for the game in which two players alternately cut off one inch or root two inches from a piece of string of length n. Player who runs out of string loses.

0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1

Offset: 0

Author

N. J. A. Sloane, based on email from R. K. Guy and Alex Fink, Aug 05 2006

Keywords

• nonn

Comments

From Michel Dekking, Feb 17 2020: (Start)

It follows from Alex Fink's remarks that (a(n)) is obtained from the sequence A276862 (removing the first 2) by mapping every 2 to 0,1 and every 3 to 0,1,2. However, the first 3 entries will be missing.

In the context of my paper "Morphic words, Beatty sequences and integer images of the Fibonacci language", this means that (a(n+3)) is obtained by decorating A006337 by the decoration delta given by delta(1) = 01, delta(2) = 012. This implies that (a(n+3)) is a morphic sequence, i.e., the letter to letter image of the fixed point of a morphism, say sigma. One obtains sigma by the 'natural' algorithm given in the "Morphic words...."-paper. In turns out that the alphabet of sigma can be chosen as {0,1,2}, and that sigma is surprisingly simple:

sigma(0) = 01, sigma(1) = 012, sigma(2) = 01.

The letter to letter map is given by the identity. In other words, if x = 010120101... is the unique fixed point of sigma, then (a(n+3)) = x. (End)

Links

• M. Dekking, Morphic words, Beatty sequences and integer images of the Fibonacci language, Theoretical Computer Science 809, 407-417 (2020).
• Alex Fink, Discussion of A119346

Crossrefs

Cf. A003151.

Formula

To get the answers, add one to sequence A003151 and then start counting from zero, but return to zero whenever you reach a member of A003151 plus one.

Added Feb 13 2020: The simplest formula is a(n) = floor(n mod (1 + sqrt 2)). - Alex Fink (see link).