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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.

%I A119346 #20 Aug 18 2020 15:40:40
%S A119346 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,
%T A119346 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,
%U A119346 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
%N 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.
%C A119346 From _Michel Dekking_, Feb 17 2020: (Start)
%C A119346 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.
%C A119346 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:
%C A119346       sigma(0) = 01, sigma(1) = 012, sigma(2) = 01.
%C A119346 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)
%H A119346 M. Dekking, <a href="https://doi.org/10.1016/j.tcs.2019.12.036">Morphic words, Beatty sequences and integer images of the Fibonacci language</a>, Theoretical Computer Science  809,  407-417 (2020).
%H A119346 Alex Fink, <a href="/A119346/a119346.txt">Discussion of A119346</a>
%F A119346 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.
%F A119346 Added Feb 13 2020: The simplest formula is a(n) = floor(n mod (1 + sqrt 2)). - Alex Fink (see link).
%Y A119346 Cf. A003151.
%K A119346 nonn
%O A119346 0,3
%A A119346 _N. J. A. Sloane_, based on email from _R. K. Guy_ and _Alex Fink_, Aug 05 2006