[seqfan] 1-t0-1 correspondence: x<=>{F_x(n)} (near-Ffibbonacci sequence for x)

Vladimir Shevelev shevelev at bgu.ac.il
Wed Jul 7 15:13:53 CEST 2010

It is not difficult to prove the following rule for reconstruction binary digits of x from [0,1] knowing the near-Fibonacci sequence for x.
Suppose that there are N known first terms of near-Fibonacci sequence {F_x(n)}. We call F_x(n) an A-term,  if F_x(n) =F_x(n-1)+F_x(n-2), and B_term, otherwise. 
1) Let F_x(N) is B_term. Instead of F_x(N), we write 11, if F_x(N)>F_x(N-1)+F_x(N-2), and 00,
  if F_x(N)<F_x(N-1)+F_x(N-2). Continuing from the right to the left, we write 1 if the previous element was 0, and 0  if the previous element was 1 iff F_x(N-1) is A-term; otherwise,
in the case of F_x(N-1)<F_x(N-2)+F_x(N-3), we write 0, and in the case of F_x(N-1)>F_x(N-2)+F_x(N-3), we write 1. This process continues up to F_x(3).
2) Let F_x(N) is A_term. Then we look for the maximal B-term F_x(M)<F_x(N). We make the same process from the right to the left for F_x(M); after that we alternate digits {0,1}beginning from F_x(M) to the right up to F_x(N).
Example. For x=1/e=.010111100... Using these digits, we get near-Fibonacci sequence: 1,1,2,3,5,9,15,25,40,64,... Since 64 is B-term and 64<25+40, then we write 00. Since 40 is A-term, then we write 1. Since 25 >15+9, then we write 1, etc. We see that all digits of x are reconstructed.

 Shevelev Vladimir‎

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