Instead of binary or decimal, the Kingdom of Leutonia uses an unusual system to represent numbers, based on the Fibonacci sequence. The Fibonacci sequence $F_0,F_1,F_2,\dots$ is defined recursively as follows $$ \begin{align*} F_0&=1\\ F_1&=1\\ F_n&=F_{n-1}+F_{n-2}\text{ for }n\ge 2 \end{align*} $$ A Leutonian number is a string of 0’s and 1’s that begins with a 1 and never has two consecutive 1’s. If $s=s_\ell s_{\ell-1}\dots s_1$ is such a string of length $\ell$, where each $s_i$ is in $\{0,1\}$, the number represented by $s$ is $n(s)=\sum\limits_{i=1}^\ell s_i\cdot F_i$.

For example, $n(1000101)=F_7+F_3+F_1=21+3+1=25$.

(a) Write out the Leutonian numbers that represent the first 12 positive integers.

(b) Prove: For every $\ell\ge1$, if $s$ is a Leutonian number of lenght $\ell$, $n(s)\ge F_\ell$

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I am trying to do (b).

That's what I did:

Basis: $p(1) : \sum_{i=1}^1 s_i \cdot F_1 \ge F_1$

$ = \sum_{i=1}^1 s_1 \cdot 1 \ge 1$

Since a leutonian number must starts with 1. $s_1 = 1$

so the base case is true.

I can't do the induction case.

  • $\begingroup$ This is just a thought, have you tried looking at a Leutonian number of length $L\gt2$ as a combination of a length $L-1$ and length $L-2$ Leutonian numbers and using the Fibanocci recurrence to do the induction case? $\endgroup$
    – Average
    Sep 25, 2014 at 5:01
  • $\begingroup$ Wikipedia: Zeckendorf's theorem. See also: math.stackexchange.com/questions/198253/… $\endgroup$ Sep 25, 2014 at 11:42
  • $\begingroup$ For the sake of completeness, you could also have added link to the original source: eecs.yorku.ca/course_archive/2014-15/F/2001/a2.pdf (Google does not return many hits for "Leutonian numbers", so I guess this might be where the problem comes from.) $\endgroup$ Sep 25, 2014 at 11:52

1 Answer 1


I'm not quite sure why you want to use induction to prove this. Can't you say that a Leutonian number $s$ of length $l$ has $s_l=1$, hence $s=\sum_{i=1}^ls_iF_i=F_l+\sum_{i=1}^{l-1}s_iF_i\geq F_l$, since each term $>0$?


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