So for example 10000 <= 1111 in baseF.

Then we proceed by induction:

Define F(n) = n’th Fibonacci number, which is 1 with n-1 0’s in baseF. (F(1) = 1, F(2) = 10, F(3) = 100, etc.)

Define G(n) = the number that is all 1’s in baseF with n 1’s. (G(1) = 1, G(2) = 11, G(3) = 111, etc.)

A) All numbers from 1 to G(1) are representable in baseF is trivially true, since G(1) = 1.

B) If all numbers from 1 to G(n) are representable in baseF, then all numbers from 1 to G(n+1) are representable.

Proof: G(n+1) = F(n+1) + G(n), so all numbers from F(n+1)+1 to F(n+1)+G(n) (=G(n+1)) are representable from the premise. And F(n+1) itself is certainly representable. So we just need to prove that the two ranges 1 — G(n) and F(n+1) — G(n+1) are overlapping ranges. i.e. that F(n+1) <= G(n) for all n>=1. (This was my statement at the start of the post.)

F(2) = G(1) = 1 (base 10)

F(3) = G(2) = 2 (base 10)

For n > 2, F(n+1) = F(n) + F(n-1) < G(n)

QED. ]]>

Assume a representation of 1’s and 0’s representing “places” low to high from right to left. So numbers are represented by …8-5-3-2-1 and you could represent 16 as 11100 .

The question then is equivalent to whether we can translate this into a numerically equivalent number with no instances of the string ’11’. Because of how Fibonacci numbers work, any string of ‘011’ can be translated to ‘100’ without changing the value of the number. Do this repetitively until there are no more instances. (This is sure to work as you can pad with zeros on the left as needed.)

So 11100 -> 011100 -> 100100

So instead of having 8+5+3 = 16 we have 13+3=16

11111 = 19

11111 -> 011111 -> 100111 -> 101001

101001 = 13+5+1 = 19

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Must think some more about “base Fibonacci”. :-)

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