# Relationship between decimal length and Fibonacci number

There are 6 single digit Fibonacci numbers. For all other number of digits in the decimal system, there are either 4 or 5 Fibonacci numbers. For example, between 10000 and 99999 there are 5:

10946 17711 28657 46368 75025

while between 10000000 and 99999999 there are 4:

14930352 24157817 39088169 63245986

My question is when the decimal place gets sufficiently large, does the ratio between those with 5 and those with 4 converge?

I ran some code and found out that from the first to the 60000th Fibonacci number, approximately 9841 decimal places have 5 Fibonacci numbers while approximately 2696 decimal places have 4 Fibonacci numbers. This puts the ratio at approximately 3.65.

Below is a list of the first 13 decimal places:

1 decimal place has 6 Fibonacci numbers

2 decimal place has 5 Fibonacci numbers

3 decimal place has 5 Fibonacci numbers

4 decimal place has 4 Fibonacci numbers

5 decimal place has 5 Fibonacci numbers

6 decimal place has 5 Fibonacci numbers

7 decimal place has 5 Fibonacci numbers

8 decimal place has 4 Fibonacci numbers

9 decimal place has 5 Fibonacci numbers

10 decimal place has 5 Fibonacci numbers

11 decimal place has 5 Fibonacci numbers

12 decimal place has 5 Fibonacci numbers

13 decimal place has 4 Fibonacci numbers

• I would guess it converges to 2×Phi but that's just a guess Commented Jun 21, 2016 at 2:55

Taking $$\phi = \frac{1 + \sqrt 5}{2},$$ your ratio is exactly $$\frac{\log {10} - 4 \log \phi}{5 \log \phi - \log {10}} \approx \frac{0.377737792}{0.103474033} \approx 3.650556386$$ The exact formula (Binet) for the Fibonacci numbers does not matter, since taking logarithms makes any constant coefficient disappear in the limit. All that matters is that the ratio rapidly approaches $$\phi.$$