Depending on what you consider the length of 0 to be, there are either 6 or 7 one-digit Fibonacci numbers. There are 5 two-digit Fibonacci numbers. There are 4 four-digit Fibonacci numbers.
Prove that for n > 1, there are always either 4 or 5 n-digit Fibonacci numbers, or find a counterexample.
For all n >= 0, the number F(n) is the closest integer to phi^n / sqrt(5) rounded to the nearest integer, so the ratio of one Fibonacci number to the next approaches phi = (1+sqrt(5))/2 ~= 1.61803398874989.
If a given Fibonacci is in the low portion of its length, say 10^k, the next Fibonacci will be the same number of digits, but at least 1.61*10^k. After that the next Fibonacci will be at least phi^2 times the original, which is about 2.62.
Here's a table of powers of phi:
1 1.61803398874989
2 2.61803398874989
3 4.23606797749979
4 6.85410196624969
5 11.0901699437495
6 17.9442719099992
7 29.0344418537486
Five iterations to the next Fibonacci is definitely sufficient to get to the next length.
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Posted by Charlie
on 2023-01-16 10:38:18 |
The start of a proof
