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.
This is a bit of a recursive argument.
Suppose for n-1 there are 4 Fibonacci numbers. The largest of them must be at least 5*10^(n-2) so the first n digit one is 1*10^(n-1).
Call them .5 and 1 for short.
The next numbers are 1.5, 2.5, 4, 6.5, 10.5
So there can't be more than 5.
Suppose instead the largest two are 1-2*epsilon and 1-epsilon.
The n digit numbers are
2-3e, 3-4e, 5-7e, 8-11e.
So there can't be fewer than 4.
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Posted by Jer
on 2023-01-16 13:25:16 |
Muddled proof
