It seems that the probability of a prime number existing between adjacent Fibonacci numbers approach

0. References

[1] https://www.chart.co.jp/subject/sugaku/suken_tsushin/91/91-4.pdf Using Bertrand Chebyshev's theorem.

[2] (Special case of ternary gradual formula)Will be attached as soon as I find the site where it is introduced.

[3] https://www.irohabook.com/fibonacci-sequence　

Definition of the Fibonacci sequence and the general term and golden ratio (the ratio of adjacent terms in the Fibonacci sequence converges to the golden ratio)

1. On the interval where 100% of the prime numbers exist

Just as there are deserts and oceans in this world, there are prime deserts and intervals where the existence of prime numbers is guaranteed in natural numbers. That's what Bertrand Chebyshev's theorem explains! [1] Bertrand made a prediction about this property, and Chebyshev proved it. The interval can be expressed by the following equation!

For example, when n=826, 2n=1652. there are prime numbers in between, for example 827. For every other natural number, there is a prime number as well!

2. what you notice when you look at the Fibonacci sequence.

Write down some of the Fibonacci numbers.

1,2,3,5,8,13,21,34,55,89,144,...

Somehow, I get the feeling that there is always a Fibonacci number between the Fibonacci numbers, including itself (I was influenced by chapter 1...). So, let's make the following hypothesis.

Hypothesis: "Most of the two adjacent Fibonacci numbers [f(n),f(n+1)] will have at least one prime number in the interval that contains them.

This hypothesis, if we say "Every two adjacent Fibonacci numbers...", is a stronger argument than Bertrand Chebyshev's theorem. However, since we are considering the case where Bertrand Chebyshev's theorem is true, I said "most...".

3. probability problems - at least one...

There is a math problem that sometimes appears in high school entrance exams, "What is the probability that at least one 00 will appear? You can try to solve this exercise. Let's try to solve this practice problem.

Question. There are two dice with the numbers 1 to 6 on them that are equally likely to come up. What is the probability that at least one of the dice is odd when thrown once at the same time?

This kind of question actually tests whether you can imagine the opposite of what is being asked. In short, you should think about "What is the probability of never getting a 00? In real life, for example, you might say, "This week, I'm not going to get any.

In real life, for example, if you think, "I'm going to go shopping at least three times this week," you would look up the days you can't go shopping before making your shopping plan, right? It's the same thing.

In this problem, the probability of getting an odd (or even) number is 1/2, so we can calculate it as follows

p=1-(1/2)^2=3/4

Did you understand about "at least"? 4.

4. fibonacci numbers and number lines

Let the nth Fibonacci number be f(n). Then plot the next two Fibonacci numbers f(n+1) and f(n+2).

We give for each Fibonacci number an interval where we can use the Bertrand-Chebyshev theorem. We will ignore 2f(n+2) since we will not be using it.

From the definition of Fibonacci numbers, we get

f(n+1)-f(n)=f(n-1)

2f(n+1)-f(n+2)=f(n+1)+f(n+1)-f(n+2)=f(n+1)-f(n)=f(n-1)

5.

5. the probability that there is a prime number in [f(n+1),2f(n)], [2f(n),f(n+2)

What I wanted to know this time was what is the probability that a prime number exists between [f(n+1),f(n+2)]. It is equal to the probability that a prime number exists in at least one of the two intervals [f(n+1),2f(n)], [2f(n),f(n+2)].

Now, did you notice the word "at least"? That's the point! In other words, this problem is

This problem is equivalent to the probability that there are no primes in the two intervals [f(n),f(n+1)],[f(n+2),2f(n+1)] at the same time.

is the same as [f(n+2),2f(n+1)]!

Simply because the probability of the existence of a prime number is higher if the number of natural numbers (=maximum-minimum+1) in the interval is greater (in fact, it is correct because the prime number theorem is monotonically increasing, which means that the probability of the existence of a prime number is higher if the number of natural numbers (=maximum-minimum+1) in the interval is greater). The probability that there is a prime number in each interval, except for 2f(n) which is obviously even,

The probability of two things happening at the same time is the product of their probabilities. Since the probability of two events occurring at the same time is the product of their respective probabilities, the probability that a prime number exists between [f(n+1),f(n+2)] is as follows

Here, we will use the well-known theorem about Fibonacci numbers [2].

However, I have annotated f(2)=1 so that you can see the evenness of the numbers.

Using this, the transformation is as follows.

This value is approximately 0.764. In percentage terms, that's 76.4%!

Note that I used the method in [3] to get the -3 power, so this probability was originally larger than 76.4%, and then gradually decreased. This means that 76.4% is the lower limit of this probability!