0

Refer to https://oeis.org/A340800 to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase. Is there an explanation for this? How can the size of primes have any influence on the last digit of the following primes?

J. M. Bergot
  • 912

3 Answers3

2

Supposing the last digits of primes form a random sequence, from the set 1,3,7,9.
Let the prime we want end in a 1. The next $n-1$ primes must end with something else, and the $n$th end with a 1. This happens with probability $$\left(\frac34\right)^{n-1}\frac14$$
So it is harder to belong in a later slot of A340800, so the first one in each slot tends to be higher.

Empy2
  • 50,853
  • 1
    The same probability occurs for any random list. So how can you claim that the improbability increases as the numbers in this random list increase? – J. M. Bergot Jan 22 '21 at 04:58
  • Each prime except $2$ and $5$ is in one of the slots, although A340800 only contains the first of each slot. As $n$ increases, numbers in the $n$th slot become rarer. If one in a million go in the $N$th slot, then it is quite likely that none of the first thousand go in that slot, by which time the early slots all have their first entry. So numbers tend to increase in A340800, but that is not always true. – Empy2 Jan 22 '21 at 05:19
  • Say we are looking for last digit 1. If it is many primes away, that means that all the last digits in between are 3,7,9 and they must be close together since there are so many of them. Why is this not true for last digit 1, but only for last digits 3,7,9? The same argument could be applied to them singly, hence there's a contradiction. – J. M. Bergot Jan 22 '21 at 06:02
0

Let $$\mathfrak P(x,n)=\Bbb P(\exists m\leq n \in \Bbb N : x+10m \text{ is prime} | x \text{ is prime})$$

The relative density of primes decreases as numbers get larger. Because of this, $\mathfrak P$ is very high for small $x$ and generally decreasing in $x$. At the same time, it is increasing in $n$ (Every time you add another potential number that might be prime, the probability increases).

One can see from this exactly what the OEIS results show, that if a prime is small, the next prime with the same last digit is very likely not much bigger, and for a large prime, you'll likely need a larger $n$ to satisfy, which implies the next prime will be further away.

Rhys Hughes
  • 12,842
  • One needs to study for higher primes the average number of primes between two primes having the same last digit to see if this number is increasing. If the average is 25 primes, then finding one with 28 primes in between is nothing exciting. If one examined all numbers between the two primes, would the composites also have the same property of having many composites not having the desired last digit? I think more data is required. – J. M. Bergot Jan 22 '21 at 05:31
  • Well, the decreasing relative density of primes directly implies that the average difference between $p_k$ and $p_{k+1}$ increases as $k\to\infty$, and this in turn means the average distance between prime subject to any other condition increases as well. – Rhys Hughes Jan 22 '21 at 10:20
  • What has to be done is to find some term in A340800 such that all terms after that are strictly increasing. Without this there is no way to avoid the randomness of these gaps. I wonder if anyone has examined the average length of gaps between primes having the same last digits as one moves higher into the primes. Applying the probability as given by Emp2 (above) does not in any way come close to what A340800 lists. – J. M. Bergot Jan 22 '21 at 23:48
0

Of course the general trend of A340800 must be increasing (although it is not monotonic). For any $M$, the maximum of $A072971(k)$ for $k \le M$ is some finite value $N$, and so if $n > N$ and $A340800(n)$ exists at all it must be $prime(k)$ for some $k > M$. The existence of $A340800(n)$ follows from the prime $k$-tuples conjecture.

Robert Israel
  • 448,999
  • Observing a(13), has four larger terms before it; a(24) has five larger terms before it; a(28) has 3 larger terms before it; a(35) has 3 larger terms before it. Clearly not monotonically increasing, which suggests that the more an event is improbable, the more trials must be made, but not necessarily true in all cases. One would have calculate the probability for all the failures between the two primes having the same last digit to judge how well the probability reflects the facts of the matter. I conclude that none of this is related to the primes as primes, but just to wild chance. – J. M. Bergot Jan 22 '21 at 07:30