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Consider the set of $k$ numbers $\{n+c_1,n+c_2,n+c_3,...,n+c_k\}$ where $c_i$ are constant positive integers and $n$ is a varying positive integer such that $n \leqslant M$. What is the probability that there are $x$ numbers in this set which are prime, where $0 \leqslant x \leqslant k$ ? If this is really hard, is there any answer to cases such as $x=0$ and $x=1$ ?

At $k=1$, we have to consider whether $n+c_1$ is prime or composite. We know that $M < n+c_1 \leqslant M+c_1$. Thus, our probability $P(k,M,c_i)$ would be: $$ P(1,M,c_1) = \frac{\pi(M+c_1)-\pi(M)}{c_1} \sim \frac{(M+c_1)\ln M-M\ln(M+c_1)}{c_1 \ln M \ln(M+c_1)}$$

Is the working for $k=1$ right? Can I directly apply Prime Number Theorem for $k \geqslant 2$ in a similar fashion. I am doubtful as we are dealing with a group of equally interspaced numbers. Any help or ideas are accepted. Thanks in advance!

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  • $\begingroup$ Probability wrt what probability distribution on $n$ ? The uniform distribution ? Then it is a matter of counting some particular $l$-uples of primes $\le M$ $\endgroup$ – reuns Jan 8 at 8:13
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It's a well known fact that the average distance between primes about as large as $n$ is $\ln n$. So the probability for a number as large as $n$ to be prime is about $1/\ln n$.

Probability that your $k$-th number is prime is therefore:

$$p_k=\frac{1}{\ln (n+c_k)}$$

Probability that your $k$-th number is not prime is:

$$\bar{p}_k=1-\frac{1}{\ln (n+c_k)}$$

For $x=0$ all numbers must be composite so the probability is:

$$P=\prod_{i=1}^k \bar{p}_i$$

For $x=1$, exactly one number has to be prime so the probability is:

$$P=p_1\bar{p}_2\bar{p}_3\dots\bar{p}_k+\bar{p}_1{p}_2\bar{p}_3\dots\bar{p}_k+\dots+\bar{p}_1\bar{p}_2\bar{p}_3\dots{p}_k$$

It's a little bit more difficult to calculate probability for $x>1$ but it's still a fairly straightforward task.

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  • $\begingroup$ This is exactly what I had initially done. But can the prime number theorem directly be applied for this question? We are dealing with a group of numbers with constant distance between them. $\endgroup$ – Haran Jan 8 at 12:59
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    $\begingroup$ This might work as a heuristic/rule of thumb, but not more strictly: the probabilities are not independent and thus cannot be plainly multiplied together. $\endgroup$ – Mees de Vries Jan 8 at 13:11
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    $\begingroup$ In particular, if $k = 2, c_2 = c_1 + 2, x = 2$, the twin prime conjecture essentially asks, "is there $c_1$ such that for any $M$ the probability equals 0". $\endgroup$ – Mees de Vries Jan 8 at 13:15

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