Number of primes in sets of $k$ positive integers












0












$begingroup$


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:
$$ P(1,M,c_1) = frac{pi(M+c_1)-pi(M)}{c_1} sim frac{(M+c_1)ln M-Mln(M+c_1)}{c_1 ln M ln(M+c_1)}$$



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!






probability number-theory prime-numbers







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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


















0












$begingroup$


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:
$$ P(1,M,c_1) = frac{pi(M+c_1)-pi(M)}{c_1} sim frac{(M+c_1)ln M-Mln(M+c_1)}{c_1 ln M ln(M+c_1)}$$



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!






probability number-theory prime-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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:
$$ P(1,M,c_1) = frac{pi(M+c_1)-pi(M)}{c_1} sim frac{(M+c_1)ln M-Mln(M+c_1)}{c_1 ln M ln(M+c_1)}$$



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!






probability number-theory prime-numbers







share|cite|improve this question











$endgroup$




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:
$$ P(1,M,c_1) = frac{pi(M+c_1)-pi(M)}{c_1} sim frac{(M+c_1)ln M-Mln(M+c_1)}{c_1 ln M ln(M+c_1)}$$



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!






probability number-theory prime-numbers




probability number-theory prime-numbers






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edited Jan 8 at 13:18







Haran

















asked Jan 8 at 7:50









HaranHaran

1,116424




1,116424












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
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






$begingroup$
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












1 Answer
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$begingroup$

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:



$$P=p_1bar{p}_2bar{p}_3dotsbar{p}_k+bar{p}_1{p}_2bar{p}_3dotsbar{p}_k+dots+bar{p}_1bar{p}_2bar{p}_3dots{p}_k$$



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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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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1 Answer
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0












$begingroup$

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:



$$P=p_1bar{p}_2bar{p}_3dotsbar{p}_k+bar{p}_1{p}_2bar{p}_3dotsbar{p}_k+dots+bar{p}_1bar{p}_2bar{p}_3dots{p}_k$$



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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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
















0












$begingroup$

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:



$$P=p_1bar{p}_2bar{p}_3dotsbar{p}_k+bar{p}_1{p}_2bar{p}_3dotsbar{p}_k+dots+bar{p}_1bar{p}_2bar{p}_3dots{p}_k$$



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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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














0












0








0





$begingroup$

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:



$$P=p_1bar{p}_2bar{p}_3dotsbar{p}_k+bar{p}_1{p}_2bar{p}_3dotsbar{p}_k+dots+bar{p}_1bar{p}_2bar{p}_3dots{p}_k$$



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






share|cite|improve this answer









$endgroup$



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:



$$P=p_1bar{p}_2bar{p}_3dotsbar{p}_k+bar{p}_1{p}_2bar{p}_3dotsbar{p}_k+dots+bar{p}_1bar{p}_2bar{p}_3dots{p}_k$$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 9:45









OldboyOldboy

8,3621936




8,3621936












  • $begingroup$
    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








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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


















  • $begingroup$
    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








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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
















$begingroup$
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






$begingroup$
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






1




1




$begingroup$
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




$begingroup$
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




1




1




$begingroup$
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




$begingroup$
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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