A conjecture on the closeness of twin primes
$begingroup$
Let $p_1$ and $p_2$ be twin primes, and let $p_1-1=a_1times b_1$ and $p_2+1=a_2times b_2$ be such that $|b_1-a_1|$ and $|b_2-a_2|$ are minimised. Similarly, let $p_1+1=p_2-1=atimes b$ be such that $|b-a|$ is minimised.
Now assume the twin prime conjecture, and let $$mathcal P_n=frac{#text{twin primes} le n:min{|b_1-a_1|,|b-a|,|b_2-a_2|}neq|b-a|}{#text{twin primes} le n}$$ What is the value of $mathcal P_infty$?
Note that by convention the pair $(2,3)$ are not twin primes.
For twin primes less than $100$, we have that $p_1=5,17$. Hence I think that if there are infinitely many such primes then they will be exceedingly sparse.
Here is a table showing $mathcal P_n$ for increasing values of $n$. begin{array}{c|c}log_{10}n&1&2&3&4&5&6&7&8&9&10\hline mathcal P_n½¼&frac{17}{35}&frac{93}{205}&frac{600}{1224}&frac{4326}{8169}&frac{31939}{58980}&frac{243876}{440312}&frac{1928700}{3424506}&frac{15661079}{27412679}\hlinetext{decimal}&small 0.5&small0.25&small0.4857&small0.4537&small0.4902&small0.5296&small0.5415&small0.5539&small0.5632&small0.5713end{array}
This can be seen more clearly in the following plot; more interesting is the behaviour after $n=10^4$.
There is a slight dip at $n=10^2,10^4$ but otherwise it looks like there is a converging increase in $mathcal P_n$.
number-theory prime-numbers conjectures prime-twins
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
$endgroup$
|
show 1 more comment
$begingroup$
Let $p_1$ and $p_2$ be twin primes, and let $p_1-1=a_1times b_1$ and $p_2+1=a_2times b_2$ be such that $|b_1-a_1|$ and $|b_2-a_2|$ are minimised. Similarly, let $p_1+1=p_2-1=atimes b$ be such that $|b-a|$ is minimised.
Now assume the twin prime conjecture, and let $$mathcal P_n=frac{#text{twin primes} le n:min{|b_1-a_1|,|b-a|,|b_2-a_2|}neq|b-a|}{#text{twin primes} le n}$$ What is the value of $mathcal P_infty$?
Note that by convention the pair $(2,3)$ are not twin primes.
For twin primes less than $100$, we have that $p_1=5,17$. Hence I think that if there are infinitely many such primes then they will be exceedingly sparse.
Here is a table showing $mathcal P_n$ for increasing values of $n$. begin{array}{c|c}log_{10}n&1&2&3&4&5&6&7&8&9&10\hline mathcal P_n½¼&frac{17}{35}&frac{93}{205}&frac{600}{1224}&frac{4326}{8169}&frac{31939}{58980}&frac{243876}{440312}&frac{1928700}{3424506}&frac{15661079}{27412679}\hlinetext{decimal}&small 0.5&small0.25&small0.4857&small0.4537&small0.4902&small0.5296&small0.5415&small0.5539&small0.5632&small0.5713end{array}
This can be seen more clearly in the following plot; more interesting is the behaviour after $n=10^4$.
There is a slight dip at $n=10^2,10^4$ but otherwise it looks like there is a converging increase in $mathcal P_n$.
number-theory prime-numbers conjectures prime-twins
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
$endgroup$
$begingroup$
Hmm, happens more often after $100$. $p_1 in {101, 107, 137}$ for example.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:12
1
$begingroup$
As long as you only want to go up to a moderate limit, modify a Sieve of Eratosthenes to record a prime factor of each number (the smallest or the largest would be natural choices). That way, you get the factorisation of every number very cheaply, and with the prime factorisation it's easy to find the factors closest to the square root. If you want to look at large numbers, it won't be so easy. Factoring in general is hard.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:23
$begingroup$
you can use this construct an artificial number system where this is true, then you will find even there it will not hold. try.
$endgroup$
– Nick
Sep 2 '18 at 19:43
1
$begingroup$
Made a very simple Sage program to find them.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:01
3
$begingroup$
Hopefully I didn't make any mistakes. It can handle up to $10^7$ easily, after which it's quite slow. At $10^7$ it fetches $31939$ passes out of $58980$ twin primes which is about 50%. It fetches a similar ratio for $10^3,10^4,cdots,10^7$.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:07
|
show 1 more comment
$begingroup$
Let $p_1$ and $p_2$ be twin primes, and let $p_1-1=a_1times b_1$ and $p_2+1=a_2times b_2$ be such that $|b_1-a_1|$ and $|b_2-a_2|$ are minimised. Similarly, let $p_1+1=p_2-1=atimes b$ be such that $|b-a|$ is minimised.
Now assume the twin prime conjecture, and let $$mathcal P_n=frac{#text{twin primes} le n:min{|b_1-a_1|,|b-a|,|b_2-a_2|}neq|b-a|}{#text{twin primes} le n}$$ What is the value of $mathcal P_infty$?
Note that by convention the pair $(2,3)$ are not twin primes.
For twin primes less than $100$, we have that $p_1=5,17$. Hence I think that if there are infinitely many such primes then they will be exceedingly sparse.
Here is a table showing $mathcal P_n$ for increasing values of $n$. begin{array}{c|c}log_{10}n&1&2&3&4&5&6&7&8&9&10\hline mathcal P_n½¼&frac{17}{35}&frac{93}{205}&frac{600}{1224}&frac{4326}{8169}&frac{31939}{58980}&frac{243876}{440312}&frac{1928700}{3424506}&frac{15661079}{27412679}\hlinetext{decimal}&small 0.5&small0.25&small0.4857&small0.4537&small0.4902&small0.5296&small0.5415&small0.5539&small0.5632&small0.5713end{array}
This can be seen more clearly in the following plot; more interesting is the behaviour after $n=10^4$.
There is a slight dip at $n=10^2,10^4$ but otherwise it looks like there is a converging increase in $mathcal P_n$.
number-theory prime-numbers conjectures prime-twins
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
$endgroup$
Let $p_1$ and $p_2$ be twin primes, and let $p_1-1=a_1times b_1$ and $p_2+1=a_2times b_2$ be such that $|b_1-a_1|$ and $|b_2-a_2|$ are minimised. Similarly, let $p_1+1=p_2-1=atimes b$ be such that $|b-a|$ is minimised.
Now assume the twin prime conjecture, and let $$mathcal P_n=frac{#text{twin primes} le n:min{|b_1-a_1|,|b-a|,|b_2-a_2|}neq|b-a|}{#text{twin primes} le n}$$ What is the value of $mathcal P_infty$?
Note that by convention the pair $(2,3)$ are not twin primes.
For twin primes less than $100$, we have that $p_1=5,17$. Hence I think that if there are infinitely many such primes then they will be exceedingly sparse.
Here is a table showing $mathcal P_n$ for increasing values of $n$. begin{array}{c|c}log_{10}n&1&2&3&4&5&6&7&8&9&10\hline mathcal P_n½¼&frac{17}{35}&frac{93}{205}&frac{600}{1224}&frac{4326}{8169}&frac{31939}{58980}&frac{243876}{440312}&frac{1928700}{3424506}&frac{15661079}{27412679}\hlinetext{decimal}&small 0.5&small0.25&small0.4857&small0.4537&small0.4902&small0.5296&small0.5415&small0.5539&small0.5632&small0.5713end{array}
This can be seen more clearly in the following plot; more interesting is the behaviour after $n=10^4$.
There is a slight dip at $n=10^2,10^4$ but otherwise it looks like there is a converging increase in $mathcal P_n$.
number-theory prime-numbers conjectures prime-twins
number-theory prime-numbers conjectures prime-twins
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
edited Dec 16 '18 at 15:44
TheSimpliFire
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
asked Sep 2 '18 at 19:03
TheSimpliFireTheSimpliFire
12.5k62460
12.5k62460
$begingroup$
Hmm, happens more often after $100$. $p_1 in {101, 107, 137}$ for example.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:12
1
$begingroup$
As long as you only want to go up to a moderate limit, modify a Sieve of Eratosthenes to record a prime factor of each number (the smallest or the largest would be natural choices). That way, you get the factorisation of every number very cheaply, and with the prime factorisation it's easy to find the factors closest to the square root. If you want to look at large numbers, it won't be so easy. Factoring in general is hard.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:23
$begingroup$
you can use this construct an artificial number system where this is true, then you will find even there it will not hold. try.
$endgroup$
– Nick
Sep 2 '18 at 19:43
1
$begingroup$
Made a very simple Sage program to find them.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:01
3
$begingroup$
Hopefully I didn't make any mistakes. It can handle up to $10^7$ easily, after which it's quite slow. At $10^7$ it fetches $31939$ passes out of $58980$ twin primes which is about 50%. It fetches a similar ratio for $10^3,10^4,cdots,10^7$.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:07
|
show 1 more comment
$begingroup$
Hmm, happens more often after $100$. $p_1 in {101, 107, 137}$ for example.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:12
1
$begingroup$
As long as you only want to go up to a moderate limit, modify a Sieve of Eratosthenes to record a prime factor of each number (the smallest or the largest would be natural choices). That way, you get the factorisation of every number very cheaply, and with the prime factorisation it's easy to find the factors closest to the square root. If you want to look at large numbers, it won't be so easy. Factoring in general is hard.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:23
$begingroup$
you can use this construct an artificial number system where this is true, then you will find even there it will not hold. try.
$endgroup$
– Nick
Sep 2 '18 at 19:43
1
$begingroup$
Made a very simple Sage program to find them.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:01
3
$begingroup$
Hopefully I didn't make any mistakes. It can handle up to $10^7$ easily, after which it's quite slow. At $10^7$ it fetches $31939$ passes out of $58980$ twin primes which is about 50%. It fetches a similar ratio for $10^3,10^4,cdots,10^7$.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:07
$begingroup$
Hmm, happens more often after $100$. $p_1 in {101, 107, 137}$ for example.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:12
$begingroup$
Hmm, happens more often after $100$. $p_1 in {101, 107, 137}$ for example.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:12
1
1
$begingroup$
As long as you only want to go up to a moderate limit, modify a Sieve of Eratosthenes to record a prime factor of each number (the smallest or the largest would be natural choices). That way, you get the factorisation of every number very cheaply, and with the prime factorisation it's easy to find the factors closest to the square root. If you want to look at large numbers, it won't be so easy. Factoring in general is hard.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:23
$begingroup$
As long as you only want to go up to a moderate limit, modify a Sieve of Eratosthenes to record a prime factor of each number (the smallest or the largest would be natural choices). That way, you get the factorisation of every number very cheaply, and with the prime factorisation it's easy to find the factors closest to the square root. If you want to look at large numbers, it won't be so easy. Factoring in general is hard.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:23
$begingroup$
you can use this construct an artificial number system where this is true, then you will find even there it will not hold. try.
$endgroup$
– Nick
Sep 2 '18 at 19:43
$begingroup$
you can use this construct an artificial number system where this is true, then you will find even there it will not hold. try.
$endgroup$
– Nick
Sep 2 '18 at 19:43
1
1
$begingroup$
Made a very simple Sage program to find them.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:01
$begingroup$
Made a very simple Sage program to find them.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:01
3
3
$begingroup$
Hopefully I didn't make any mistakes. It can handle up to $10^7$ easily, after which it's quite slow. At $10^7$ it fetches $31939$ passes out of $58980$ twin primes which is about 50%. It fetches a similar ratio for $10^3,10^4,cdots,10^7$.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:07
$begingroup$
Hopefully I didn't make any mistakes. It can handle up to $10^7$ easily, after which it's quite slow. At $10^7$ it fetches $31939$ passes out of $58980$ twin primes which is about 50%. It fetches a similar ratio for $10^3,10^4,cdots,10^7$.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:07
|
show 1 more comment
1 Answer
1
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oldest
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$begingroup$
Twin primes other than $3,5$ are of the form $6kpm1$, which makes $p_1-1 = 6k-2$, $p_2 + 1 = 6k+2$, and $p_1+1 = p_2-1 = 6k$. Define $f(x) = min{|a-b|}$ where $a b = x$, and you are asking are there infinitely many $k$ where $6kpm1$ are both primes, and $f(6k-2) < f(6k)$ or $f(6k+2) < f(6k)$.
Experimentally, about $54%$ of the twin primes are "close", not the $33frac13%$ that one would naively expect. Still, most likely infinitely many.
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
$endgroup$
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
add a comment |
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1 Answer
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$begingroup$
Twin primes other than $3,5$ are of the form $6kpm1$, which makes $p_1-1 = 6k-2$, $p_2 + 1 = 6k+2$, and $p_1+1 = p_2-1 = 6k$. Define $f(x) = min{|a-b|}$ where $a b = x$, and you are asking are there infinitely many $k$ where $6kpm1$ are both primes, and $f(6k-2) < f(6k)$ or $f(6k+2) < f(6k)$.
Experimentally, about $54%$ of the twin primes are "close", not the $33frac13%$ that one would naively expect. Still, most likely infinitely many.
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
$endgroup$
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
add a comment |
$begingroup$
Twin primes other than $3,5$ are of the form $6kpm1$, which makes $p_1-1 = 6k-2$, $p_2 + 1 = 6k+2$, and $p_1+1 = p_2-1 = 6k$. Define $f(x) = min{|a-b|}$ where $a b = x$, and you are asking are there infinitely many $k$ where $6kpm1$ are both primes, and $f(6k-2) < f(6k)$ or $f(6k+2) < f(6k)$.
Experimentally, about $54%$ of the twin primes are "close", not the $33frac13%$ that one would naively expect. Still, most likely infinitely many.
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
$endgroup$
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
add a comment |
$begingroup$
Twin primes other than $3,5$ are of the form $6kpm1$, which makes $p_1-1 = 6k-2$, $p_2 + 1 = 6k+2$, and $p_1+1 = p_2-1 = 6k$. Define $f(x) = min{|a-b|}$ where $a b = x$, and you are asking are there infinitely many $k$ where $6kpm1$ are both primes, and $f(6k-2) < f(6k)$ or $f(6k+2) < f(6k)$.
Experimentally, about $54%$ of the twin primes are "close", not the $33frac13%$ that one would naively expect. Still, most likely infinitely many.
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
$endgroup$
Twin primes other than $3,5$ are of the form $6kpm1$, which makes $p_1-1 = 6k-2$, $p_2 + 1 = 6k+2$, and $p_1+1 = p_2-1 = 6k$. Define $f(x) = min{|a-b|}$ where $a b = x$, and you are asking are there infinitely many $k$ where $6kpm1$ are both primes, and $f(6k-2) < f(6k)$ or $f(6k+2) < f(6k)$.
Experimentally, about $54%$ of the twin primes are "close", not the $33frac13%$ that one would naively expect. Still, most likely infinitely many.
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
edited Sep 9 '18 at 10:39
TheSimpliFire
12.5k62460
12.5k62460
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
answered Sep 3 '18 at 21:39
gnasher729gnasher729
5,9871028
5,9871028
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
add a comment |
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
For what batch size is twin prime closeness avereged at 54%?
$endgroup$
– Nick
Sep 29 '18 at 1:45
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
$begingroup$
Should this be posted as a comment?
$endgroup$
– TheSimpliFire
Dec 16 '18 at 15:11
add a comment |
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$begingroup$
Hmm, happens more often after $100$. $p_1 in {101, 107, 137}$ for example.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:12
1
$begingroup$
As long as you only want to go up to a moderate limit, modify a Sieve of Eratosthenes to record a prime factor of each number (the smallest or the largest would be natural choices). That way, you get the factorisation of every number very cheaply, and with the prime factorisation it's easy to find the factors closest to the square root. If you want to look at large numbers, it won't be so easy. Factoring in general is hard.
$endgroup$
– Daniel Fischer♦
Sep 2 '18 at 19:23
$begingroup$
you can use this construct an artificial number system where this is true, then you will find even there it will not hold. try.
$endgroup$
– Nick
Sep 2 '18 at 19:43
1
$begingroup$
Made a very simple Sage program to find them.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:01
3
$begingroup$
Hopefully I didn't make any mistakes. It can handle up to $10^7$ easily, after which it's quite slow. At $10^7$ it fetches $31939$ passes out of $58980$ twin primes which is about 50%. It fetches a similar ratio for $10^3,10^4,cdots,10^7$.
$endgroup$
– Yong Hao Ng
Sep 3 '18 at 5:07