Mixing
Capturing all prime numbers and only prime numbers greater than 3
0
$begingroup$
Please advice whether below is correct, anything new or useful?
where n, a and b are integers ≥ 1
For $6n - 1$ all numbers will be prime numbers except where $n= 6ab + b - a$
For $6n + 1$ all will be prime numbers except where $n= 6ab ±(b + a)$
I believe the above 2 will capture all the primes above 3.
This is my first post so forgive me if I have made any errors or you feel I am wasting your time.
prime-numbers
asked Jan 18 at 10:43
K. HussainK. Hussain
13
$endgroup$
add a comment |
0
$begingroup$
Please advice whether below is correct, anything new or useful?
where n, a and b are integers ≥ 1
For $6n - 1$ all numbers will be prime numbers except where $n= 6ab + b - a$
For $6n + 1$ all will be prime numbers except where $n= 6ab ±(b + a)$
I believe the above 2 will capture all the primes above 3.
This is my first post so forgive me if I have made any errors or you feel I am wasting your time.
prime-numbers
asked Jan 18 at 10:43
K. HussainK. Hussain
13
$endgroup$
$begingroup$
What you write somehow makes sense, but it has no argument. You are stating simply a result, which is indeed false. It is very hard to find a formula which captures all prime numbers.
$endgroup$
– Crostul
Jan 18 at 16:34
$begingroup$
If $m=6n-1$ is composite then $m=pq$ where $p=1 mod 6$ and $q=-1 mod 6$ (or vice versa). Let $p=6a+1$ and $q=6b-1$ and your results follow. Similarly if $m=6n+1$ is composite then $m=pq$ where $p=q=1 mod 6$ or $p=q=-1 mod 6$ and your second result follows.
$endgroup$
– gandalf61
Jan 18 at 16:43
$begingroup$
@Crostul mentions results are false without saying why and @ gandalf61 gives explanation of how I got the above result although i used different way as I am not familiar with mod. Thank you. So is it correct? if not can anyone provide a number that it will not work for. If correct is it anything new or useful as I have not come across it before?
$endgroup$
– K. Hussain
Jan 18 at 17:13
$begingroup$
If you exclude 2 and 3 multiples, than evey composite can be expressed as (6b-1)*(6a-1)=36ab-6a-6b+1, or (6b+1)*(6a-1)=36ab+6a-6b-1, or (6b-1)*(6a+1)=36ab-6a+6b-1, or (6b+1)*(6a+1)=36ab+6a+6b+1. So any number not multiple of 2 or 3 and not in one of these forms is prime. Don't know if it is usefull, but it is not new.
$endgroup$
– Collag3n
Jan 18 at 18:34
$begingroup$
@Collag3n thanks that is how I got the above. So it is not new. thanks
$endgroup$
– K. Hussain
Jan 18 at 18:50
add a comment |
0
0
0
$begingroup$
Please advice whether below is correct, anything new or useful?
where n, a and b are integers ≥ 1
For $6n - 1$ all numbers will be prime numbers except where $n= 6ab + b - a$
For $6n + 1$ all will be prime numbers except where $n= 6ab ±(b + a)$
I believe the above 2 will capture all the primes above 3.
This is my first post so forgive me if I have made any errors or you feel I am wasting your time.
prime-numbers
asked Jan 18 at 10:43
K. HussainK. Hussain
13
$endgroup$
Please advice whether below is correct, anything new or useful?
where n, a and b are integers ≥ 1
For $6n - 1$ all numbers will be prime numbers except where $n= 6ab + b - a$
For $6n + 1$ all will be prime numbers except where $n= 6ab ±(b + a)$
I believe the above 2 will capture all the primes above 3.
This is my first post so forgive me if I have made any errors or you feel I am wasting your time.
prime-numbers
prime-numbers
asked Jan 18 at 10:43
K. HussainK. Hussain
13
asked Jan 18 at 10:43
K. HussainK. Hussain
13
edited Jan 18 at 15:36
K. Hussain
asked Jan 18 at 10:43
K. HussainK. Hussain
13
asked Jan 18 at 10:43
K. HussainK. Hussain
13
asked Jan 18 at 10:43
K. HussainK. Hussain
13
13
$begingroup$
What you write somehow makes sense, but it has no argument. You are stating simply a result, which is indeed false. It is very hard to find a formula which captures all prime numbers.
$endgroup$
– Crostul
Jan 18 at 16:34
$begingroup$
If $m=6n-1$ is composite then $m=pq$ where $p=1 mod 6$ and $q=-1 mod 6$ (or vice versa). Let $p=6a+1$ and $q=6b-1$ and your results follow. Similarly if $m=6n+1$ is composite then $m=pq$ where $p=q=1 mod 6$ or $p=q=-1 mod 6$ and your second result follows.
$endgroup$
– gandalf61
Jan 18 at 16:43
$begingroup$
@Crostul mentions results are false without saying why and @ gandalf61 gives explanation of how I got the above result although i used different way as I am not familiar with mod. Thank you. So is it correct? if not can anyone provide a number that it will not work for. If correct is it anything new or useful as I have not come across it before?
$endgroup$
– K. Hussain
Jan 18 at 17:13
$begingroup$
If you exclude 2 and 3 multiples, than evey composite can be expressed as (6b-1)*(6a-1)=36ab-6a-6b+1, or (6b+1)*(6a-1)=36ab+6a-6b-1, or (6b-1)*(6a+1)=36ab-6a+6b-1, or (6b+1)*(6a+1)=36ab+6a+6b+1. So any number not multiple of 2 or 3 and not in one of these forms is prime. Don't know if it is usefull, but it is not new.
$endgroup$
– Collag3n
Jan 18 at 18:34
$begingroup$
@Collag3n thanks that is how I got the above. So it is not new. thanks
$endgroup$
– K. Hussain
Jan 18 at 18:50
add a comment |
$begingroup$
What you write somehow makes sense, but it has no argument. You are stating simply a result, which is indeed false. It is very hard to find a formula which captures all prime numbers.
$endgroup$
– Crostul
Jan 18 at 16:34
$begingroup$
If $m=6n-1$ is composite then $m=pq$ where $p=1 mod 6$ and $q=-1 mod 6$ (or vice versa). Let $p=6a+1$ and $q=6b-1$ and your results follow. Similarly if $m=6n+1$ is composite then $m=pq$ where $p=q=1 mod 6$ or $p=q=-1 mod 6$ and your second result follows.
$endgroup$
– gandalf61
Jan 18 at 16:43
$begingroup$
@Crostul mentions results are false without saying why and @ gandalf61 gives explanation of how I got the above result although i used different way as I am not familiar with mod. Thank you. So is it correct? if not can anyone provide a number that it will not work for. If correct is it anything new or useful as I have not come across it before?
$endgroup$
– K. Hussain
Jan 18 at 17:13
$begingroup$
If you exclude 2 and 3 multiples, than evey composite can be expressed as (6b-1)*(6a-1)=36ab-6a-6b+1, or (6b+1)*(6a-1)=36ab+6a-6b-1, or (6b-1)*(6a+1)=36ab-6a+6b-1, or (6b+1)*(6a+1)=36ab+6a+6b+1. So any number not multiple of 2 or 3 and not in one of these forms is prime. Don't know if it is usefull, but it is not new.
$endgroup$
– Collag3n
Jan 18 at 18:34
$begingroup$
@Collag3n thanks that is how I got the above. So it is not new. thanks
$endgroup$
– K. Hussain
Jan 18 at 18:50
$begingroup$
What you write somehow makes sense, but it has no argument. You are stating simply a result, which is indeed false. It is very hard to find a formula which captures all prime numbers.
$endgroup$
– Crostul
Jan 18 at 16:34
$begingroup$
What you write somehow makes sense, but it has no argument. You are stating simply a result, which is indeed false. It is very hard to find a formula which captures all prime numbers.
$endgroup$
– Crostul
Jan 18 at 16:34
$begingroup$
If $m=6n-1$ is composite then $m=pq$ where $p=1 mod 6$ and $q=-1 mod 6$ (or vice versa). Let $p=6a+1$ and $q=6b-1$ and your results follow. Similarly if $m=6n+1$ is composite then $m=pq$ where $p=q=1 mod 6$ or $p=q=-1 mod 6$ and your second result follows.
$endgroup$
– gandalf61
Jan 18 at 16:43
$begingroup$
If $m=6n-1$ is composite then $m=pq$ where $p=1 mod 6$ and $q=-1 mod 6$ (or vice versa). Let $p=6a+1$ and $q=6b-1$ and your results follow. Similarly if $m=6n+1$ is composite then $m=pq$ where $p=q=1 mod 6$ or $p=q=-1 mod 6$ and your second result follows.
$endgroup$
– gandalf61
Jan 18 at 16:43
$begingroup$
@Crostul mentions results are false without saying why and @ gandalf61 gives explanation of how I got the above result although i used different way as I am not familiar with mod. Thank you. So is it correct? if not can anyone provide a number that it will not work for. If correct is it anything new or useful as I have not come across it before?
$endgroup$
– K. Hussain
Jan 18 at 17:13
$begingroup$
@Crostul mentions results are false without saying why and @ gandalf61 gives explanation of how I got the above result although i used different way as I am not familiar with mod. Thank you. So is it correct? if not can anyone provide a number that it will not work for. If correct is it anything new or useful as I have not come across it before?
$endgroup$
– K. Hussain
Jan 18 at 17:13
$begingroup$
If you exclude 2 and 3 multiples, than evey composite can be expressed as (6b-1)*(6a-1)=36ab-6a-6b+1, or (6b+1)*(6a-1)=36ab+6a-6b-1, or (6b-1)*(6a+1)=36ab-6a+6b-1, or (6b+1)*(6a+1)=36ab+6a+6b+1. So any number not multiple of 2 or 3 and not in one of these forms is prime. Don't know if it is usefull, but it is not new.
$endgroup$
– Collag3n
Jan 18 at 18:34
$begingroup$
If you exclude 2 and 3 multiples, than evey composite can be expressed as (6b-1)*(6a-1)=36ab-6a-6b+1, or (6b+1)*(6a-1)=36ab+6a-6b-1, or (6b-1)*(6a+1)=36ab-6a+6b-1, or (6b+1)*(6a+1)=36ab+6a+6b+1. So any number not multiple of 2 or 3 and not in one of these forms is prime. Don't know if it is usefull, but it is not new.
$endgroup$
– Collag3n
Jan 18 at 18:34
$begingroup$
@Collag3n thanks that is how I got the above. So it is not new. thanks
$endgroup$
– K. Hussain
Jan 18 at 18:50
$begingroup$
@Collag3n thanks that is how I got the above. So it is not new. thanks
$endgroup$
– K. Hussain
Jan 18 at 18:50
add a comment |
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$begingroup$
What you write somehow makes sense, but it has no argument. You are stating simply a result, which is indeed false. It is very hard to find a formula which captures all prime numbers.
$endgroup$
– Crostul
Jan 18 at 16:34
$begingroup$
If $m=6n-1$ is composite then $m=pq$ where $p=1 mod 6$ and $q=-1 mod 6$ (or vice versa). Let $p=6a+1$ and $q=6b-1$ and your results follow. Similarly if $m=6n+1$ is composite then $m=pq$ where $p=q=1 mod 6$ or $p=q=-1 mod 6$ and your second result follows.
$endgroup$
– gandalf61
Jan 18 at 16:43
$begingroup$
@Crostul mentions results are false without saying why and @ gandalf61 gives explanation of how I got the above result although i used different way as I am not familiar with mod. Thank you. So is it correct? if not can anyone provide a number that it will not work for. If correct is it anything new or useful as I have not come across it before?
$endgroup$
– K. Hussain
Jan 18 at 17:13
$begingroup$
If you exclude 2 and 3 multiples, than evey composite can be expressed as (6b-1)*(6a-1)=36ab-6a-6b+1, or (6b+1)*(6a-1)=36ab+6a-6b-1, or (6b-1)*(6a+1)=36ab-6a+6b-1, or (6b+1)*(6a+1)=36ab+6a+6b+1. So any number not multiple of 2 or 3 and not in one of these forms is prime. Don't know if it is usefull, but it is not new.
$endgroup$
– Collag3n
Jan 18 at 18:34
$begingroup$
@Collag3n thanks that is how I got the above. So it is not new. thanks
$endgroup$
– K. Hussain
Jan 18 at 18:50