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What is the form of odd prime numbers that their totients are co-prime with 5?
I'm trying to generate odd prime numbers such that their totients are co-prime with $5$. I want to know if what I'm doing is correct, or can be proven.
- Assume $2, 3$ and $5$ are primes
- Check if $10i - 3, 10i - 1, 10i + 3$ for $i$ greater or equal to $1$
now I did the following to get those forms and I wonder if the approach is correct:
Assume every integer is in the form of $5n-2, 5n-1, 5n, 5n+1, 5n+2$
substitute each of them with $k$ in $2k+1,$ and get $10n-3, 10n-1, 10n+1, 10n+3, 10n+5$
remove the forms that don't form prime for n in positive integers
and have the forms reduced to $10n-3, 10n-1, 10n+1, 10n+3$totient of a prime $p$ is $p-1,$ so subtract $1$ from each of the forms
and see if they're divisible by $5$ for every positive integer $n,
10n-4, 10n-2, 10n+2$.check numbers of the form $10i-3, 10i-1$ and $10i+3$ for primality.
prime-numbers divisibility greatest-common-divisor totient-function primality-test
edited Nov 21 '18 at 15:02
Klangen
1,65911334
asked Jan 2 '17 at 15:42
Nae
1115
|
show 2 more comments
I'm trying to generate odd prime numbers such that their totients are co-prime with $5$. I want to know if what I'm doing is correct, or can be proven.
- Assume $2, 3$ and $5$ are primes
- Check if $10i - 3, 10i - 1, 10i + 3$ for $i$ greater or equal to $1$
now I did the following to get those forms and I wonder if the approach is correct:
Assume every integer is in the form of $5n-2, 5n-1, 5n, 5n+1, 5n+2$
substitute each of them with $k$ in $2k+1,$ and get $10n-3, 10n-1, 10n+1, 10n+3, 10n+5$
remove the forms that don't form prime for n in positive integers
and have the forms reduced to $10n-3, 10n-1, 10n+1, 10n+3$totient of a prime $p$ is $p-1,$ so subtract $1$ from each of the forms
and see if they're divisible by $5$ for every positive integer $n,
10n-4, 10n-2, 10n+2$.check numbers of the form $10i-3, 10i-1$ and $10i+3$ for primality.
prime-numbers divisibility greatest-common-divisor totient-function primality-test
edited Nov 21 '18 at 15:02
Klangen
1,65911334
asked Jan 2 '17 at 15:42
Nae
1115
Your title doesn't match the body as it does not mention the totient. The title is easy to answer-any odd number that doesn't end in $5$. The totient of a prime $p$ is $p-1$, so you are asking for primes that are not $5k+1$. The easiest approach to this is just to pick numbers not $1 bmod 5$ and test for primality, but it appears you have other criteria.
– Ross Millikan
Jan 2 '17 at 15:46
@RossMillikan I updated the title, thanks. My cirteria are number to be odd, prime, and have its totient coprime with 5. So if there is a reduced for of numbers I'd appreciate that.
– Nae
Jan 2 '17 at 15:53
You are checking numbers that end in $7,9,1$ for primality, but not numbers that end in $3$. There is nothing wrong with those that end in $3$-as $13$ and $23$ seem to meet your requirement. I think you should just pick non $1 bmod 5$ numbers and check for primality. You can improve your odds by prefiltering for small factors, which is essentially what your ending digit does. You can pick $k bmod {210}$, for example with $k$ coprime to $2,3,5,7$ and generate numbers of the form $210m+k$
– Ross Millikan
Jan 2 '17 at 16:03
@RossMillikan I'm sorry, I've made a mistake on my last line, 10i+1 should be 10i+3 instead, and thus include 13, 23 etc. I don't understand the last part of your comment with k mod...
– Nae
Jan 2 '17 at 16:10
The effect of what you are doing is to preselect numbers that are not divisible by $2$ or $5$ and are not $1 bmod 5$. That will increase the density of acceptable primes in the numbers you test. You can do better still by making sure that the numbers you test are not multiples of $3$ or $7$ either, which is the point of my working $bmod {210}$
– Ross Millikan
Jan 2 '17 at 16:19
|
show 2 more comments
I'm trying to generate odd prime numbers such that their totients are co-prime with $5$. I want to know if what I'm doing is correct, or can be proven.
- Assume $2, 3$ and $5$ are primes
- Check if $10i - 3, 10i - 1, 10i + 3$ for $i$ greater or equal to $1$
now I did the following to get those forms and I wonder if the approach is correct:
Assume every integer is in the form of $5n-2, 5n-1, 5n, 5n+1, 5n+2$
substitute each of them with $k$ in $2k+1,$ and get $10n-3, 10n-1, 10n+1, 10n+3, 10n+5$
remove the forms that don't form prime for n in positive integers
and have the forms reduced to $10n-3, 10n-1, 10n+1, 10n+3$totient of a prime $p$ is $p-1,$ so subtract $1$ from each of the forms
and see if they're divisible by $5$ for every positive integer $n,
10n-4, 10n-2, 10n+2$.check numbers of the form $10i-3, 10i-1$ and $10i+3$ for primality.
prime-numbers divisibility greatest-common-divisor totient-function primality-test
edited Nov 21 '18 at 15:02
Klangen
1,65911334
asked Jan 2 '17 at 15:42
Nae
1115
I'm trying to generate odd prime numbers such that their totients are co-prime with $5$. I want to know if what I'm doing is correct, or can be proven.
- Assume $2, 3$ and $5$ are primes
- Check if $10i - 3, 10i - 1, 10i + 3$ for $i$ greater or equal to $1$
now I did the following to get those forms and I wonder if the approach is correct:
Assume every integer is in the form of $5n-2, 5n-1, 5n, 5n+1, 5n+2$
substitute each of them with $k$ in $2k+1,$ and get $10n-3, 10n-1, 10n+1, 10n+3, 10n+5$
remove the forms that don't form prime for n in positive integers
and have the forms reduced to $10n-3, 10n-1, 10n+1, 10n+3$totient of a prime $p$ is $p-1,$ so subtract $1$ from each of the forms
and see if they're divisible by $5$ for every positive integer $n,
10n-4, 10n-2, 10n+2$.check numbers of the form $10i-3, 10i-1$ and $10i+3$ for primality.
prime-numbers divisibility greatest-common-divisor totient-function primality-test
prime-numbers divisibility greatest-common-divisor totient-function primality-test
edited Nov 21 '18 at 15:02
Klangen
1,65911334
asked Jan 2 '17 at 15:42
Nae
1115
edited Nov 21 '18 at 15:02
Klangen
1,65911334
asked Jan 2 '17 at 15:42
Nae
1115
edited Nov 21 '18 at 15:02
Klangen
1,65911334
edited Nov 21 '18 at 15:02
Klangen
1,65911334
edited Nov 21 '18 at 15:02
Klangen
1,65911334
1,65911334
asked Jan 2 '17 at 15:42
Nae
1115
asked Jan 2 '17 at 15:42
Nae
1115
asked Jan 2 '17 at 15:42
Nae
1115
1115
Your title doesn't match the body as it does not mention the totient. The title is easy to answer-any odd number that doesn't end in $5$. The totient of a prime $p$ is $p-1$, so you are asking for primes that are not $5k+1$. The easiest approach to this is just to pick numbers not $1 bmod 5$ and test for primality, but it appears you have other criteria.
– Ross Millikan
Jan 2 '17 at 15:46
@RossMillikan I updated the title, thanks. My cirteria are number to be odd, prime, and have its totient coprime with 5. So if there is a reduced for of numbers I'd appreciate that.
– Nae
Jan 2 '17 at 15:53
You are checking numbers that end in $7,9,1$ for primality, but not numbers that end in $3$. There is nothing wrong with those that end in $3$-as $13$ and $23$ seem to meet your requirement. I think you should just pick non $1 bmod 5$ numbers and check for primality. You can improve your odds by prefiltering for small factors, which is essentially what your ending digit does. You can pick $k bmod {210}$, for example with $k$ coprime to $2,3,5,7$ and generate numbers of the form $210m+k$
– Ross Millikan
Jan 2 '17 at 16:03
@RossMillikan I'm sorry, I've made a mistake on my last line, 10i+1 should be 10i+3 instead, and thus include 13, 23 etc. I don't understand the last part of your comment with k mod...
– Nae
Jan 2 '17 at 16:10
The effect of what you are doing is to preselect numbers that are not divisible by $2$ or $5$ and are not $1 bmod 5$. That will increase the density of acceptable primes in the numbers you test. You can do better still by making sure that the numbers you test are not multiples of $3$ or $7$ either, which is the point of my working $bmod {210}$
– Ross Millikan
Jan 2 '17 at 16:19
|
show 2 more comments
Your title doesn't match the body as it does not mention the totient. The title is easy to answer-any odd number that doesn't end in $5$. The totient of a prime $p$ is $p-1$, so you are asking for primes that are not $5k+1$. The easiest approach to this is just to pick numbers not $1 bmod 5$ and test for primality, but it appears you have other criteria.
– Ross Millikan
Jan 2 '17 at 15:46
@RossMillikan I updated the title, thanks. My cirteria are number to be odd, prime, and have its totient coprime with 5. So if there is a reduced for of numbers I'd appreciate that.
– Nae
Jan 2 '17 at 15:53
You are checking numbers that end in $7,9,1$ for primality, but not numbers that end in $3$. There is nothing wrong with those that end in $3$-as $13$ and $23$ seem to meet your requirement. I think you should just pick non $1 bmod 5$ numbers and check for primality. You can improve your odds by prefiltering for small factors, which is essentially what your ending digit does. You can pick $k bmod {210}$, for example with $k$ coprime to $2,3,5,7$ and generate numbers of the form $210m+k$
– Ross Millikan
Jan 2 '17 at 16:03
@RossMillikan I'm sorry, I've made a mistake on my last line, 10i+1 should be 10i+3 instead, and thus include 13, 23 etc. I don't understand the last part of your comment with k mod...
– Nae
Jan 2 '17 at 16:10
The effect of what you are doing is to preselect numbers that are not divisible by $2$ or $5$ and are not $1 bmod 5$. That will increase the density of acceptable primes in the numbers you test. You can do better still by making sure that the numbers you test are not multiples of $3$ or $7$ either, which is the point of my working $bmod {210}$
– Ross Millikan
Jan 2 '17 at 16:19
Your title doesn't match the body as it does not mention the totient. The title is easy to answer-any odd number that doesn't end in $5$. The totient of a prime $p$ is $p-1$, so you are asking for primes that are not $5k+1$. The easiest approach to this is just to pick numbers not $1 bmod 5$ and test for primality, but it appears you have other criteria.
– Ross Millikan
Jan 2 '17 at 15:46
Your title doesn't match the body as it does not mention the totient. The title is easy to answer-any odd number that doesn't end in $5$. The totient of a prime $p$ is $p-1$, so you are asking for primes that are not $5k+1$. The easiest approach to this is just to pick numbers not $1 bmod 5$ and test for primality, but it appears you have other criteria.
– Ross Millikan
Jan 2 '17 at 15:46
@RossMillikan I updated the title, thanks. My cirteria are number to be odd, prime, and have its totient coprime with 5. So if there is a reduced for of numbers I'd appreciate that.
– Nae
Jan 2 '17 at 15:53
@RossMillikan I updated the title, thanks. My cirteria are number to be odd, prime, and have its totient coprime with 5. So if there is a reduced for of numbers I'd appreciate that.
– Nae
Jan 2 '17 at 15:53
You are checking numbers that end in $7,9,1$ for primality, but not numbers that end in $3$. There is nothing wrong with those that end in $3$-as $13$ and $23$ seem to meet your requirement. I think you should just pick non $1 bmod 5$ numbers and check for primality. You can improve your odds by prefiltering for small factors, which is essentially what your ending digit does. You can pick $k bmod {210}$, for example with $k$ coprime to $2,3,5,7$ and generate numbers of the form $210m+k$
– Ross Millikan
Jan 2 '17 at 16:03
You are checking numbers that end in $7,9,1$ for primality, but not numbers that end in $3$. There is nothing wrong with those that end in $3$-as $13$ and $23$ seem to meet your requirement. I think you should just pick non $1 bmod 5$ numbers and check for primality. You can improve your odds by prefiltering for small factors, which is essentially what your ending digit does. You can pick $k bmod {210}$, for example with $k$ coprime to $2,3,5,7$ and generate numbers of the form $210m+k$
– Ross Millikan
Jan 2 '17 at 16:03
@RossMillikan I'm sorry, I've made a mistake on my last line, 10i+1 should be 10i+3 instead, and thus include 13, 23 etc. I don't understand the last part of your comment with k mod...
– Nae
Jan 2 '17 at 16:10
@RossMillikan I'm sorry, I've made a mistake on my last line, 10i+1 should be 10i+3 instead, and thus include 13, 23 etc. I don't understand the last part of your comment with k mod...
– Nae
Jan 2 '17 at 16:10
The effect of what you are doing is to preselect numbers that are not divisible by $2$ or $5$ and are not $1 bmod 5$. That will increase the density of acceptable primes in the numbers you test. You can do better still by making sure that the numbers you test are not multiples of $3$ or $7$ either, which is the point of my working $bmod {210}$
– Ross Millikan
Jan 2 '17 at 16:19
The effect of what you are doing is to preselect numbers that are not divisible by $2$ or $5$ and are not $1 bmod 5$. That will increase the density of acceptable primes in the numbers you test. You can do better still by making sure that the numbers you test are not multiples of $3$ or $7$ either, which is the point of my working $bmod {210}$
– Ross Millikan
Jan 2 '17 at 16:19
|
show 2 more comments
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Your title doesn't match the body as it does not mention the totient. The title is easy to answer-any odd number that doesn't end in $5$. The totient of a prime $p$ is $p-1$, so you are asking for primes that are not $5k+1$. The easiest approach to this is just to pick numbers not $1 bmod 5$ and test for primality, but it appears you have other criteria.
– Ross Millikan
Jan 2 '17 at 15:46
@RossMillikan I updated the title, thanks. My cirteria are number to be odd, prime, and have its totient coprime with 5. So if there is a reduced for of numbers I'd appreciate that.
– Nae
Jan 2 '17 at 15:53
You are checking numbers that end in $7,9,1$ for primality, but not numbers that end in $3$. There is nothing wrong with those that end in $3$-as $13$ and $23$ seem to meet your requirement. I think you should just pick non $1 bmod 5$ numbers and check for primality. You can improve your odds by prefiltering for small factors, which is essentially what your ending digit does. You can pick $k bmod {210}$, for example with $k$ coprime to $2,3,5,7$ and generate numbers of the form $210m+k$
– Ross Millikan
Jan 2 '17 at 16:03
@RossMillikan I'm sorry, I've made a mistake on my last line, 10i+1 should be 10i+3 instead, and thus include 13, 23 etc. I don't understand the last part of your comment with k mod...
– Nae
Jan 2 '17 at 16:10
The effect of what you are doing is to preselect numbers that are not divisible by $2$ or $5$ and are not $1 bmod 5$. That will increase the density of acceptable primes in the numbers you test. You can do better still by making sure that the numbers you test are not multiples of $3$ or $7$ either, which is the point of my working $bmod {210}$
– Ross Millikan
Jan 2 '17 at 16:19