Mixing
Are there further primes of the form $varphi(n)^{varphi(varphi(n))}+1$?
$begingroup$
For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.
According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.
Do further primes $f(n)$ exist ?
number-theory elementary-number-theory prime-numbers totient-function
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
$endgroup$
|
show 6 more comments
$begingroup$
For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.
According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.
Do further primes $f(n)$ exist ?
number-theory elementary-number-theory prime-numbers totient-function
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
$endgroup$
1
$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22
2
$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24
1
$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19
1
$begingroup$
Then I triedSelect[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &]in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22
1
$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58
|
show 6 more comments
$begingroup$
For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.
According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.
Do further primes $f(n)$ exist ?
number-theory elementary-number-theory prime-numbers totient-function
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
$endgroup$
For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.
According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.
Do further primes $f(n)$ exist ?
number-theory elementary-number-theory prime-numbers totient-function
number-theory elementary-number-theory prime-numbers totient-function
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
asked Jul 26 '18 at 16:06
PeterPeter
47.3k1039128
47.3k1039128
1
$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22
2
$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24
1
$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19
1
$begingroup$
Then I triedSelect[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &]in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22
1
$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58
|
show 6 more comments
1
$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22
2
$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24
1
$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19
1
$begingroup$
Then I triedSelect[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &]in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22
1
$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58
1
1
$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22
$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22
2
2
$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24
$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24
1
1
$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19
$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19
1
1
$begingroup$
Then I tried
Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22
$begingroup$
Then I tried
Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22
1
1
$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58
$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58
|
show 6 more comments
1 Answer
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$begingroup$
This is mostly just a summary of what I have found that is to big for a comment.
Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
$$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
We also have that
$$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.
If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
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add a comment |
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1 Answer
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$begingroup$
This is mostly just a summary of what I have found that is to big for a comment.
Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
$$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
We also have that
$$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.
If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
$endgroup$
add a comment |
$begingroup$
This is mostly just a summary of what I have found that is to big for a comment.
Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
$$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
We also have that
$$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.
If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
$endgroup$
add a comment |
$begingroup$
This is mostly just a summary of what I have found that is to big for a comment.
Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
$$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
We also have that
$$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.
If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
$endgroup$
This is mostly just a summary of what I have found that is to big for a comment.
Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
$$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
We also have that
$$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.
If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
answered Jan 8 at 21:37
coDE_RPcoDE_RP
599
599
add a comment |
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1
$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22
2
$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24
1
$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19
1
$begingroup$
Then I tried
Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &]in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22
1
$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58