Can we expect infinite many primes $p$ equal to the start of $frac(sqrtp)$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are there infinitely many primes of the form $kcdot 2^n +1$?Are there infinitely many primes $p$ such that $frac(p-1)! +1p$ is prime?Does $S(n)$ contain infinite many primes?Can it be proven that infinite many primes can be formed only using two distinct digits?How many of the primes can be subdivided into concatenations of smaller primes?Question about prime numbers starting with a given digits.Are there infinite many squares with a decimal expansion having no consecutive equal digits?Primes formed by concatenating the mersenne numbers from $2^2-1$ to $2^n-1$A race between ec-primes and mersenne-primes. Who will win in the long run?Can we prove that infinite many primes begin with any given digitstring?
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Can we expect infinite many primes $p$ equal to the start of $frac(sqrtp)$?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are there infinitely many primes of the form $kcdot 2^n +1$?Are there infinitely many primes $p$ such that $frac(p-1)! +1p$ is prime?Does $S(n)$ contain infinite many primes?Can it be proven that infinite many primes can be formed only using two distinct digits?How many of the primes can be subdivided into concatenations of smaller primes?Question about prime numbers starting with a given digits.Are there infinite many squares with a decimal expansion having no consecutive equal digits?Primes formed by concatenating the mersenne numbers from $2^2-1$ to $2^n-1$A race between ec-primes and mersenne-primes. Who will win in the long run?Can we prove that infinite many primes begin with any given digitstring?
0
$begingroup$
The following routine searches for prime numbers $ p $ equal to the beginning of the fractional part of the decimal expansion of $ sqrtp $ :
? forprime(p=1,4*10^9,s=length(digits(p));x=frac(sqrt(p));if(truncate(x*10^s)==p
,print1(p," ")))
5711 8053139
?
- Can we expect infinite many primes with the given property ?
- What is the next prime number with the given property ?
number-theory elementary-number-theory prime-numbers decimal-expansion
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
$endgroup$
add a comment |
0
$begingroup$
The following routine searches for prime numbers $ p $ equal to the beginning of the fractional part of the decimal expansion of $ sqrtp $ :
? forprime(p=1,4*10^9,s=length(digits(p));x=frac(sqrt(p));if(truncate(x*10^s)==p
,print1(p," ")))
5711 8053139
?
- Can we expect infinite many primes with the given property ?
- What is the next prime number with the given property ?
number-theory elementary-number-theory prime-numbers decimal-expansion
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
$endgroup$
1
$begingroup$
The first question is unlikely to get a solid answer.
$endgroup$
– Cameron Buie
Mar 31 at 14:35
1
$begingroup$
You are supposed to understand $lfloor frac10^ssqrtp rfloor = p$ means $p^3/2 = 10^s +O(sqrtp)$. So this is implied by most prime gap conjectures.
$endgroup$
– reuns
Mar 31 at 16:07
$begingroup$
I do not expect a proof for finite or infinite many primes, a heuristic (like in the case of the Wieferich primes) would be utterly sufficient.
$endgroup$
– Peter
Mar 31 at 16:09
2
$begingroup$
For now the answer is "we don't know" and "yes" if you replace $sqrtp$ by $p^theta$ for some $theta > 0.525$.
$endgroup$
– reuns
Mar 31 at 16:11
1
$begingroup$
Another downvote without any reason ...
$endgroup$
– Peter
Apr 1 at 7:55
add a comment |
0
0
0
1
$begingroup$
The following routine searches for prime numbers $ p $ equal to the beginning of the fractional part of the decimal expansion of $ sqrtp $ :
? forprime(p=1,4*10^9,s=length(digits(p));x=frac(sqrt(p));if(truncate(x*10^s)==p
,print1(p," ")))
5711 8053139
?
- Can we expect infinite many primes with the given property ?
- What is the next prime number with the given property ?
number-theory elementary-number-theory prime-numbers decimal-expansion
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
$endgroup$
The following routine searches for prime numbers $ p $ equal to the beginning of the fractional part of the decimal expansion of $ sqrtp $ :
? forprime(p=1,4*10^9,s=length(digits(p));x=frac(sqrt(p));if(truncate(x*10^s)==p
,print1(p," ")))
5711 8053139
?
- Can we expect infinite many primes with the given property ?
- What is the next prime number with the given property ?
number-theory elementary-number-theory prime-numbers decimal-expansion
number-theory elementary-number-theory prime-numbers decimal-expansion
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
asked Mar 31 at 14:26
PeterPeter
49.2k1240138
49.2k1240138
1
$begingroup$
The first question is unlikely to get a solid answer.
$endgroup$
– Cameron Buie
Mar 31 at 14:35
1
$begingroup$
You are supposed to understand $lfloor frac10^ssqrtp rfloor = p$ means $p^3/2 = 10^s +O(sqrtp)$. So this is implied by most prime gap conjectures.
$endgroup$
– reuns
Mar 31 at 16:07
$begingroup$
I do not expect a proof for finite or infinite many primes, a heuristic (like in the case of the Wieferich primes) would be utterly sufficient.
$endgroup$
– Peter
Mar 31 at 16:09
2
$begingroup$
For now the answer is "we don't know" and "yes" if you replace $sqrtp$ by $p^theta$ for some $theta > 0.525$.
$endgroup$
– reuns
Mar 31 at 16:11
1
$begingroup$
Another downvote without any reason ...
$endgroup$
– Peter
Apr 1 at 7:55
add a comment |
1
$begingroup$
The first question is unlikely to get a solid answer.
$endgroup$
– Cameron Buie
Mar 31 at 14:35
1
$begingroup$
You are supposed to understand $lfloor frac10^ssqrtp rfloor = p$ means $p^3/2 = 10^s +O(sqrtp)$. So this is implied by most prime gap conjectures.
$endgroup$
– reuns
Mar 31 at 16:07
$begingroup$
I do not expect a proof for finite or infinite many primes, a heuristic (like in the case of the Wieferich primes) would be utterly sufficient.
$endgroup$
– Peter
Mar 31 at 16:09
2
$begingroup$
For now the answer is "we don't know" and "yes" if you replace $sqrtp$ by $p^theta$ for some $theta > 0.525$.
$endgroup$
– reuns
Mar 31 at 16:11
1
$begingroup$
Another downvote without any reason ...
$endgroup$
– Peter
Apr 1 at 7:55
1
1
$begingroup$
The first question is unlikely to get a solid answer.
$endgroup$
– Cameron Buie
Mar 31 at 14:35
$begingroup$
The first question is unlikely to get a solid answer.
$endgroup$
– Cameron Buie
Mar 31 at 14:35
1
1
$begingroup$
You are supposed to understand $lfloor frac10^ssqrtp rfloor = p$ means $p^3/2 = 10^s +O(sqrtp)$. So this is implied by most prime gap conjectures.
$endgroup$
– reuns
Mar 31 at 16:07
$begingroup$
You are supposed to understand $lfloor frac10^ssqrtp rfloor = p$ means $p^3/2 = 10^s +O(sqrtp)$. So this is implied by most prime gap conjectures.
$endgroup$
– reuns
Mar 31 at 16:07
$begingroup$
I do not expect a proof for finite or infinite many primes, a heuristic (like in the case of the Wieferich primes) would be utterly sufficient.
$endgroup$
– Peter
Mar 31 at 16:09
$begingroup$
I do not expect a proof for finite or infinite many primes, a heuristic (like in the case of the Wieferich primes) would be utterly sufficient.
$endgroup$
– Peter
Mar 31 at 16:09
2
2
$begingroup$
For now the answer is "we don't know" and "yes" if you replace $sqrtp$ by $p^theta$ for some $theta > 0.525$.
$endgroup$
– reuns
Mar 31 at 16:11
$begingroup$
For now the answer is "we don't know" and "yes" if you replace $sqrtp$ by $p^theta$ for some $theta > 0.525$.
$endgroup$
– reuns
Mar 31 at 16:11
1
1
$begingroup$
Another downvote without any reason ...
$endgroup$
– Peter
Apr 1 at 7:55
$begingroup$
Another downvote without any reason ...
$endgroup$
– Peter
Apr 1 at 7:55
add a comment |
0
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1
$begingroup$
The first question is unlikely to get a solid answer.
$endgroup$
– Cameron Buie
Mar 31 at 14:35
1
$begingroup$
You are supposed to understand $lfloor frac10^ssqrtp rfloor = p$ means $p^3/2 = 10^s +O(sqrtp)$. So this is implied by most prime gap conjectures.
$endgroup$
– reuns
Mar 31 at 16:07
$begingroup$
I do not expect a proof for finite or infinite many primes, a heuristic (like in the case of the Wieferich primes) would be utterly sufficient.
$endgroup$
– Peter
Mar 31 at 16:09
2
$begingroup$
For now the answer is "we don't know" and "yes" if you replace $sqrtp$ by $p^theta$ for some $theta > 0.525$.
$endgroup$
– reuns
Mar 31 at 16:11
1
$begingroup$
Another downvote without any reason ...
$endgroup$
– Peter
Apr 1 at 7:55