A recreational problem The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manaraprime factors of numbers formed by primorialsAll the small primes close together yet againSimple quadratic, crazy question part 2Can every odd prime $pne 11$ be the smallest prime factor of a carmichael-number with $3$ prime factors?Is the product of consecutive primes in $(a, b)[n]$ $=$ $1$ $pmod ab$?Pythagorean triples that “survive” Euler's totient functionA question about a certain type of primesPrimes of the form $p^2+p+41$

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A recreational problem



The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manaraprime factors of numbers formed by primorialsAll the small primes close together yet againSimple quadratic, crazy question part 2Can every odd prime $pne 11$ be the smallest prime factor of a carmichael-number with $3$ prime factors?Is the product of consecutive primes in $(a, b)[n]$ $=$ $1$ $pmod ab$?Pythagorean triples that “survive” Euler's totient functionA question about a certain type of primesPrimes of the form $p^2+p+41$










1












$begingroup$


Let be $rad(n)$ the radical of a integer $n$. So for $n=2^2cdot3^4cdot7^11$, $rad(n)=2cdot3cdot7$, you take the prime factors without considering the exponents.



Now for $n=6$, you have $rad(6!+1)=7cdot103$. If you concatenate base ten from the smallest to the greatest prime factor, you get $7||103$=$7103$, which is a prime.



Now consider $rad(n!+1)$, it must have at least two distinct prime factors.
Are there other primes other than $7103$, which are the concatenation base ten of the prime factors (from the smallest to the greatest) of $rad(n!+1)$?



Other examples found with Pari: for $n=19$ and $n=23$.



If you consider $rad(n!-1)$ instead I did not yet found up to prime $67$ an $n$ which is prime and does the job!



The solutions for $rad(n!-1)$ so far found are for $n=15,21,22,25$ which are all semi-primes.










share|cite|improve this question











$endgroup$











  • $begingroup$
    @Arthur can you find other examples?
    $endgroup$
    – user660792
    Apr 8 at 8:06






  • 2




    $begingroup$
    I don't understand your claim, "it must have at least two distinct prime factors". Lots of numbers of the form $n!+1$ are prime: oeis.org/A002981
    $endgroup$
    – Ethan MacBrough
    Apr 8 at 8:10






  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Apr 8 at 8:12






  • 1




    $begingroup$
    @Ethan MacBrough n!+1 must not be prime otherwise you cannot concatenate at least two elemets
    $endgroup$
    – user660792
    Apr 8 at 8:15






  • 1




    $begingroup$
    $n=19$ gives the prime $711713311273363831$.
    $endgroup$
    – Chrystomath
    Apr 8 at 10:29















1












$begingroup$


Let be $rad(n)$ the radical of a integer $n$. So for $n=2^2cdot3^4cdot7^11$, $rad(n)=2cdot3cdot7$, you take the prime factors without considering the exponents.



Now for $n=6$, you have $rad(6!+1)=7cdot103$. If you concatenate base ten from the smallest to the greatest prime factor, you get $7||103$=$7103$, which is a prime.



Now consider $rad(n!+1)$, it must have at least two distinct prime factors.
Are there other primes other than $7103$, which are the concatenation base ten of the prime factors (from the smallest to the greatest) of $rad(n!+1)$?



Other examples found with Pari: for $n=19$ and $n=23$.



If you consider $rad(n!-1)$ instead I did not yet found up to prime $67$ an $n$ which is prime and does the job!



The solutions for $rad(n!-1)$ so far found are for $n=15,21,22,25$ which are all semi-primes.










share|cite|improve this question











$endgroup$











  • $begingroup$
    @Arthur can you find other examples?
    $endgroup$
    – user660792
    Apr 8 at 8:06






  • 2




    $begingroup$
    I don't understand your claim, "it must have at least two distinct prime factors". Lots of numbers of the form $n!+1$ are prime: oeis.org/A002981
    $endgroup$
    – Ethan MacBrough
    Apr 8 at 8:10






  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Apr 8 at 8:12






  • 1




    $begingroup$
    @Ethan MacBrough n!+1 must not be prime otherwise you cannot concatenate at least two elemets
    $endgroup$
    – user660792
    Apr 8 at 8:15






  • 1




    $begingroup$
    $n=19$ gives the prime $711713311273363831$.
    $endgroup$
    – Chrystomath
    Apr 8 at 10:29













1












1








1





$begingroup$


Let be $rad(n)$ the radical of a integer $n$. So for $n=2^2cdot3^4cdot7^11$, $rad(n)=2cdot3cdot7$, you take the prime factors without considering the exponents.



Now for $n=6$, you have $rad(6!+1)=7cdot103$. If you concatenate base ten from the smallest to the greatest prime factor, you get $7||103$=$7103$, which is a prime.



Now consider $rad(n!+1)$, it must have at least two distinct prime factors.
Are there other primes other than $7103$, which are the concatenation base ten of the prime factors (from the smallest to the greatest) of $rad(n!+1)$?



Other examples found with Pari: for $n=19$ and $n=23$.



If you consider $rad(n!-1)$ instead I did not yet found up to prime $67$ an $n$ which is prime and does the job!



The solutions for $rad(n!-1)$ so far found are for $n=15,21,22,25$ which are all semi-primes.










share|cite|improve this question











$endgroup$




Let be $rad(n)$ the radical of a integer $n$. So for $n=2^2cdot3^4cdot7^11$, $rad(n)=2cdot3cdot7$, you take the prime factors without considering the exponents.



Now for $n=6$, you have $rad(6!+1)=7cdot103$. If you concatenate base ten from the smallest to the greatest prime factor, you get $7||103$=$7103$, which is a prime.



Now consider $rad(n!+1)$, it must have at least two distinct prime factors.
Are there other primes other than $7103$, which are the concatenation base ten of the prime factors (from the smallest to the greatest) of $rad(n!+1)$?



Other examples found with Pari: for $n=19$ and $n=23$.



If you consider $rad(n!-1)$ instead I did not yet found up to prime $67$ an $n$ which is prime and does the job!



The solutions for $rad(n!-1)$ so far found are for $n=15,21,22,25$ which are all semi-primes.







number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 8 at 10:18

























asked Apr 8 at 8:03







user660792


















  • $begingroup$
    @Arthur can you find other examples?
    $endgroup$
    – user660792
    Apr 8 at 8:06






  • 2




    $begingroup$
    I don't understand your claim, "it must have at least two distinct prime factors". Lots of numbers of the form $n!+1$ are prime: oeis.org/A002981
    $endgroup$
    – Ethan MacBrough
    Apr 8 at 8:10






  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Apr 8 at 8:12






  • 1




    $begingroup$
    @Ethan MacBrough n!+1 must not be prime otherwise you cannot concatenate at least two elemets
    $endgroup$
    – user660792
    Apr 8 at 8:15






  • 1




    $begingroup$
    $n=19$ gives the prime $711713311273363831$.
    $endgroup$
    – Chrystomath
    Apr 8 at 10:29
















  • $begingroup$
    @Arthur can you find other examples?
    $endgroup$
    – user660792
    Apr 8 at 8:06






  • 2




    $begingroup$
    I don't understand your claim, "it must have at least two distinct prime factors". Lots of numbers of the form $n!+1$ are prime: oeis.org/A002981
    $endgroup$
    – Ethan MacBrough
    Apr 8 at 8:10






  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Apr 8 at 8:12






  • 1




    $begingroup$
    @Ethan MacBrough n!+1 must not be prime otherwise you cannot concatenate at least two elemets
    $endgroup$
    – user660792
    Apr 8 at 8:15






  • 1




    $begingroup$
    $n=19$ gives the prime $711713311273363831$.
    $endgroup$
    – Chrystomath
    Apr 8 at 10:29















$begingroup$
@Arthur can you find other examples?
$endgroup$
– user660792
Apr 8 at 8:06




$begingroup$
@Arthur can you find other examples?
$endgroup$
– user660792
Apr 8 at 8:06




2




2




$begingroup$
I don't understand your claim, "it must have at least two distinct prime factors". Lots of numbers of the form $n!+1$ are prime: oeis.org/A002981
$endgroup$
– Ethan MacBrough
Apr 8 at 8:10




$begingroup$
I don't understand your claim, "it must have at least two distinct prime factors". Lots of numbers of the form $n!+1$ are prime: oeis.org/A002981
$endgroup$
– Ethan MacBrough
Apr 8 at 8:10




1




1




$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Apr 8 at 8:12




$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Apr 8 at 8:12




1




1




$begingroup$
@Ethan MacBrough n!+1 must not be prime otherwise you cannot concatenate at least two elemets
$endgroup$
– user660792
Apr 8 at 8:15




$begingroup$
@Ethan MacBrough n!+1 must not be prime otherwise you cannot concatenate at least two elemets
$endgroup$
– user660792
Apr 8 at 8:15




1




1




$begingroup$
$n=19$ gives the prime $711713311273363831$.
$endgroup$
– Chrystomath
Apr 8 at 10:29




$begingroup$
$n=19$ gives the prime $711713311273363831$.
$endgroup$
– Chrystomath
Apr 8 at 10:29










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