How long does it take for this sequence to obtain this loop?
$begingroup$
For positive integers $m,n$, define a sequence $S_m(n)$ so that $S_m(1)=m$, $S_m(n+1)=S_m(n)^2-1$ if $S_m(n)$ is prime, and $S_m(n+1)$ is the greatest prime factor of $S_m(n)$ otherwise. It is clear that, regardless of $m$, this sequence always gets caught in the infinite loop $2,3,8,2,3,dots$, because for any prime $p$, $p^2-1=(p+1)(p-1)$ and if $pneq 2$ then the greatest prime factor of this term is at most $lceil p/2rceil<p$.
What is less clear, is the rate at which the sequence becomes stuck into that loop. If we define $t(m)$ to be the smallest integer $n$ so that $S_m(n)=2$, then how does $t(m)$ grow? By some numerical testing I've found that $t(m)<20$ for all $m<10000$, which seems to suggest a logarithmic growth speed. In addition, for any positive integer $N$, is it always true that there exists an $m$ so that $t(m)>N$? Any thoughts are appreciated!
sequences-and-series elementary-number-theory
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
$endgroup$
add a comment |
$begingroup$
For positive integers $m,n$, define a sequence $S_m(n)$ so that $S_m(1)=m$, $S_m(n+1)=S_m(n)^2-1$ if $S_m(n)$ is prime, and $S_m(n+1)$ is the greatest prime factor of $S_m(n)$ otherwise. It is clear that, regardless of $m$, this sequence always gets caught in the infinite loop $2,3,8,2,3,dots$, because for any prime $p$, $p^2-1=(p+1)(p-1)$ and if $pneq 2$ then the greatest prime factor of this term is at most $lceil p/2rceil<p$.
What is less clear, is the rate at which the sequence becomes stuck into that loop. If we define $t(m)$ to be the smallest integer $n$ so that $S_m(n)=2$, then how does $t(m)$ grow? By some numerical testing I've found that $t(m)<20$ for all $m<10000$, which seems to suggest a logarithmic growth speed. In addition, for any positive integer $N$, is it always true that there exists an $m$ so that $t(m)>N$? Any thoughts are appreciated!
sequences-and-series elementary-number-theory
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
$endgroup$
add a comment |
$begingroup$
For positive integers $m,n$, define a sequence $S_m(n)$ so that $S_m(1)=m$, $S_m(n+1)=S_m(n)^2-1$ if $S_m(n)$ is prime, and $S_m(n+1)$ is the greatest prime factor of $S_m(n)$ otherwise. It is clear that, regardless of $m$, this sequence always gets caught in the infinite loop $2,3,8,2,3,dots$, because for any prime $p$, $p^2-1=(p+1)(p-1)$ and if $pneq 2$ then the greatest prime factor of this term is at most $lceil p/2rceil<p$.
What is less clear, is the rate at which the sequence becomes stuck into that loop. If we define $t(m)$ to be the smallest integer $n$ so that $S_m(n)=2$, then how does $t(m)$ grow? By some numerical testing I've found that $t(m)<20$ for all $m<10000$, which seems to suggest a logarithmic growth speed. In addition, for any positive integer $N$, is it always true that there exists an $m$ so that $t(m)>N$? Any thoughts are appreciated!
sequences-and-series elementary-number-theory
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
$endgroup$
For positive integers $m,n$, define a sequence $S_m(n)$ so that $S_m(1)=m$, $S_m(n+1)=S_m(n)^2-1$ if $S_m(n)$ is prime, and $S_m(n+1)$ is the greatest prime factor of $S_m(n)$ otherwise. It is clear that, regardless of $m$, this sequence always gets caught in the infinite loop $2,3,8,2,3,dots$, because for any prime $p$, $p^2-1=(p+1)(p-1)$ and if $pneq 2$ then the greatest prime factor of this term is at most $lceil p/2rceil<p$.
What is less clear, is the rate at which the sequence becomes stuck into that loop. If we define $t(m)$ to be the smallest integer $n$ so that $S_m(n)=2$, then how does $t(m)$ grow? By some numerical testing I've found that $t(m)<20$ for all $m<10000$, which seems to suggest a logarithmic growth speed. In addition, for any positive integer $N$, is it always true that there exists an $m$ so that $t(m)>N$? Any thoughts are appreciated!
sequences-and-series elementary-number-theory
sequences-and-series elementary-number-theory
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
asked Dec 20 '18 at 10:31
YiFanYiFan
3,9931627
3,9931627
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recording the following observation to get the ball rolling.
Assume that $ell>2$ is a Sophie Germain prime, in other words, a prime such that $p=2ell+1$ is also a prime. In that case
$$
(p+1)(p-1)=(2ell+2)cdot2ell=2^3cdotfrac{ell+1}2ell
$$
implying that $ell$ is the largest prime factor of $p^2-1$. So if $S_m(n)=p$ then $S_m(n+2)=ell$.
A Cunningham chain of length $k$ is a sequence of iterated Sophie Germain primes
$$p_1<p_2<p_3<cdots<p_k$$
such that $p_{i+1}=2p_i+1$ for all $i$. For example, $5<11<23<47$ is Cunningham sequence of length $4$. Iterating the argument of the previous paragraph tells us that if $S_m(n)=p_k$ for some $m,n$, then $S_m(n+2(k-1))=p_1$. This implies that if a Cunngham chain of length $k$ exists, then it takes about $2k$ steps for your sequence to come down to $p_1$.
According to the cited Wikipedia page it is an open conjucture that arbitrarily long Cunningham chains exist. In light of the above that conjecture would imply that the function $t$ is unbounded.
Of course, there is no need to have longer and longer Cunningham chains for $t$ to be unbounded. I just felt that this connection may be interesting. It is, of course, easy to show that the set of prime factors of numbers of the form $p^2-1$ is unbounded, but I couldn't see how unboundedness of $t$ would follow from such a simple fact.
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
1
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
add a comment |
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1 Answer
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$begingroup$
Recording the following observation to get the ball rolling.
Assume that $ell>2$ is a Sophie Germain prime, in other words, a prime such that $p=2ell+1$ is also a prime. In that case
$$
(p+1)(p-1)=(2ell+2)cdot2ell=2^3cdotfrac{ell+1}2ell
$$
implying that $ell$ is the largest prime factor of $p^2-1$. So if $S_m(n)=p$ then $S_m(n+2)=ell$.
A Cunningham chain of length $k$ is a sequence of iterated Sophie Germain primes
$$p_1<p_2<p_3<cdots<p_k$$
such that $p_{i+1}=2p_i+1$ for all $i$. For example, $5<11<23<47$ is Cunningham sequence of length $4$. Iterating the argument of the previous paragraph tells us that if $S_m(n)=p_k$ for some $m,n$, then $S_m(n+2(k-1))=p_1$. This implies that if a Cunngham chain of length $k$ exists, then it takes about $2k$ steps for your sequence to come down to $p_1$.
According to the cited Wikipedia page it is an open conjucture that arbitrarily long Cunningham chains exist. In light of the above that conjecture would imply that the function $t$ is unbounded.
Of course, there is no need to have longer and longer Cunningham chains for $t$ to be unbounded. I just felt that this connection may be interesting. It is, of course, easy to show that the set of prime factors of numbers of the form $p^2-1$ is unbounded, but I couldn't see how unboundedness of $t$ would follow from such a simple fact.
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
1
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
add a comment |
$begingroup$
Recording the following observation to get the ball rolling.
Assume that $ell>2$ is a Sophie Germain prime, in other words, a prime such that $p=2ell+1$ is also a prime. In that case
$$
(p+1)(p-1)=(2ell+2)cdot2ell=2^3cdotfrac{ell+1}2ell
$$
implying that $ell$ is the largest prime factor of $p^2-1$. So if $S_m(n)=p$ then $S_m(n+2)=ell$.
A Cunningham chain of length $k$ is a sequence of iterated Sophie Germain primes
$$p_1<p_2<p_3<cdots<p_k$$
such that $p_{i+1}=2p_i+1$ for all $i$. For example, $5<11<23<47$ is Cunningham sequence of length $4$. Iterating the argument of the previous paragraph tells us that if $S_m(n)=p_k$ for some $m,n$, then $S_m(n+2(k-1))=p_1$. This implies that if a Cunngham chain of length $k$ exists, then it takes about $2k$ steps for your sequence to come down to $p_1$.
According to the cited Wikipedia page it is an open conjucture that arbitrarily long Cunningham chains exist. In light of the above that conjecture would imply that the function $t$ is unbounded.
Of course, there is no need to have longer and longer Cunningham chains for $t$ to be unbounded. I just felt that this connection may be interesting. It is, of course, easy to show that the set of prime factors of numbers of the form $p^2-1$ is unbounded, but I couldn't see how unboundedness of $t$ would follow from such a simple fact.
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
1
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
add a comment |
$begingroup$
Recording the following observation to get the ball rolling.
Assume that $ell>2$ is a Sophie Germain prime, in other words, a prime such that $p=2ell+1$ is also a prime. In that case
$$
(p+1)(p-1)=(2ell+2)cdot2ell=2^3cdotfrac{ell+1}2ell
$$
implying that $ell$ is the largest prime factor of $p^2-1$. So if $S_m(n)=p$ then $S_m(n+2)=ell$.
A Cunningham chain of length $k$ is a sequence of iterated Sophie Germain primes
$$p_1<p_2<p_3<cdots<p_k$$
such that $p_{i+1}=2p_i+1$ for all $i$. For example, $5<11<23<47$ is Cunningham sequence of length $4$. Iterating the argument of the previous paragraph tells us that if $S_m(n)=p_k$ for some $m,n$, then $S_m(n+2(k-1))=p_1$. This implies that if a Cunngham chain of length $k$ exists, then it takes about $2k$ steps for your sequence to come down to $p_1$.
According to the cited Wikipedia page it is an open conjucture that arbitrarily long Cunningham chains exist. In light of the above that conjecture would imply that the function $t$ is unbounded.
Of course, there is no need to have longer and longer Cunningham chains for $t$ to be unbounded. I just felt that this connection may be interesting. It is, of course, easy to show that the set of prime factors of numbers of the form $p^2-1$ is unbounded, but I couldn't see how unboundedness of $t$ would follow from such a simple fact.
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
Recording the following observation to get the ball rolling.
Assume that $ell>2$ is a Sophie Germain prime, in other words, a prime such that $p=2ell+1$ is also a prime. In that case
$$
(p+1)(p-1)=(2ell+2)cdot2ell=2^3cdotfrac{ell+1}2ell
$$
implying that $ell$ is the largest prime factor of $p^2-1$. So if $S_m(n)=p$ then $S_m(n+2)=ell$.
A Cunningham chain of length $k$ is a sequence of iterated Sophie Germain primes
$$p_1<p_2<p_3<cdots<p_k$$
such that $p_{i+1}=2p_i+1$ for all $i$. For example, $5<11<23<47$ is Cunningham sequence of length $4$. Iterating the argument of the previous paragraph tells us that if $S_m(n)=p_k$ for some $m,n$, then $S_m(n+2(k-1))=p_1$. This implies that if a Cunngham chain of length $k$ exists, then it takes about $2k$ steps for your sequence to come down to $p_1$.
According to the cited Wikipedia page it is an open conjucture that arbitrarily long Cunningham chains exist. In light of the above that conjecture would imply that the function $t$ is unbounded.
Of course, there is no need to have longer and longer Cunningham chains for $t$ to be unbounded. I just felt that this connection may be interesting. It is, of course, easy to show that the set of prime factors of numbers of the form $p^2-1$ is unbounded, but I couldn't see how unboundedness of $t$ would follow from such a simple fact.
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
answered Dec 20 '18 at 14:43
Jyrki LahtonenJyrki Lahtonen
109k13169372
109k13169372
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
1
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
add a comment |
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
1
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
$begingroup$
Very interesting, thanks. I haven't thought of the connection to Cunningham chains before.
$endgroup$
– YiFan
Dec 21 '18 at 5:20
1
1
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
$begingroup$
To be honest, I'm a bit disappointed about my inability to prove that $t$ is undounded. It should (?) be possible to prove that a large enough class of primes appears as values of $S_m(3)$, inductively argue that the set of values of $S_m(2k+1)$ is not bounded for any $k$, but... no cigar.
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 5:36
add a comment |
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