Mixing
For which primes does this iterated function act transitively? (Sort of a finite analogue of Collatz conjecture.)
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Background: I was trying to prove something having to do with cyclic group actions on matroids and was able to show that what I want holds if a particular elementary-looking number-theoretic property holds, but upon closer inspection it holds for some primes and not others and now I am wondering if this "elementary" problem is known/studied somewhere or perhaps it is an open problem--I have no idea since I only stumbled into it entirely accidentally. Even learning what area it should be considered, what potential tools there are for studying it, and what else it might be related to (even just keywords to search for!) would be helpful.
The problem statement: Let $p$ be an odd prime and consider the permutation of the set $1,ldots,fracp-12$ defined by sending $n$ to $fracn2$ if $n$ is even and sending it to $fracp-n2$ if $n$ is odd. For which primes $p$ does this permutation have a single orbit?
Here are some examples of iterating this function:
$p=7$) $1 mapsto 3 mapsto 2 mapsto 1$, so it is transitive.
$p=11$) $1 mapsto 5 mapsto 3 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=13$) $1 mapsto 6 mapsto 3 mapsto 5 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=17$) $1 mapsto 8 mapsto 4 mapsto 2 mapsto 1$: here there are two orbits of size 4.
This reminds at least loosely in spirit of the Collatz conjecture but I presume (and hope!) it is much simpler since for each $p$ there are only finitely many numbers to consider. We tried some computer experiments and found lots of primes for which there is a single orbit and also lots for which there is not; we put the sequence of "good" primes in the OEIS and did not find any matches, which makes me suspect there is not something simple like a congruence condition on the prime, but I don't know any number theory so I really have no idea.
nt.number-theory prime-numbers permutations collatz-conjecture
asked 5 hours ago
Noah Giansiracusa
7343711
add a comment |Â
up vote
5
down vote
favorite
Background: I was trying to prove something having to do with cyclic group actions on matroids and was able to show that what I want holds if a particular elementary-looking number-theoretic property holds, but upon closer inspection it holds for some primes and not others and now I am wondering if this "elementary" problem is known/studied somewhere or perhaps it is an open problem--I have no idea since I only stumbled into it entirely accidentally. Even learning what area it should be considered, what potential tools there are for studying it, and what else it might be related to (even just keywords to search for!) would be helpful.
The problem statement: Let $p$ be an odd prime and consider the permutation of the set $1,ldots,fracp-12$ defined by sending $n$ to $fracn2$ if $n$ is even and sending it to $fracp-n2$ if $n$ is odd. For which primes $p$ does this permutation have a single orbit?
Here are some examples of iterating this function:
$p=7$) $1 mapsto 3 mapsto 2 mapsto 1$, so it is transitive.
$p=11$) $1 mapsto 5 mapsto 3 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=13$) $1 mapsto 6 mapsto 3 mapsto 5 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=17$) $1 mapsto 8 mapsto 4 mapsto 2 mapsto 1$: here there are two orbits of size 4.
This reminds at least loosely in spirit of the Collatz conjecture but I presume (and hope!) it is much simpler since for each $p$ there are only finitely many numbers to consider. We tried some computer experiments and found lots of primes for which there is a single orbit and also lots for which there is not; we put the sequence of "good" primes in the OEIS and did not find any matches, which makes me suspect there is not something simple like a congruence condition on the prime, but I don't know any number theory so I really have no idea.
nt.number-theory prime-numbers permutations collatz-conjecture
asked 5 hours ago
Noah Giansiracusa
7343711
2
Reverse it. You are finding primes where 2 is (is not) a primitive root mod p. Gerhard "Now You Search The Literature" Paseman, 2018.10.11.
– Gerhard Paseman
5 hours ago
Upon reflection (pun intended) it is a little more complicated. I still think primitive root applies. Gerhard "Not Ready For Palindromic Signature" Paseman, 2018.10.11.
– Gerhard Paseman
4 hours ago
@GerhardPaseman: already the example $p = 7$ shows that this is not what's going on. The function acts transitively modulo $7$, but $2$ is obviously not a primitive root.
– R. van Dobben de Bruyn
4 hours ago
1
The length of the orbit is the order of 2 modulo p if that order is odd, otherwise it's half the order.
– Felipe Voloch
4 hours ago
Link to the sequence in OEIS?
– Gerry Myerson
21 mins ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Background: I was trying to prove something having to do with cyclic group actions on matroids and was able to show that what I want holds if a particular elementary-looking number-theoretic property holds, but upon closer inspection it holds for some primes and not others and now I am wondering if this "elementary" problem is known/studied somewhere or perhaps it is an open problem--I have no idea since I only stumbled into it entirely accidentally. Even learning what area it should be considered, what potential tools there are for studying it, and what else it might be related to (even just keywords to search for!) would be helpful.
The problem statement: Let $p$ be an odd prime and consider the permutation of the set $1,ldots,fracp-12$ defined by sending $n$ to $fracn2$ if $n$ is even and sending it to $fracp-n2$ if $n$ is odd. For which primes $p$ does this permutation have a single orbit?
Here are some examples of iterating this function:
$p=7$) $1 mapsto 3 mapsto 2 mapsto 1$, so it is transitive.
$p=11$) $1 mapsto 5 mapsto 3 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=13$) $1 mapsto 6 mapsto 3 mapsto 5 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=17$) $1 mapsto 8 mapsto 4 mapsto 2 mapsto 1$: here there are two orbits of size 4.
This reminds at least loosely in spirit of the Collatz conjecture but I presume (and hope!) it is much simpler since for each $p$ there are only finitely many numbers to consider. We tried some computer experiments and found lots of primes for which there is a single orbit and also lots for which there is not; we put the sequence of "good" primes in the OEIS and did not find any matches, which makes me suspect there is not something simple like a congruence condition on the prime, but I don't know any number theory so I really have no idea.
nt.number-theory prime-numbers permutations collatz-conjecture
asked 5 hours ago
Noah Giansiracusa
7343711
Background: I was trying to prove something having to do with cyclic group actions on matroids and was able to show that what I want holds if a particular elementary-looking number-theoretic property holds, but upon closer inspection it holds for some primes and not others and now I am wondering if this "elementary" problem is known/studied somewhere or perhaps it is an open problem--I have no idea since I only stumbled into it entirely accidentally. Even learning what area it should be considered, what potential tools there are for studying it, and what else it might be related to (even just keywords to search for!) would be helpful.
The problem statement: Let $p$ be an odd prime and consider the permutation of the set $1,ldots,fracp-12$ defined by sending $n$ to $fracn2$ if $n$ is even and sending it to $fracp-n2$ if $n$ is odd. For which primes $p$ does this permutation have a single orbit?
Here are some examples of iterating this function:
$p=7$) $1 mapsto 3 mapsto 2 mapsto 1$, so it is transitive.
$p=11$) $1 mapsto 5 mapsto 3 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=13$) $1 mapsto 6 mapsto 3 mapsto 5 mapsto 4 mapsto 2 mapsto 1$, so it is transitive.
$p=17$) $1 mapsto 8 mapsto 4 mapsto 2 mapsto 1$: here there are two orbits of size 4.
This reminds at least loosely in spirit of the Collatz conjecture but I presume (and hope!) it is much simpler since for each $p$ there are only finitely many numbers to consider. We tried some computer experiments and found lots of primes for which there is a single orbit and also lots for which there is not; we put the sequence of "good" primes in the OEIS and did not find any matches, which makes me suspect there is not something simple like a congruence condition on the prime, but I don't know any number theory so I really have no idea.
nt.number-theory prime-numbers permutations collatz-conjecture
nt.number-theory prime-numbers permutations collatz-conjecture
asked 5 hours ago
Noah Giansiracusa
7343711
asked 5 hours ago
Noah Giansiracusa
7343711
asked 5 hours ago
Noah Giansiracusa
7343711
asked 5 hours ago
Noah Giansiracusa
7343711
asked 5 hours ago
Noah Giansiracusa
7343711
7343711
2
Reverse it. You are finding primes where 2 is (is not) a primitive root mod p. Gerhard "Now You Search The Literature" Paseman, 2018.10.11.
– Gerhard Paseman
5 hours ago
Upon reflection (pun intended) it is a little more complicated. I still think primitive root applies. Gerhard "Not Ready For Palindromic Signature" Paseman, 2018.10.11.
– Gerhard Paseman
4 hours ago
@GerhardPaseman: already the example $p = 7$ shows that this is not what's going on. The function acts transitively modulo $7$, but $2$ is obviously not a primitive root.
– R. van Dobben de Bruyn
4 hours ago
1
The length of the orbit is the order of 2 modulo p if that order is odd, otherwise it's half the order.
– Felipe Voloch
4 hours ago
Link to the sequence in OEIS?
– Gerry Myerson
21 mins ago
add a comment |Â
2
Reverse it. You are finding primes where 2 is (is not) a primitive root mod p. Gerhard "Now You Search The Literature" Paseman, 2018.10.11.
– Gerhard Paseman
5 hours ago
Upon reflection (pun intended) it is a little more complicated. I still think primitive root applies. Gerhard "Not Ready For Palindromic Signature" Paseman, 2018.10.11.
– Gerhard Paseman
4 hours ago
@GerhardPaseman: already the example $p = 7$ shows that this is not what's going on. The function acts transitively modulo $7$, but $2$ is obviously not a primitive root.
– R. van Dobben de Bruyn
4 hours ago
1
The length of the orbit is the order of 2 modulo p if that order is odd, otherwise it's half the order.
– Felipe Voloch
4 hours ago
Link to the sequence in OEIS?
– Gerry Myerson
21 mins ago
2
2
Reverse it. You are finding primes where 2 is (is not) a primitive root mod p. Gerhard "Now You Search The Literature" Paseman, 2018.10.11.
– Gerhard Paseman
5 hours ago
Reverse it. You are finding primes where 2 is (is not) a primitive root mod p. Gerhard "Now You Search The Literature" Paseman, 2018.10.11.
– Gerhard Paseman
5 hours ago
Upon reflection (pun intended) it is a little more complicated. I still think primitive root applies. Gerhard "Not Ready For Palindromic Signature" Paseman, 2018.10.11.
– Gerhard Paseman
4 hours ago
Upon reflection (pun intended) it is a little more complicated. I still think primitive root applies. Gerhard "Not Ready For Palindromic Signature" Paseman, 2018.10.11.
– Gerhard Paseman
4 hours ago
@GerhardPaseman: already the example $p = 7$ shows that this is not what's going on. The function acts transitively modulo $7$, but $2$ is obviously not a primitive root.
– R. van Dobben de Bruyn
4 hours ago
@GerhardPaseman: already the example $p = 7$ shows that this is not what's going on. The function acts transitively modulo $7$, but $2$ is obviously not a primitive root.
– R. van Dobben de Bruyn
4 hours ago
1
1
The length of the orbit is the order of 2 modulo p if that order is odd, otherwise it's half the order.
– Felipe Voloch
4 hours ago
The length of the orbit is the order of 2 modulo p if that order is odd, otherwise it's half the order.
– Felipe Voloch
4 hours ago
Link to the sequence in OEIS?
– Gerry Myerson
21 mins ago
Link to the sequence in OEIS?
– Gerry Myerson
21 mins ago
add a comment |Â
1 Answer
1
active
oldest
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up vote
4
down vote
It's easier to work with the inverse permutation. Starting from $1$ and working backwards we end up with a sequence $1=2^0to pm 2^1to pm 2^2to cdots topm 2^fracp-12=1$. Where the signs are uniquely determined so that $pm 2^rin 1,2,dots,fracp-12$. This cycle covers all numbers in $1,2,dots,fracp-12$ exactly once if and only if $2^rneq pm 1pmodp$ for any $1le rle fracp-32$.
Proposition: This happens precisely when either $2$ is a primitive root for $p$, or if $p=7pmod 8$ and $2$ has order $fracp-12$.
Proof: In the situation that $2^fracp-12=-1pmodp$ we have that $2^r$ for $r=1,2,cdots, p-1$ forms a complete system of residues, therefore $2$ is a primitive root for $p$. We are left with the case where $2^fracp-12=1pmodp$. This means that $2$ has order $fracp-12$ in $mathbb Z/pmathbb Z$ and moreover $-1$ is not of the form $2^rpmodp$. In turn, this implies $2$ is the square of a primitive root, and so $2^r$ covers all quadratic residues which means $-1$ is not a quadratic residue. Since $left(frac2pright)=1$ and $left(frac-1pright)=-1$ we have $p=7pmod 8$.
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It's easier to work with the inverse permutation. Starting from $1$ and working backwards we end up with a sequence $1=2^0to pm 2^1to pm 2^2to cdots topm 2^fracp-12=1$. Where the signs are uniquely determined so that $pm 2^rin 1,2,dots,fracp-12$. This cycle covers all numbers in $1,2,dots,fracp-12$ exactly once if and only if $2^rneq pm 1pmodp$ for any $1le rle fracp-32$.
Proposition: This happens precisely when either $2$ is a primitive root for $p$, or if $p=7pmod 8$ and $2$ has order $fracp-12$.
Proof: In the situation that $2^fracp-12=-1pmodp$ we have that $2^r$ for $r=1,2,cdots, p-1$ forms a complete system of residues, therefore $2$ is a primitive root for $p$. We are left with the case where $2^fracp-12=1pmodp$. This means that $2$ has order $fracp-12$ in $mathbb Z/pmathbb Z$ and moreover $-1$ is not of the form $2^rpmodp$. In turn, this implies $2$ is the square of a primitive root, and so $2^r$ covers all quadratic residues which means $-1$ is not a quadratic residue. Since $left(frac2pright)=1$ and $left(frac-1pright)=-1$ we have $p=7pmod 8$.
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
add a comment |Â
up vote
4
down vote
It's easier to work with the inverse permutation. Starting from $1$ and working backwards we end up with a sequence $1=2^0to pm 2^1to pm 2^2to cdots topm 2^fracp-12=1$. Where the signs are uniquely determined so that $pm 2^rin 1,2,dots,fracp-12$. This cycle covers all numbers in $1,2,dots,fracp-12$ exactly once if and only if $2^rneq pm 1pmodp$ for any $1le rle fracp-32$.
Proposition: This happens precisely when either $2$ is a primitive root for $p$, or if $p=7pmod 8$ and $2$ has order $fracp-12$.
Proof: In the situation that $2^fracp-12=-1pmodp$ we have that $2^r$ for $r=1,2,cdots, p-1$ forms a complete system of residues, therefore $2$ is a primitive root for $p$. We are left with the case where $2^fracp-12=1pmodp$. This means that $2$ has order $fracp-12$ in $mathbb Z/pmathbb Z$ and moreover $-1$ is not of the form $2^rpmodp$. In turn, this implies $2$ is the square of a primitive root, and so $2^r$ covers all quadratic residues which means $-1$ is not a quadratic residue. Since $left(frac2pright)=1$ and $left(frac-1pright)=-1$ we have $p=7pmod 8$.
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It's easier to work with the inverse permutation. Starting from $1$ and working backwards we end up with a sequence $1=2^0to pm 2^1to pm 2^2to cdots topm 2^fracp-12=1$. Where the signs are uniquely determined so that $pm 2^rin 1,2,dots,fracp-12$. This cycle covers all numbers in $1,2,dots,fracp-12$ exactly once if and only if $2^rneq pm 1pmodp$ for any $1le rle fracp-32$.
Proposition: This happens precisely when either $2$ is a primitive root for $p$, or if $p=7pmod 8$ and $2$ has order $fracp-12$.
Proof: In the situation that $2^fracp-12=-1pmodp$ we have that $2^r$ for $r=1,2,cdots, p-1$ forms a complete system of residues, therefore $2$ is a primitive root for $p$. We are left with the case where $2^fracp-12=1pmodp$. This means that $2$ has order $fracp-12$ in $mathbb Z/pmathbb Z$ and moreover $-1$ is not of the form $2^rpmodp$. In turn, this implies $2$ is the square of a primitive root, and so $2^r$ covers all quadratic residues which means $-1$ is not a quadratic residue. Since $left(frac2pright)=1$ and $left(frac-1pright)=-1$ we have $p=7pmod 8$.
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
It's easier to work with the inverse permutation. Starting from $1$ and working backwards we end up with a sequence $1=2^0to pm 2^1to pm 2^2to cdots topm 2^fracp-12=1$. Where the signs are uniquely determined so that $pm 2^rin 1,2,dots,fracp-12$. This cycle covers all numbers in $1,2,dots,fracp-12$ exactly once if and only if $2^rneq pm 1pmodp$ for any $1le rle fracp-32$.
Proposition: This happens precisely when either $2$ is a primitive root for $p$, or if $p=7pmod 8$ and $2$ has order $fracp-12$.
Proof: In the situation that $2^fracp-12=-1pmodp$ we have that $2^r$ for $r=1,2,cdots, p-1$ forms a complete system of residues, therefore $2$ is a primitive root for $p$. We are left with the case where $2^fracp-12=1pmodp$. This means that $2$ has order $fracp-12$ in $mathbb Z/pmathbb Z$ and moreover $-1$ is not of the form $2^rpmodp$. In turn, this implies $2$ is the square of a primitive root, and so $2^r$ covers all quadratic residues which means $-1$ is not a quadratic residue. Since $left(frac2pright)=1$ and $left(frac-1pright)=-1$ we have $p=7pmod 8$.
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
answered 4 hours ago
Gjergji Zaimi
59.8k3154296
59.8k3154296
add a comment |Â
add a comment |Â
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2
Reverse it. You are finding primes where 2 is (is not) a primitive root mod p. Gerhard "Now You Search The Literature" Paseman, 2018.10.11.
– Gerhard Paseman
5 hours ago
Upon reflection (pun intended) it is a little more complicated. I still think primitive root applies. Gerhard "Not Ready For Palindromic Signature" Paseman, 2018.10.11.
– Gerhard Paseman
4 hours ago
@GerhardPaseman: already the example $p = 7$ shows that this is not what's going on. The function acts transitively modulo $7$, but $2$ is obviously not a primitive root.
– R. van Dobben de Bruyn
4 hours ago
1
The length of the orbit is the order of 2 modulo p if that order is odd, otherwise it's half the order.
– Felipe Voloch
4 hours ago
Link to the sequence in OEIS?
– Gerry Myerson
21 mins ago