How many arithmetic progressions are necessary to form an arbitrary subset of $mathbb{Z}_n$?
$begingroup$
I couldn't find a way to ask the question in under 150 characters, so I've rewritten it in a way that makes more sense.
Given an integer $n$, what is the minimum $k$ such that for any subset $X subseteq mathbb{Z}/nmathbb{Z}$, there exists a set of order at most $k$ of arithmetic progressions whose union is exactly equal to $X$?
I am interested in the group ($mathbb{Z}/n mathbb{Z}$, +), so I would like to consider an arithmetic progression to be a set ${a, a+d, a+2d, dots, a+kd}$, where addition is calculated mod $n$. For example, in $mathbb{Z}/12 mathbb{Z}$, I would consider $9, 11, 1, 3$ to be an arithmetic progression of length 4.
To give some examples of what I am thinking about, I will consider some subsets of $mathbb{Z}/24 mathbb{Z}$. The set ${0, 1, 2, dots, 11}$ can be considered to be a single arithmetic progression. The set ${0,3,4,5,20,22}$ can be considered to the the union of the two arithmetic progressions $(3,4,5)$ and $(20,22,0)$. The set ${1,2,4,8,16}$ is not a union of two arithmetic progressions, but we can consider it to the union of three arithmetic progressions, namely $(1,2)$, $(2,4)$, and $(8,16)$. What I have shown is that given some subset of $mathbb{Z}/24 mathbb{Z}$, we may need three (or possibly more) arithmetic progressions in order to form a union exactly equal to that subset. So for $n=24$, $k geq 3$.
Is there a general way to calculate $k$ for general $n$? Are there any results or conjectures that are related to this kind of question in any way?
Any help would be appreciated. Thank you.
combinatorics number-theory
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
$endgroup$
add a comment |
$begingroup$
I couldn't find a way to ask the question in under 150 characters, so I've rewritten it in a way that makes more sense.
Given an integer $n$, what is the minimum $k$ such that for any subset $X subseteq mathbb{Z}/nmathbb{Z}$, there exists a set of order at most $k$ of arithmetic progressions whose union is exactly equal to $X$?
I am interested in the group ($mathbb{Z}/n mathbb{Z}$, +), so I would like to consider an arithmetic progression to be a set ${a, a+d, a+2d, dots, a+kd}$, where addition is calculated mod $n$. For example, in $mathbb{Z}/12 mathbb{Z}$, I would consider $9, 11, 1, 3$ to be an arithmetic progression of length 4.
To give some examples of what I am thinking about, I will consider some subsets of $mathbb{Z}/24 mathbb{Z}$. The set ${0, 1, 2, dots, 11}$ can be considered to be a single arithmetic progression. The set ${0,3,4,5,20,22}$ can be considered to the the union of the two arithmetic progressions $(3,4,5)$ and $(20,22,0)$. The set ${1,2,4,8,16}$ is not a union of two arithmetic progressions, but we can consider it to the union of three arithmetic progressions, namely $(1,2)$, $(2,4)$, and $(8,16)$. What I have shown is that given some subset of $mathbb{Z}/24 mathbb{Z}$, we may need three (or possibly more) arithmetic progressions in order to form a union exactly equal to that subset. So for $n=24$, $k geq 3$.
Is there a general way to calculate $k$ for general $n$? Are there any results or conjectures that are related to this kind of question in any way?
Any help would be appreciated. Thank you.
combinatorics number-theory
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
$endgroup$
$begingroup$
It might give a first idea to write down where the first jumps in $k$ occur, up to $n=12$ or so.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
1
$begingroup$
Conjecture: An $X$ with maximal $k$ is given by the first $k+1$ entries of $(0,1,3,6,10,15, ...)$, as soon as the distance of the last entry to $n$ is greater than all distances that occur before. This gives $k$ to be the largest number such that $frac{k(k+1)}{2} +k le n$, which is $k = lfloor sqrt{2n+frac94} -frac32 rfloor$.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
add a comment |
$begingroup$
I couldn't find a way to ask the question in under 150 characters, so I've rewritten it in a way that makes more sense.
Given an integer $n$, what is the minimum $k$ such that for any subset $X subseteq mathbb{Z}/nmathbb{Z}$, there exists a set of order at most $k$ of arithmetic progressions whose union is exactly equal to $X$?
I am interested in the group ($mathbb{Z}/n mathbb{Z}$, +), so I would like to consider an arithmetic progression to be a set ${a, a+d, a+2d, dots, a+kd}$, where addition is calculated mod $n$. For example, in $mathbb{Z}/12 mathbb{Z}$, I would consider $9, 11, 1, 3$ to be an arithmetic progression of length 4.
To give some examples of what I am thinking about, I will consider some subsets of $mathbb{Z}/24 mathbb{Z}$. The set ${0, 1, 2, dots, 11}$ can be considered to be a single arithmetic progression. The set ${0,3,4,5,20,22}$ can be considered to the the union of the two arithmetic progressions $(3,4,5)$ and $(20,22,0)$. The set ${1,2,4,8,16}$ is not a union of two arithmetic progressions, but we can consider it to the union of three arithmetic progressions, namely $(1,2)$, $(2,4)$, and $(8,16)$. What I have shown is that given some subset of $mathbb{Z}/24 mathbb{Z}$, we may need three (or possibly more) arithmetic progressions in order to form a union exactly equal to that subset. So for $n=24$, $k geq 3$.
Is there a general way to calculate $k$ for general $n$? Are there any results or conjectures that are related to this kind of question in any way?
Any help would be appreciated. Thank you.
combinatorics number-theory
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
$endgroup$
I couldn't find a way to ask the question in under 150 characters, so I've rewritten it in a way that makes more sense.
Given an integer $n$, what is the minimum $k$ such that for any subset $X subseteq mathbb{Z}/nmathbb{Z}$, there exists a set of order at most $k$ of arithmetic progressions whose union is exactly equal to $X$?
I am interested in the group ($mathbb{Z}/n mathbb{Z}$, +), so I would like to consider an arithmetic progression to be a set ${a, a+d, a+2d, dots, a+kd}$, where addition is calculated mod $n$. For example, in $mathbb{Z}/12 mathbb{Z}$, I would consider $9, 11, 1, 3$ to be an arithmetic progression of length 4.
To give some examples of what I am thinking about, I will consider some subsets of $mathbb{Z}/24 mathbb{Z}$. The set ${0, 1, 2, dots, 11}$ can be considered to be a single arithmetic progression. The set ${0,3,4,5,20,22}$ can be considered to the the union of the two arithmetic progressions $(3,4,5)$ and $(20,22,0)$. The set ${1,2,4,8,16}$ is not a union of two arithmetic progressions, but we can consider it to the union of three arithmetic progressions, namely $(1,2)$, $(2,4)$, and $(8,16)$. What I have shown is that given some subset of $mathbb{Z}/24 mathbb{Z}$, we may need three (or possibly more) arithmetic progressions in order to form a union exactly equal to that subset. So for $n=24$, $k geq 3$.
Is there a general way to calculate $k$ for general $n$? Are there any results or conjectures that are related to this kind of question in any way?
Any help would be appreciated. Thank you.
combinatorics number-theory
combinatorics number-theory
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
asked Dec 24 '18 at 5:29
Peter BradshawPeter Bradshaw
430211
430211
$begingroup$
It might give a first idea to write down where the first jumps in $k$ occur, up to $n=12$ or so.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
1
$begingroup$
Conjecture: An $X$ with maximal $k$ is given by the first $k+1$ entries of $(0,1,3,6,10,15, ...)$, as soon as the distance of the last entry to $n$ is greater than all distances that occur before. This gives $k$ to be the largest number such that $frac{k(k+1)}{2} +k le n$, which is $k = lfloor sqrt{2n+frac94} -frac32 rfloor$.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
add a comment |
$begingroup$
It might give a first idea to write down where the first jumps in $k$ occur, up to $n=12$ or so.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
1
$begingroup$
Conjecture: An $X$ with maximal $k$ is given by the first $k+1$ entries of $(0,1,3,6,10,15, ...)$, as soon as the distance of the last entry to $n$ is greater than all distances that occur before. This gives $k$ to be the largest number such that $frac{k(k+1)}{2} +k le n$, which is $k = lfloor sqrt{2n+frac94} -frac32 rfloor$.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
$begingroup$
It might give a first idea to write down where the first jumps in $k$ occur, up to $n=12$ or so.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
$begingroup$
It might give a first idea to write down where the first jumps in $k$ occur, up to $n=12$ or so.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
1
1
$begingroup$
Conjecture: An $X$ with maximal $k$ is given by the first $k+1$ entries of $(0,1,3,6,10,15, ...)$, as soon as the distance of the last entry to $n$ is greater than all distances that occur before. This gives $k$ to be the largest number such that $frac{k(k+1)}{2} +k le n$, which is $k = lfloor sqrt{2n+frac94} -frac32 rfloor$.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
$begingroup$
Conjecture: An $X$ with maximal $k$ is given by the first $k+1$ entries of $(0,1,3,6,10,15, ...)$, as soon as the distance of the last entry to $n$ is greater than all distances that occur before. This gives $k$ to be the largest number such that $frac{k(k+1)}{2} +k le n$, which is $k = lfloor sqrt{2n+frac94} -frac32 rfloor$.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
add a comment |
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$begingroup$
It might give a first idea to write down where the first jumps in $k$ occur, up to $n=12$ or so.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54
1
$begingroup$
Conjecture: An $X$ with maximal $k$ is given by the first $k+1$ entries of $(0,1,3,6,10,15, ...)$, as soon as the distance of the last entry to $n$ is greater than all distances that occur before. This gives $k$ to be the largest number such that $frac{k(k+1)}{2} +k le n$, which is $k = lfloor sqrt{2n+frac94} -frac32 rfloor$.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 18:54