Let $A=\{a_s+n_s\mathbb Z\}_{s=1}^k$ be a finite system of residue classes, where $a_s$ and $n_s>0$ are integers. For a positive integer $m$, if $A$ covers each integer at least $m$ times then we call $A$ an $m$-cover of $\mathbb Z$. It is easy to see that $\sum_{s=1}^k\frac1{n_s}\ge m$ if $A$ is an $m$-cover of $\mathbb Z$.

In 1989 Ming-Zhi Zhang showed that if $A$ is a cover of $\mathbb Z$ then $\sum_{s\in I}\frac1{n_s}$ is a positive integer for some $I\subseteq\{1,\ldots,k\}$. Later I extended this in various ways; for example, I proved in my 1996 TAMS paper that if $A$ is an $m$-cover of $\mathbb Z$ then there are at least $m$ positive integers in the form $\sum_{s\in I}\frac1{n_s}$ with $\emptyset\not=I\subseteq\{1,\ldots,k\}$.

I have formulated the following conjecture.

CONJECTURE 1. If $A=\{a_s+n_s\mathbb Z\}_{s=1}^k$ is an $m$-cover of $\mathbb Z$, then $$\sum_{s\in I}\frac1{n_s}=m\qquad\text{for some}\ I\subseteq\{1,\ldots,k\}.$$

Perhaps, this can be extended to covers of groups by finitely many left cosets.

CONJECTURE 2. Let $\mathcal A=\{a_iG_i\}_{i=1}^k$ be a finite system of left cosets in a group $G$, where $a_1,\ldots,a_k$ are elements of $G$ and $G_1,\ldots,G_k$ are subgroups of $G$. If $\mathcal A$ covers every element of $G$ at least $m$ times (where $m$ is a positive integer), then $$\sum_{i\in I}\frac1{[G:G_i]}=m\quad\text{for some}\ I\subseteq\{1,\ldots,k\}.$$

Question. Can one prove or disprove Conjectures 1 and 2?

Your comments are welcome!

  • $\begingroup$ Forgive my ignorance, why when $A$ is a $m$-cover of $\mathbb{Z}$, $\sum_{s=1}^{k} \frac{1}{n_{s}} = m$ may not hold? It seem can be proved by a counting argument. $\endgroup$
    – katago
    Dec 19, 2020 at 16:00
  • 8
    $\begingroup$ I doubt I'm the only one that thinks it's inappropriate how you use mathoverflow to advertise your personal conjectures. $\endgroup$ Dec 19, 2020 at 17:14
  • 1
    $\begingroup$ Conjectures 1-2 here are new problems, not old conjectures in my published papers. Except for reference-request type questions, each OP may express his/her positive or negative opinion on potential answer to the question he/she asks. I don't think positive or negative opinions (as formal or informal conjectures) should be banned. $\endgroup$ Dec 20, 2020 at 1:02

1 Answer 1


Definition[Perfect m-cevering]: Let $A=\left\{a_{s}+n_{s} \mathbb{Z}\right\}_{s=1}^{k}$ be a finite system of residue classes, where $a_{s}$ and $n_{s}>0$ are integers. For a positive integer $m$, if $A$ covers each integer exactly $m$ times then we call $A$ an $m$ cover of $\mathbb{Z}$. It is easy to see that $\sum_{s=1}^{k} \frac{1}{n_{s}} = m$ if $A$ is an $m$ -cover of $\mathbb{Z}$.

Toy model(corresponding to the case $A=\left\{a_{s}+n_{s} \mathbb{Z}\right\}_{s=1}^{k}$, $n_s=2^{t_s}$):

Consider a Binary tree we can assign 0 or 1 on its every nodes, $$ f: V\left(T_{tree}\right) \rightarrow\{0,1\} $$ and for every leaf $v$, we define $$ L(v)=\sum_{n \text { ancertor } \atop \text { of } v} f(n) $$ then, $$ \sum_{n \text { ancertor } \atop \text { of } v} f(n) \geq M>0 $$ $\forall v \in leaf$, take the node which is maximum(in the sense it is the nearest to the root in all the ancestor of the leaf), s.t.

$$n_{\nu}=\min_{n \text { ancertor } \atop \text { of } v} \quad d(\mathrm{root}, \mathrm{n}) $$

and drop all the leaf which is a descendant of , then we get a set $\left\{n_{1}, n_{2}, \cdots, k_{t}\right\}$

and every leaf is in some and unique in it, then we can prove the set is exact a $1$ -cover of $\mathbb{Z}$, and as a corollary exists $I$, $\sum_{a_{i} \in I \subset\left\{1, 2, \cdots, k\right.\}} \frac{1}{2^{a_{i}}}=1$

Remark: And in fact the above argument already proved if $A=\left\{a_{s}+n_{s} \mathbb{Z}\right\}_{s=1}^{k}$ is a $m$-covering of $\mathbb{Z}$, then there exist subset $A_k$ of $A$ as a perfect $k$-covering of $\mathbb{Z}, \forall 1\leq k\leq m$.

Claim in general, $A$ is a $m$-covering of $\mathbb{Z}$, then there may not have a subset $A_k$ of $A$ consist of a perfect $k$-covering of $\mathbb{Z}$.

In the general case, the key point is $a_{i}\mathbb{Z}+b_i$ is finite, so there is only finitely many prime $p$ involved, and for every $p$ there is a $p$-parts tree, and we can similarly consider the construction as above but now to make the ancestor of the leaf is the nearest one to the root in all tree and for every tree the ancestor of the leaf nearest to the root, and now the choice is not unique. But all such ancestors consist of a partially ordered set and we can always choose a maximum one in the partially ordered set.

and the result will not relate to the number theory structure, in fact, $2k, 3k+2, 6k+1, 6k+3$ will not have a subset which is a perfect covering of $\mathbb{Z}$. But related to something just about counting, in fact in the case, there is no perfect covering we can see the small fragments part is much more than the relatively large part and will offset the effect of the intersections of the relatively large parts.

In fact, establish some Karamata's Inequality type inequality will declare conjecture 1 and conjecture 2 is true.

So I think these two new conjecture are all true and a complete proof is very close if one carefully check the number

$$g(w) \triangleq \#\left\{a_{s} \mid a_{s}=\prod_{i=1}^{k} p_{i}^{w(i)}\right\}$$

for all $w=(w(1), \cdots, w(k)) \in\{0,1\}^{k}$, where $w(i) \in\{0.1\}$, and $p_1,...,p_k$ is the prime occur in $\Pi_j a_j$, and prove some Karamata's Inequality on $g(w)$.

  • $\begingroup$ For any exact $m$-cover $A=\{a_s+n_s\mathbb Z\}_{s=1}^k$, I proved in a 1992 paper [Israel J. Math.] that for each $n=1,\ldots,m$ we have $\sum_{s\in I}\frac1{n_s}=n$ for some $I\subseteq\{1,\ldots,k\}$. $\endgroup$ Dec 20, 2020 at 8:01
  • $\begingroup$ Zhi-Wei Sun: thanks, I will check this. $\endgroup$
    – katago
    Dec 20, 2020 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.