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Background: I am currently working on a research project on the existence of certain spanning sets for finite abelian groups, $G$. In particular, I am looking at the existence of perfect $s$-bases of size $m$ for $G$.

We call $A = \{a_i \}_{i=1}^m \subseteq G$ an $s$-basis for $G$ if

$$[0,s]A = \left\{ \sum_{i=1}^m \lambda_i a_i \ \Bigg| \ \lambda_i \in \mathbb{N} \text{ and } \sum_{i=1}^m \lambda_i, \ a_i \in A \right\} = G.$$

Essentially, every element of $G$ can be made as a sum of at most $s$ many elements of $A$ including repetition of elements. For example, if we take $G = \mathbb{Z}/3\mathbb{Z}$ and $A = G \backslash \{0\}$ then $A$ is a 1-basis for $G.$

We call an $s$-basis for $G$ perfect if

$$|G| = \binom{m+s}{s}.$$

This definition essentially implies that every element of $G$ has a unique sum representation of elements in $A$ up to commutativity.

In particular, there is a conjecture that states:

There does not exist perfect $s$-bases of size $m$ for any finite abelian group $G$, unless $s = 1$ and $m=1.$

Checking the cases for $s=1$ and $m=1$ are simple. In particular, the conjecture has been proven for when $s=2$ and $s=3$ by Bela Bajnok in his paper and I extended his proof to check the cases for $s=4$ and $s=5.$

His proof technique involves finding bounds on the value of $m$ and then systematically checking those $m$ values for perfect $s$-bases in each abelian group of the appropriate sizes. It gets messy and difficulty to generalize his technique so I am working on hopefully finding a neater proof technique for some cases.

TL:DR: I have reduced my problem to my question above. I have shown that for $s>1$ we need that

$$|G| = \binom{m+s}{s} = k(s+1)$$

for some $k \in \mathbb{N}.$ Notice that when $m=1$ we simply get $|G| = s+1$ so the conjecture still holds. Additionally, if there is a way to classify values of $m$ such that

$$\frac{\binom{m+s}{s}}{s+1} \in \mathbb{N}$$

then we can reduce the problem to checking only groups of certain sizes.

Personal Background: I am currently a rising fourth-year undergraduate. I am familiar and comfortable with typical undergraduate material and have studied a bit of graduate complex analysis and algebra.

N. Geis
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    $\text{gcd}(m,s+1) = 1$ is a sufficient condition. – Aryabhata Aug 03 '17 at 20:22
  • @Aryabhata: is it $\gcd(m,s+1)$ or $\gcd(m,(s+1)!)$ ? – G Cab Aug 03 '17 at 20:46
  • Thank you for the response @Aryabhata. Could explain how that condition is sufficient? I cannot see that. – N. Geis Aug 03 '17 at 20:47
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    NGeis: If $r$ is a rational such that $ar$ and $br$ are both integers for $\text{gcd}(a,b) = 1$, then $r$ is an integer. @GCab: It is not $(s+1)!$. – Aryabhata Aug 03 '17 at 21:26
  • @Aryabhata: Thank you for clarifying. I see how $\gcd(m, s+1) = 1$ is a sufficient condition. I will assume $m$ satisfies that condition for the next day or so and see where that will lead me. – N. Geis Aug 03 '17 at 22:40
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    @N.Geis The method mentioned in the first comment is discussed at length in this answer. – Bill Dubuque Aug 03 '17 at 22:44
  • If $s+1$ is a prime number then you can check that $\text{gcd}(m,s+1) =1$ is also a necessary condition, since other wise $\text{gcd}(m+i,s+1) = 1$ for all $1\leq i\leq s$. If $s+1$ is not prime, then $s$ is factored into smaller factors and I have a feeling that this condition is not necessary, but I have not yet had a idea how a necessary condition looks like , and I will add more comments when I get something new. – Siming Tu Aug 05 '17 at 14:33

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