I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product, in some order. The small Davenport constant $d(G)$ is the maximal length of a one-free sequence, i.e., $d(G)+1=D(G)$.

In particular, for $A_n$ I want to get a lower bound by computing a general family of one-product free sequences for each $n$. For $A_3$ we know $d(A_3)=2$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $D(A_4)=4$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $D(A_5)=8$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known. Moreover, in these three cases $d(A_n)=2^{n-2}$, is this coincidence or could be proved to be some lower/upper bound?

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general and $n=6$ and $n=7$ in particular. One of them, looking at the examples, could be for $n$ odd getting $n-1$ copies of some set of $n$-cycles. For $n$ even, one possible idea could be taking maximal sequence of $A_{n-1}$ and taking another copy conjugating all elements by some $g$ not fixing $n$.

Does anyone come up with some idea to get one-free sequences in $A_n$? Apart from the obvious one of taking $m-1$ times an element of order $m$. Similarly, any idea for getting upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!

I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product, in some order. The small Davenport constant $d(G)$ is the maximal length of a one-free sequence, i.e., $d(G)+1=D(G)$.

In particular, for $A_n$ I want to get a lower bound by computing a general family of one-product free sequences for each $n$. For $A_3$ we know $d(A_3)=2$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $d(A_4)=4$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $d(A_5)=8$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known. Moreover, in these three cases $d(A_n)=2^{n-2}$, is this coincidence or could be proved to be some lower/upper bound?

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general and $n=6$ and $n=7$ in particular. One of them, looking at the examples, could be for $n$ odd getting $n-1$ copies of some set of $n$-cycles. For $n$ even, one possible idea could be taking maximal sequence of $A_{n-1}$ and taking another copy conjugating all elements by some $g$ not fixing $n$.

Does anyone come up with some idea to get one-free sequences in $A_n$? Apart from the obvious one of taking $m-1$ times an element of order $m$. Similarly, any idea for getting upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!

I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product.

In particular for $A_n$ I want to get a lower bound by computing a general example of one-product free sequences. For $A_3$ we know $D(A_3)=3$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $D(A_4)=5$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $D(A_5)=9$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known.

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general and $n=5$ in particular. One of them, looking at the examples, for $n$ odd is getting n-1 copies of a particular set (no repetitions, each element at most once) of n-cycles satisfying the following:

1. All the possible products obtained by multiplication of (some subset) the elements are ALL n-cycles.

2. Two (distinct) elements $g_1, g_2$ cannot be in the same group generated by some n-cycle, i.e., one element cannot be a power of other.

3. The product of two elements cannot be a power of a third one

The last two properties are because when we take n-1 copies of this sets, we don't want to get identity. In total, our sequence would have length $(n-1)*l$, where $l$ is the maximal length of such a set.

Does anyone come up with some idea to get how could this maximal sets look like or what could be the maximal length of such sets? Similarly, any idea for getting lower/upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!

I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product, in some order. The small Davenport constant $d(G)$ is the maximal length of a one-free sequence, i.e., $d(G)+1=D(G)$.

In particular, for $A_n$ I want to get a lower bound by computing a general family of one-product free sequences for each $n$. For $A_3$ we know $d(A_3)=2$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $D(A_4)=4$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $D(A_5)=8$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known. Moreover, in these three cases $d(A_n)=2^{n-2}$, is this coincidence or could be proved to be some lower/upper bound?

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general and $n=6$ and $n=7$ in particular. One of them, looking at the examples, could be for $n$ odd getting $n-1$ copies of some set of $n$-cycles. For $n$ even, one possible idea could be taking maximal sequence of $A_{n-1}$ and taking another copy conjugating all elements by some $g$ not fixing $n$.

Does anyone come up with some idea to get one-free sequences in $A_n$? Apart from the obvious one of taking $m-1$ times an element of order $m$. Similarly, any idea for getting upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!

I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product.

In particular for $A_n$ I want to get a lower bound by computing a general example of one-product free sequences. For $A_3$ we know $D(A_3)=3$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $D(A_4)=5$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $D(A_5)=9$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known.

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general. One of them, looking at the examples, for $n$ odd is getting n-1 copies of a particular set (no repetitions, each element at most once) of n-cycles satisfying the following:

1. All the possible products obtained by multiplication of (some subset) the elements are ALL n-cycles.

2. Two (distinct) elements $g_1, g_2$ cannot be in the same group generated by some n-cycle, i.e., one element cannot be a power of other.

3. The product of two elements cannot be a power of a third one

The last two properties are because when we take n-1 copies of this sets, we don't want to get identity. In total, our sequence would have length $(n-1)*l$, where $l$ is the maximal length of such a set.

Does anyone come up with some idea to get how could this maximal sets look like or what could be the maximal length of such sets? Similarly, any idea for getting lower/upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!

I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product.

In particular for $A_n$ I want to get a lower bound by computing a general example of one-product free sequences. For $A_3$ we know $D(A_3)=3$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $D(A_4)=5$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $D(A_5)=9$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known.

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general and $n=5$ in particular. One of them, looking at the examples, for $n$ odd is getting n-1 copies of a particular set (no repetitions, each element at most once) of n-cycles satisfying the following:

1. All the possible products obtained by multiplication of (some subset) the elements are ALL n-cycles.

2. Two (distinct) elements $g_1, g_2$ cannot be in the same group generated by some n-cycle, i.e., one element cannot be a power of other.

3. The product of two elements cannot be a power of a third one

The last two properties are because when we take n-1 copies of this sets, we don't want to get identity. In total, our sequence would have length $(n-1)*l$, where $l$ is the maximal length of such a set.

Does anyone come up with some idea to get how could this maximal sets look like or what could be the maximal length of such sets? Similarly, any idea for getting lower/upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!