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Let's we have finite field $F_q$ for some prime $q=2^M-1$.

I am looking for special sequence {$a_{i}$, $i \in {1,..,q-1}$}, ($\{a_{1},...,a_{q-1}\}=F_q/\{0\}$) with the following properties:

$r_{1}=a_1$ = 2;

Let's ${r_{s}} = \sum_{k=1}^s{a_{k}}$

then:

$\{r_{1},...,r_{q}\} =F_q$.

Thanks.

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    $\begingroup$ You must have $a_i=0$ for some $i$, and then you'll have $r_i=r_{i-1}$. $\endgroup$
    – Gerry Myerson
    Mar 28, 2021 at 22:09
  • $\begingroup$ Correct, excluded zero $\endgroup$
    – Alexander
    Mar 29, 2021 at 8:26

1 Answer 1

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This paper uses the polynomial method and other techniques to study the problem.

Hicks, Ollis and Schmitt, Distinct Partial Sums in Cyclic Groups: Polynomial Method and Constructive Approaches

The following conjecture is stated there.

Conjecture 1.2. (Archdeacon, Dinitz, Mattern and Stinson) For any cyclic group $\mathbb Z_n$ and any $k-$subset $A ⊆ \mathbb Z_n\setminus\{0\}$, it is possible to find an ordering of the elements of $A$ such that no two of its partial sums are equal for $1≤i<j≤k.$

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