I have a task:   

Find all $n\ \epsilon \ N, \ n > 1$ for which a permutation $a\_1,\ a\_2,\ ...,\ an\ $ of numbers $ 0,1, ..., n - 1$ exists such that $a\_1,\ a\_1+a\_2,\ ...,\ a\_1+a\_2+\ ...\ +an\ $ form a CRS $mod\ n$.


So far I've come to the conclusion that $a\_1$ must be $0$ because otherwise there would be two equal numbers from the listed above (so it won't be a CRS) and I think that odd numbers don't form CRS because $n\*(n-1)/2 \ \ (mod\ n)$ is also $0$ so again won't form CRS. My assumption is that all **even** numbers can form CRS and satisfy the conditions above since I have found such permutations for numbers 2,4,6,8, but I don't know how to prove it and if it is right at all.


Would appreciate some help: first - whether I have made any mistakes so far, and second - with proof if I am right.


Thanks in advance!



Complete residue system modulo n (permutation of numbers 0 to n-1) such that
Your question is about *sequenceable groups*, [introduced in 1961](http://msp.org/pjm/1961/11-4/pjm-v11-n4-p14-p.pdf) by Basil Gordon. 


A finite group is called *sequenceable* if its elements can be written as a sequence $(g\_1,g\_2,\dotsc,g\_n)$ so that all the partial products $g\_1,g\_1g\_2,\dotsc,g\_1g\_2\dotsb g\_n$ are pairwise distinct. Gordon has shown that a finite *abelian* group is sequenceable if and only if it contains exactly one involution (element of order $2$). It is easily seen that this condition is necessary, and Gordon proved that it is also sufficient.


A great review of this subject can be found in [this paper](http://www.combinatorics.org/ojs/index.php/eljc/article/viewFile/DS10/pdf) by Matt Ollis; see also more recent papers by the same author, such as [this one](http://www.mathnet.ru/links/d7fd4cccd8e6599bd0c8091f6047bd12/adm545.pdf).


