TITLE: Complete residue system modulo n (permutation of numbers 0 to n-1) such that
QUESTION [6 upvotes]: 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!

REPLY [11 votes]: Your question is about sequenceable groups, introduced in 1961 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 by Matt Ollis; see also more recent papers by the same author, such as this one.