# Complete residue system modulo n (permutation of numbers 0 to n-1) such that

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.

• You say that odd numbers don't work but prime numbers do work...? – Greg Martin Jul 17 '17 at 23:29
• Sorry, meant to say even – vixenn Jul 18 '17 at 18:44

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.

• Thank you for the materials! I've seen similar problem, with products and not sums, but I don't understand it well enough to be able to apply it to my problem with the sums. – vixenn Jul 18 '17 at 20:49
• @vixenn: Seva completely answered your question (his product is your sum), so I suggest that you accept his answer officially (so that it turns green). – GH from MO Jul 19 '17 at 10:03
• Reading the materials a little bit more thoroughly, I think I understand it better now. Thanks for the help, @Seva :) – vixenn Jul 19 '17 at 20:32