TITLE: Self-avoiding walk on $\mathbb{Z}$
QUESTION [18 upvotes]: How many sequences $a_1,a_2,a_3,\dotsc$, satisfy:
i) $a_1=0$
ii) ($a_{n+1}=a_n-n$ or $a_{n+1}=a_n+n$)
iii) $a_i\neq a_j$ for $i\neq j$
iiii) $\mathbb{Z}=\{a_i\}_{i>0}$
Are the two alternating sequences the only solutions?
$a_1,a_2,a_3,..=0,1,-1,2,-2,3,-3,4,...$
or
$a_1,a_2,a_3,..=0,-1,1,-2,2,-3,...$
Is there a sequence satsifying i),iii),iiii) and ii) ($a_{n+1}=a_n-n^2$ or $a_{n+1}=a_n+n^2$) ?

REPLY [4 votes]: Only uncountably many. See this link to MO.