Problems regarding $\{x_n \}$ defined by $x_1=1$; $x_n$ is the smallest distinct natural number such that $x_1+...+x_n$ is divisible by $n$. Let me denote a sequence of distinct natural numbers by $x_n$ whose terms are determined as follows:
$x_1$ is $1$ and $x_2$ is the smallest distinct natural number $n$ such that $x_1+x_2$ is divisible by $2$, which is $3$. Similarly $x_n$ is the smallest distinct natural number such that $x_1+...+x_n$ is divisible by $n$. Thus $x_1=1$, $x_2=3$, $x_3=2$.
Now the problem is to show that
(i) $x_n$ is surjective
(ii) $x_{(x_n)}$ is $n$, for example $x_2=3$ and $x_3=2$

Not an answer, but it might help somebody.  I can't prove this, it is observation.  With $\phi = \frac {1+\sqrt 5}2$, the naturals greater than $1$ are partitioned into pairs $\lceil n\phi \rceil$ and $\lceil n\phi^2 \rceil$.  This is a partitioning as $\frac 1\phi + \frac 1{\phi^2}=1$.  Then $a_{\lceil n\phi \rceil}=\lceil n\phi^2 \rceil$ and $a_{\lceil n\phi^2 \rceil}=\lceil n\phi \rceil$.  The gap between them the elements of the pair is  $n$, as $\phi^2=1+\phi$.  I can't relate this to the divisibility test.
