Palindromic Patterns of Greatest Divisors $\leq k$

We work with the positive integers $\in \mathbf{Z^{+}}$.
The list length is the least common multiple for all numbers $\leq k$: $$\text{rlcm}(1)\equiv 1; \text{rlcm}(2)\equiv 2; \text{rlcm}(n) \equiv \text{lcm}(\text{rlcm}(n - 1), n)$$ The greatest divisor $\leq k$ is a definition that gives a unique identity to the $1$ (one) because $1$ is the greatest divisor $\leq k$ for numbers that are co-prime to all values $\leq k$. The greatest divisor routine: $$\text{maxd}(1,n)\equiv1;\text{maxd}(k,n)\equiv \begin{cases} k &\text{for }0= n \text{mod} k \\ \text{maxd}(k-1,n) &\text{for } 0\neq n \text{mod} k \end{cases}$$

List of greatest divisors $\leq k=4$. List length is lcm$(1,\dots,k)$.
$1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 4$

Five replications of the above list:
$1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 4, 1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 4, 1, 2, 3, 4, 1, 3,$
$1, 4, 3, 2, 1, 4, 1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 4, 1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 4$

List of greatest divisors $\leq k=5$. List length is lcm$(1,\dots,k)$. $1, 2, 3, 4, 5, 3, 1, 4, 3, 5, 1, 4, 1, 2, 5, 4, 1, 3, 1, 5, 3, 2, 1, 4, 5, 2, 3, 4, 1, 5,$
$1, 4, 3, 2, 5, 4, 1, 2, 3, 5, 1, 3, 1, 4, 5, 2, 1, 4, 1, 5, 3, 4, 1, 3, 5, 4, 3, 2, 1, 5$

List of greatest divisors $\leq k=6$. List length is lcm$(1,\dots,k)$. $1, 2, 3, 4, 5, 6, 1, 4, 3, 5, 1, 6, 1, 2, 5, 4, 1, 6, 1, 5, 3, 2, 1, 6, 5, 2, 3, 4, 1, 6,$
$1, 4, 3, 2, 5, 6, 1, 2, 3, 5, 1, 6, 1, 4, 5, 2, 1, 6, 1, 5, 3, 4, 1, 6, 5, 4, 3, 2, 1, 6$

When you remove the last element of each list it will read the same in both directions.
Note-1: the first $k$ numbers of each list are the divisors $\leq k$.
Note-2: there is at least one $1$ between each $k$ in its respective list. (i.e., a co-prime in every segment of size $k$).

How can I prove that these finite lists repeat forever?
How can I prove that there is at least one $1$ in each segment of size $k$?

Edit As to the answer by Gerry Myerson: each k-segment has a multiple of k as the last element (i.e., no derangement.) I knew this, but didn't realize that it must be an assumption for examining segments within the symetric sequence. (Looking at 17-seg with 13-k sequence.)

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Maybe I am just slow, but I think there could be another sentence or two explaining what is going on; how do you construct these lists again? – Per Alexandersson Aug 10 '14 at 12:29
@PerAlexandersson, math.stackexchange.com/q/867135/28555 points to a post that explains how I found it. mathematica.stackexchange.com/q/48452/973 points to the Mathematica code for the recursive routine. – Fred Kline Aug 10 '14 at 13:11
@PerAlexandersson, math.stackexchange.com/q/886041/28555 shows another identity (conjecture) based on the same palindromic divisor sequence. – Fred Kline Aug 10 '14 at 13:16

Let $k=13$. The segment of length $17$ from $2184$ to $2200$, inclusive, has no $1$; every integer in that range has a prime divisor not exceeding $13$.
Actually, it is not a counter-example. See modified OP. $2193$ is a multiple of 17, inside the 17-seg which shows it is a derangement. – Fred Kline Aug 10 '14 at 4:31