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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 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 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 at 13:16

1 Answer 1

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 $13$.

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+1, Nice counter-example! We still have palindromic list for 13, which means we have two ranges of that size. I'll try to drop the second requirement to see where that heads. I'm not too optimistic. –  Fred Kline Jan 21 '13 at 11:42
    
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 at 4:31

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