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3
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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 $k$?
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2
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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 $k$?
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1
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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 $k$?
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