In regards to repeating decimals and cyclic numbers: I understand that many of them, while multiplying by certain integers will produce a number with the same variant of digits, however my question is in regards to the patterns already occurring within some of them.
Many fractions that convert into a number with repeating decimals, will exhibit a pattern within the repetend. This pattern is made of a consecutive list of numbers growing by the same multiple.
example: $1/31=0.032258064516129...$., the repetend is $032258064516129$ which can be broken into the numbers $032258$, $064516$, $129032$ that are related by multiples of 2: $$032258∗2=064516, \quad 064516∗2=129032.$$
example: $1/13=0.076923....$, the repetend is $076923$ which can be broken into $0769$, $2307$ that are related by multiplying by 3: $$0769∗3=2307$$
example: $1/17=0.0588235294117647....$, the repetend is $0588235294117647$ which can be broken into the numbers $5882352941$, $11764705882$ that are related by multiplying by 2: $$5882352941∗2=11764705882$$
example: $1/19=0.052631578947368421...$, the repetend is $052631578947368421$ which can be broken into the numbers $05263$ , $15789$ , $47367$ , $142101$ (think of the initial 1 in 142101 being removed from the 8 in the preceding 47368) that are related by multiples of 3: $$05263∗3=15789, \quad 15789∗3=47367, \quad 47367∗3=142101$$
example: $1/21=0.047619...$, the repetend is $047619$ which can be broken into the numbers $0476$, $1904$ that are related by multiplying by 4: $$0476∗4=1904$$
What are these multiples? That is, given 1/$n$ with a repeating decimal, what multiple connects the different pieces of the repetend? (And how long are those pieces?)
The first thing to think about is finding the repeating decimal by long division. Taking $1/17$ for example, your first division starts with a remainder $1$. You append a $0$ to it and try to divide $17$ into $10$. It goes $0$ times, which gives the leading $0$ in the repeat. You then pass a remainder of $10$ to the next digit. You pull down another $0$, getting $100$, divide $17$ into it getting $5$ with a remainder of $15$. The $5$ is the second digit of the repeat and the $15$ goes on for the next digit.
If you have a full length repeat like $17$, you must use each of the remainders from $1$ to $17-1$ once. When you repeat a remainder you have finished the repeat. For $13$, which has only a six digit repeat you only use six remainders. This explains all the relationships you show.
After the $n^{th}$ digit the remainder is $10^n \mod 17$. We can find that $10^{10} \equiv 2 \pmod {17}$ That means after $10$ digits you pass a remainder of $2$ to the next digit. The repeat from there will clearly be twice the repeat starting from the front, where the remainder is $1$.