Why is $12$ special in relation to $6n\pm1$?

Question

If you take the numbers of the form $6n\pm1$ and arrange them into $12$ columns, i.e., $$5,7,11,13,17,19,23,25,29,35,37,41,\ldots,$$

all the columns have digital roots that are the same throughout the whole column. So all numbers under the $5$ column have digital root $5$, $7=7$, $11=2$, $13=4$, $17=8$, $19=1$, $23=5$, $25=7$, $29=2$, $31=4$, $35=8$, $37=1$, so the pattern repeats.

Also why does each number in the column give the number of rows to count down to give numbers divisible by itself? So for $5$ we count down $5$ rows and we get the next number divisible by $5,$ the same for $7,$ etc. And also for every number in the column, e.g., for $41$ we count down $41$ rows to get the next number divisible by $41.$

It’s a kind of prime sieve but why is it so ‘self-referential’ for want of a better word? Is it something to do with the number $12$?

Answer 1

When you arrange the numbers this way the difference between the numbers in neighboring rows is $36=4\cdot 9$. Adding a multiple of $9$ to a number does not change the digital root, which explains why all the numbers in a column have the same digital root. When you count down $k$ rows below $k$ the number you find is $k+36k=37k$ which is divisible by $k$. As $k$ has no factors of $2$ or $3$ this will be the next multiple of $k$.

Answer 2

The first point (digital roots) - actually you get the same digital roots if you put them into six columns as well, so it is $6$, not $12$, that is relevant.

The reason this works is that the digital root is the same as the congruence class mod $9$. Two numbers of the form $6n\pm1$ are in the same congruence class mod $3$ if they both have a $+1$ or both have a $-1$, and so you need an even number of columns to get the $+1$s and $-1$s to line up. To get the same congruence class mod $9$ you also need the $n$s in each column to be congruent mod $3$, i.e. you need a number of columns that are divisible by $3$.

Essentially this is just saying that if $m=n+3k$ then $6m+1\equiv 6n+1$ mod $9$, and these two numbers are $6k$ terms apart in your sequence (and similarly for $-1$).

For the second point, this works just because columns are arithmetic progressions, and the step ($36$) is coprime to the elements. The numbers in the arithmetic progression $an+b$ which are divisible by $5$ will come every $5$ terms, provided $5$ and $a$ are coprime, because they come when $n\equiv -ba^{-1}$ mod $5$, and $a^{-1}$ mod $5$ exists and is unique.