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The multiplication graphs $p/q$ – with an edge between $a,b$ iff $ap \equiv b\operatorname{mod} q$ – for some twin primes $p$ and $q = p+2$ reveal an astonishing pattern:

  • $(p+0)/(p+1)$ and $(p+1)/(p+2)$ have a seeming $2$-fold rotational symmetry.
    (This is per se not astonishing.)

  • $(p+0)/(p+2)$ always has a seeming $3$-fold rotational symmetry.
    (This is per se astonishing.)

  • $(p-1)/(p+2)$ and $(p+0)/(p+3)$ have a seeming $4$-fold rotational symmetry

Note that in each case the symmetry is only a seeming but also an obvious one.

Note further that one might try to generalize this to (for appropriate $p$)

  • $(p+n)/(p+m)$ have a seeming $|n-m|+1$-fold rotational symmetry

I wonder: What does this mean?

How do $2$, $3$, $4$ come into play for arbitrary twin primes?

Now I know that the question must be (thanks to Gerhard Paseman):

How do $2$, $3$, $4$ come into play for arbitrary $6k\pm 1$ pairs?


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The same pattern arises when choosing $r=6\cdot 41$ and $p = r-1$, $q = r+1$:

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Note that neither $p=245$ nor $q=247$ are prime: $245 = 5\cdot 7^2$ and $247 = 13\cdot 19$.

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    $\begingroup$ Twin primes (with one exception) are 6k+-1. What do your graphs look like for other values of k? Gerhard "Thinks It's About Small Primes" Paseman, 2018.11.19. $\endgroup$ – Gerhard Paseman Nov 19 '18 at 18:08
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To explain what seems to be there you need to describe it more exactly. It might help to draw an edge from $i$ to $j$ starting a little after $i$ because the edges do have a direction.

For the graphs $n/(n+1)$ there is an edge from $i$ to $n+1-i$ (so we could in this case consider them undirected.) This gives a reflective symmetry of order $2$ but it can't be a purely rotational symmetry unless $n+1$ is even (in which case it is). It is true that the edges are all horizontal (if $0$ is at the top) so in the case that $n+1$ is odd, overlaying the diagram with itself rotated a half turn puts each rotated edge in a position half way between two parallel former edges of similar length.

So you need to be more specific about what you mean (in your answer below) by $a+b$-fold symmetry when the number of points $an+b$ is not a multiple of $a+b.$


I will look at the graphs $n/(an+b)$ in the case that $a$ and $b$ are relatively prime and $an+b$ is a multiple of $a+b.$ Then $n \bmod (a+b) \equiv 1$ so

$$n/(an+b)={\Big(}p(a+b)+1{\Big)}/{\Big(}(ap+1)(a+b){\Big)}.$$ Here the $(a+b)$ fold rotation would take $i$ to $i+ap+1.$

Note that $$ n(ap+1)={\Big(}p(a+b)+1{\Big)}(ap+1) \equiv ap+1 \bmod {\Big(}(ap+1)(a+b){\Big)} $$

So the edge from $i$ to $ni$ does indeed rotate to an edge from $i+ap+1$ to $ni+n(ap+1) \equiv ni+ap+1.$


The case that $\gcd(a,b)=g \gt 1$ is similar.

LATER

My advice wasn't to draw more graphs, it was to be more specific about the exact nature of the seeming approximate similarity. Ideally expressing it symbolically and using arithmetic/number theory to explain it.

Here is a case that should give sufficient detail to allow a more general analysis. I'll start with diagrams, describe a feature, and then explain it arithmetically. The actual history was in the other direction but it doesn't make as good a story.

I will look at the exact or "almost" $5$-fold symmetry For $n/(2n+3).$ It is better to do $n/(n+2)$ or $n/(2n+1)$ first but this case illustrates an additional feature.

Here, first is the case $16/35$ which is typical for $(5q+1)/(10q+5)$

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The small outward segments at $0,7,14,21,28$ correspond to the "edges $(i,i)$ in those five cases. And we already understand the exact symmetry from a rotation of $\frac{7}{35}=\frac15.$

A good illustration of what you see is $17/37.$ The middle picture superimposes in red a rotation of $\frac15$ which looks approximately symmetric and on the right a rotation of $2/5$ which is not so symmetric.

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But then there is $14/31.$ Here the rotation by $\frac15$ is not as close to a symmetry as the rotation by $2/5.$

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But what does that mean?The complete edge list is

$[0, 0], [1, 14], [2, 28], [3, 11], [4, 25], [5, 8], [6, 22], [7, 5], [8, 19], [9, 2]$

$ [10, 16], [11, 30], [12, 13], [13, 27], [14, 10], [15, 24], [16, 7], [17, 21], [18, 4], [19, 18], [20, 1]$

$[21,15], [22, 29], [23, 12], [24, 26], [25, 9], [26, 23], [27, 6], [28, 20], [29, 3], [30, 17]$

There is an angle made by the edges $[5,8]$ and $[8,19]$ . Moving by $\frac25$ of a complete rotation would increase everything by $\frac{62}5=12.4$ and move that angle to $[17.4,20.4]$ and $[20.4,31.4]=[20.4,0.4]$ The closest actual edges are $[17,21]$ and $[20,1].$ These cross internally though close to the boundary.

This is not an isolated example. One can see that the edges add: $[6,22]+[7,5]=[13,27]. $ It is easy to see why. There is only one fixed point $[0,0]$ and the next closest $[12,13]$ and $[19,18].$ So an angle $[i,j],[j,p]$ rotates to $[i+12.4,j+12.4],[j+12.4,p+12.4]$ which is fairly close to actual (crossing) edges $[i+12,j+13],[j+12,p+13].$ It could be argued that it is slightly better to advance by $12.5$ which is $\frac{12.5}{31}=0.4032258$ of a rotation rather that $\frac25=0.4.$

So here is my conclusion for the case $n/m=n/(an+b)$ which I will limit, for now, to the case that $\gcd(n,m)=1.$ Then there is a unique $0 \lt k \lt m$ so that the edge $[k,nk]=[k,k+1] \bmod m.$ The "best" rotation is by $\frac{k+0.5}{m}$ (Rotating that much in the opposite direction is just as good due to the reflective symmetry.) Evidently this is pretty close one of $\frac{c}{a+b}$ and $c$ depends on the congruence class of $n \bmod {a+b}.$ The next step would be to specify how that works and prove it.

In the case $n/(2n+3):$

  • When $n=5q+2,$ such as $17/37$ above, one has $n/(2n+3)=(5q+2)/(10q+7)$ and for $k=2q+1,$we have $nk=10q^2+9q+2 \equiv 2q+2 \bmod m.$

  • When $n=5q+4,$ such as $14/31$ above, one has $n/(2n+3)=(5q+4)/(10q+11)$ and for $k=4q+4,$ $nk=20q^2+36q+16 \equiv 14q+16 \equiv 4q+5 \bmod m.$

The other two congruence classes, $5q$ and $5q+3,$ are similar.

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  • $\begingroup$ I followed your advice in a separate answer. $\endgroup$ – Hans-Peter Stricker Nov 21 '18 at 8:38
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The question has an unexpected answer (at least to me). It turns out, that all multiplication graphs of the form $n/(an + b)$ have a seeming $a + b$-fold rotation symmetry. Especially, graphs of the form $n/(n + b)$ have a seeming $b + 1$-fold rotation symmetry.

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Thus

  • all graphs of the form $n/(n+2)$ have a seeming $3$-fold symmetry

  • all graphs of the form $n/(n+1)$ have a seeming $2$-fold symmetry

  • all graphs of the form $(n-1)/n$ have a seeming $2$-fold symmetry

  • all graphs of the form $(n-1)/(n+2)$ have a seeming $4$-fold symmetry

  • all graphs of the form $n/(n+3)$ have a seeming $4$-fold symmetry

This "explains" the pictures in the question, but has nothing to do with prime numbers and prime twins at all!

What remains to be explained and understood: Why is this so? Why do all multiplication graphs of the form $n/(an + b)$ have a seeming $a + b$-fold rotation symmetry?

$a=2$, $b=1$

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$a=2$, $b=2$

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$a=3$, $b=1$

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Following Aaron's advice I show here the graphs $n/(2n+1)$ and $n/(n + 2)$ which have an "approximate" $3$-fold rotational symmetry first for small $n$, then for large $n$, and for $n=3k + 1$.

You don't see the approximate symmetry for small $n$, except an exact one for the relatively rare cases $n=3k + 1$, i.e. for $4/6$ and $7/9$, resp. for $4/9$ and $7/15$:

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For larger $n$ the approximate symmetry becomes obvious while it still holds that only the graphs with $n=3k + 1$ are truly symmetric (which for large $n$ cannot be checked visually anymore):

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Here are the first graphs with $n = 3k+1$ which can visually be checked to be truly symmetric:

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