First, I would like to introduce a peculiar way to display the prime numbers (greater than $9$) by means of the ten they belong to ($x$-axis), and their ending digit ($y$-axis).
Here's an example of such layout, reporting the first $25$ primes.
(You might be interested in the following conjecture, which is based on this two-dimensional representation of prime numbers).
We can now organize the primes (except $17, 19,29$) in a series of parallelograms, defined by four points corresponding to the four integers $N+1$, $N+11$, $N+49$, and $N+59$, where $N=0,3,6, 9\ldots$ denotes an increasing number of tens ($N$-parallelogram).
As it was conjectured (and then shown) in this post, on the edge of each of these parallelograms, we can find at most $7$ primes.
The red crosses in the picture below indicate the "missing primes" for each $N$-parallelogram, i.e. the integers that lie on the $N$-parallelogram (in one of the eight positions in which we could locate a prime), but that are not prime numbers.
Now, we can easily see that the missing primes divisible by $7$ are located in a well defined position on each $N$-parallelogram, as one can easily verify in the following scheme:
Each gray segment, indeed, connects four missing primes divisible by $7$. For instance, the first segment from the left connects $49, 77, 133, 161$.
Similarly, we can recognize the missing primes divisible by $17$ in correspondence of the green segments in the following picture (again, four missing primes for each segment).
For instance, the first segment from the left connects the missing primes $119, 187, 323, 391$.
It is clear that this scheme can be generalized, always yielding to a neatly organized structure (somehow cylindrical) of missing primes.
Exactly here comes my question.
Given $N$, is there an elementary way to determine the exact number of missing primes, and their position, on the $N$-parallelogram?
I tried to use the interesting comments and the clever answers related to this post and also to this one, but I could not go far. Therefore, I will be very thankful for any suggestion.
I apologize in case of confusion and/or naivety, and I will ask you also to improve the correctness of this question.
Thanks again!
You are showing the Sieve of Eratosthenes in a new geometry. Your eight dots correspond to numbers not divisible by $2$ or $5$. Your lines connecting numbers divisble by $7$ are that stage of the sieve. The straight lines come because if you move over two tens and up one you get a new multiple of $7$ because $3 \cdot 7=21$l. You haven't dealt with multiples of $3$ that I can see. The lines for $17$ come because $3\cdot 17=51$ and so on.