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Given the series of prime numbers greater than 9, we can organize them in four rows, according to their last digit ($d=1,3,7$ or $9$), and in $k=1,2,3\ldots$ columns corresponding to the $k$-multiple of $10$ we have to add to those four digits in order to obtain a prime number. Therefore, each prime is identified by a single point $P(k,d)$.

I illustrate this representation in the following scheme.

enter image description here

For instance, in correspondence of the column $k=15$ ($x$-axis), we find two points on the rows $d=1$ and $d=7$ ($y$-axis), because $15\cdot 10+1=151$ and $15\cdot 10+7=157$ are primes.

Within this representation of prime numbers, we can introduce $6$ sinusoidal functions of $k$ such that each prime is intercepted by one and only one of these $6$ functions:

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These $6$ sinusoidal functions are:

$$ \color{green}{ f_0(k)}=5+4\cos(\frac{\pi (k -0)}{3}),\,\,\, \color{blue}{ f_1(k)}=5+4\cos(\frac{\pi (k-1)}{3}), $$ $$ \color{orange}{ g_3(k)}=5+8\cos(\frac{\pi (k-3) }{3}),\,\,\, \color{purple}{ g_4(k)}=5+8\cos(\frac{\pi (k-4)}{3}), $$ $$ \color{brown}{ u_3(k)}=5+2\cos(\frac{\pi (k-3) }{3}),\,\,\, \color{grey}{ u_4(k)}=5+2\cos(\frac{\pi (k-4)}{3}). $$

We can underline the association between primes and functions by coloring the primes according to the (unique) function that pass through them, as shown in these pictures:

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My question is:

Can we reduce the number of these sinusoidal functions and still attain the property that each prime is reached by one and only one function?

The question is motivated by the fact that each ten (i.e. each $k$-multiple of $10$) can give rise to $0,1,2,3$ or $4$ primes, and therefore I somehow suspect that $5$ sinusoidal functions should be enough.

Thank you for your help and suggestions! Sorry for imprecision and trivialities.

NOTE: This post is a wrap of this one, in which the question was unclear.

FINAL NOTE: The answer to this question is YES (thanks to user Empy2):

Four sinusoidal functions are enough to target all prime numbers only once (i.e. each prime number is reached by one and only one wave). The four functions are $$ f_{\pm}(k)=6\pm 2\sqrt{3}\cos\left[\frac{\pi}{3}\left(k−\frac{3}{2}\right) \right], $$

$$ g_{\pm}(k)=4\pm 2\sqrt{3}\cos\left[\frac{\pi}{3}\left(k−\frac{1}{2}\right) \right]. $$

We can represent these $4$ functions as follows:

enter image description here

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There are sine-waves going through $$19,29,36,43,53,66,79,89,...\\ 13, 23, 36,49,59,66,73,83,...\\ 17,24,31,41,54,67,77,84,...\\ 11,24,37,47,54,61,71,84,...$$

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  • $\begingroup$ Thanks. But I didn't get your point. Do you mean that we can use only $4$ waves to target all the primes, each prime being targeted by only one of these four waves? $\endgroup$ – user559615 Sep 27 '18 at 6:52
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    $\begingroup$ Exactly. Multiples of 6 appear twice, and odd multiples of 3 don't appear at all. $\endgroup$ – Empy2 Sep 27 '18 at 7:03
  • $\begingroup$ Looks great! Please, can you help me to explicit these $4$ waves as a function of $k$? $\endgroup$ – user559615 Sep 27 '18 at 7:07
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    $\begingroup$ $6\pn2\sqrt3\cos((x-1,5)\pi/3)$ and $4\pm2\sqrt3\cos((x-0.5)\pi/3)$, I think. $\endgroup$ – Empy2 Sep 27 '18 at 7:11
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    $\begingroup$ Sometimes all four are needed, like 101,103,107,109 $\endgroup$ – Empy2 Sep 27 '18 at 7:18

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