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[Dec 3 2023] I am currently studying analytic number theory and my teacher suggested to ask here if the following sum

$$S(x) = \sum_{p} x^p = x^2 + x^3 + x^5 + x^7 + ...$$

Where $p$ is a prime number is known and if it has an asymptotic behaviour.

Motivation:

If you square the sum

$S(x)^2 = (\sum_{p} x^p)^2$

You get

$S(x)^2 = \sum_{n=p+q} a_{n}x^{n}$

Where $p$ and $q$ are primes and $a_{n}$ is an integer.

It is not hard to see that Goldbach's conjecture is equivalent to claim that the $2k^{th}$ derivative of $S^2$ at $x=0$ is different than $0$, for all positive integer $k$.

[Mar 10 2025] (edit)

Question:

Can we find an expression that approximates arbitrarily well $S(x)$ for special values of $x \in \mathbb{D}$?

That is, a simpler function (in the sense that is easier to evaluate) $A(x)$ such that $$S(x)=A(x)+a(x)$$

With $a(x)$ acting as an error function. Something that tends to $0$ as $x$ tends to one of those "special values".

As an example, it'd be interesting to see the behaviour of $S(x)$ as $x$ approaches a root of unity.

Thank you

Najdorf
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  • Asymptotic for what $x$? Are you assuming that $|x| < 1$ ? – Nilotpal Sinha Dec 06 '23 at 15:49
  • Yes. If the absolute value of $x$ is equal or greater than $1$ then the sum does not converge. The thing is that it looks like there shoud be an expression (or at least an approximation) to that infinite sum, right? – Najdorf Dec 06 '23 at 22:19
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    $S(x)$ has the unit circle as natural boundary, so it cannot be continued at any point there - not sure though if this has any relevance to what you are interested in as that seems to be more for $x$ near zero – Conrad Dec 07 '23 at 04:39
  • Yes, for $x$ near zero. – Najdorf Dec 07 '23 at 07:19
  • I can understand the motivation for the Goldbach conjecture, but why in the world would you want an approximation? – Supernerd411 Mar 10 '25 at 11:40
  • $S(x)=x^2+O(x^3)$. – Gerry Myerson Mar 10 '25 at 11:41
  • @GerryMyerson that only works for $x$ near $0$. – Najdorf Mar 10 '25 at 11:44
  • @Supernerd411 because I'm a super nerd for prime numbers – Najdorf Mar 10 '25 at 11:44
  • @Najdorf I can relate. I have an unhealthy obsession with pi. – Supernerd411 Mar 10 '25 at 11:45
  • @Najdorf But what would you do with an approximation? – Supernerd411 Mar 10 '25 at 11:46
  • Well, Najdorf, you were the one who wrote "for $x$ near zero". So, make up your mind – what are you really interested in? Also, have a look at the earlier question about this series, https://math.stackexchange.com/questions/602706/generating-function-for-the-characteristic-function-of-primes and by the way, I love your variation in the Sicilian Defense. – Gerry Myerson Mar 10 '25 at 11:48
  • @Supernerd411 I'm currently doing my degree thesis (not sure how it's called in english) on the circle method and I'm focusing on the partition problem. There, a functional equation is used to approximate the partition function for values near the roots of unity. I was wondering if this very idea could be extrapolated using this other infinite sum – Najdorf Mar 10 '25 at 11:51
  • @Najdorf Isn't the partition function just defined for natural numbers? Is there an extension? – Supernerd411 Mar 10 '25 at 11:54
  • @GerryMyerson thanks, I'll check the link. And yes, when I first asked the question I thought it was interesting to check the behaviour near $0$ but my motivations have slightly changed. See my long comment to "supernerd411". Also lately I'm playing the Caro-Kann more often, so maybe I should make up my mind with the name I use here as well! – Najdorf Mar 10 '25 at 11:57
  • @Supernerd411 There is a generating function of the form $$\sum_{n} p(n)x^{n}$$ defined at the unit disk – Najdorf Mar 10 '25 at 12:00
  • But the values the partition function is taking are still $n$ which are natural aren't they? I'm sorry if there's something obvious here which I am not understanding :) – Supernerd411 Mar 10 '25 at 12:02
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    Btw Najdorf I like Dragon better. Yours is too theory based. – Supernerd411 Mar 10 '25 at 12:04
  • @Supernerd411 nono, I might be the one not explaining it correctly. The function I put there can be seen as a function of $x$, but it looks so hard to evaluate. But what happens if $x$ is near a root of unity? well, that expression can be aproximated by a simpler function with extreme presition. Hope I'm making myself clear! Glad to try to explain it differently – Najdorf Mar 10 '25 at 12:06
  • @Supernerd411 +1, the dragon is more conceptual – Najdorf Mar 10 '25 at 12:07
  • I am confused because you said that you use a functional equation to approximate the partition function for values near the roots of equity. But as far as I know, the partition function is only defined for natural numbers. Also, I love the line in the Caro-Kann Advanced in which you push c5. – Supernerd411 Mar 10 '25 at 12:14
  • yes my mistake. I use a functional equation to approximate the "generating partition function", that is, approximating that weird sum near special values of x. And yes, that line is cool but I hate it for black – Najdorf Mar 10 '25 at 12:20
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    Ah ok. Doubt cleared :) Also, HOW DARE YOU?! – Supernerd411 Mar 10 '25 at 12:23
  • @Najdorf Try making a continued fraction for this generating function. I believe prime gaps will also come into play here. – Supernerd411 Mar 10 '25 at 12:54

1 Answers1

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For Question 1, Since every primes $\ge 5$ are of the form $6k \pm 1$, by summing up the geometric sequences $x^{6k-1} + x^{6k+1}$ for $k = 1,2,\ldots, \infty$ and adding $x^{2} + x^{3}$, and taking advantage of the fact the the density of primes among the first few numbers of these form is high we get

$$ \sum_{p \ge 2} x^p = \frac{1 + x + x^3 + x^5 -x^6 - x^7}{x^4(x - x^7)} + O(x^{25}) $$

Nilotpal Sinha
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