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I have been playing with the following function: $$ f(x)=\frac{\pi x (1-x^2)}{\sin\pi x}\prod_{k=2}^\infty \frac{\sin(\pi x/k)}{\pi x/k} $$ It is hard to get correct numerical values. I'll start with the basic. Can you confirm that (1) we have absolute convergence, (2) the limit exists if $x$ is an integer, (3) if $x$ is a composite number then $f(x)=0$, and if $x$ is prime, $f(x)\neq 0$?

For simplicity, let's consider the absolute value of $f(x)$. Now the interesting part. If $x$ is not too close to a composite number, it sounds like $$ |f(x)|\sim \exp(-\lambda x) $$ for some $\lambda >0$, possibly $\lambda\approx 4.5$. Is there an asymptotic formula that can be easily derived? Since $f(x)=0$ if $x$ is composite, that formula would be valid only for some values of $x$, for instance if the fractional part is within some range, or if $x$ is prime, which seems to be where the formula works best.

If this was true, you could approximately compute the number of primes $<n$ as $$\pi(n)\approx \sum_{k=2}^n e^{\lambda k} |f(k)|. $$ The formula is useless for computational purposes, but I am wondering if it might have some theoretical interest. Anyway, my question is this: can you get some asymptotic formula as $x\rightarrow\infty$ depending on the fractional part of $x$ or if $x$ is prime? Mine might not be correct. Even better, what is the value of $\lambda$ assuming it ever exists?

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    $\begingroup$ (1) follows from the Taylor approximation $\sin t=t+O(|t|^3)$ while (2) and (3) follow from (1) and from $\sin(\pi z)$ having a simple pole at each integer. You might be interested Alain Connes' paper "Around Wilson's theorem". $\endgroup$
    – Ofir Gorodetsky
    Commented Jan 24, 2023 at 15:09
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    $\begingroup$ Related to this math.stackexchange.com/q/3529718/789323 ? $\endgroup$
    – bambi
    Commented Apr 8, 2023 at 5:58

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I believe that the doubly infinite product $\prod _{k=1} ^\infty \prod_{m=1} ^\infty \left( 1 - \frac {x^2} {k^2 m^2} \right)$ is absolutely convergent and equals $f(x) \frac {\sin^2 (\pi x)} {(\pi x)^2 (1 - x^2)}$. Is that wrong? I am just using $\frac {\sin(\pi x)} {\pi x} = \prod_{m=1} ^\infty \left(1 - \frac {x^2} {m^2} \right)$ here.

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  • $\begingroup$ Writing $\sin(\pi x)/(\pi x)$ as an infinite product, and grouping each factor $1/[1-(x/k)^2]$ with each factor $\sin(\pi x/k)/(\pi x /k)$, may help understand what is going on. $\endgroup$
    – Vincent Granville
    Commented Jan 24, 2023 at 13:19
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    $\begingroup$ The doubly infinite product just equals $\prod_{n=1}^\infty ( 1 - \frac{x^2}{n^2})^{d(n)}$ where $d(n)$ is the number of positive divisors of $n$, of course. $\endgroup$
    – Michael Rieck
    Commented Jan 24, 2023 at 14:16
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    $\begingroup$ I haven't double-checked your formula, I assume it is correct; I expect it to be something like that. Thank you for your answer! But I have a more difficult time dealing with the asymptotic formula, valid for primes it seems, probably for many non-integer numbers, but of course not for composite integer numbers. Hoping to get some help with that. $\endgroup$
    – Vincent Granville
    Commented Jan 24, 2023 at 18:14
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    $\begingroup$ Vincent, perhaps this will be helpful: Let $F(z) = \prod_{n=1}^\infty (1-\frac{z^2}{n^2})^{d(n)} = \sin^2(\pi z) f(z) / [ (\pi z)^2 (1-z^2) ]$. Then, $F'(z) / F(z) = \frac{d}{dz} \sum_{n=1}^\infty d(n) \ln(1-\frac{z^2}{n^2}) = 2 z \sum_{n=1}^\infty \frac{d(n)}{z^2-n^2}$. $\endgroup$
    – Michael Rieck
    Commented Jan 25, 2023 at 14:12

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