( I've been thinking about asking this for a long time . Though this is not rigorous; It can be thought of as heuristic or extraction of information from different viewpoint.)

We know "super-regularized"( the term coined by authors of the paper : http://cds.cern.ch/record/630829/files/sis-2003-264.pdf) product of primes $4π²$ i.e.

$$\infty \# = \prod_{k=1}^\infty p_k = 4\pi^2$$

Now , by operating by $\log$ and converting product to summation we get :

$$\sum_{k=1}^\infty \log(p_k)$$ .

In their paper they considered (for strange reasons somehow) $1$ as prime.

So, $G(k)=p_{(k+1)}= p(k+1)$ can be considered our new $n$th prime function ( "most natural" $n$th prime function with all nice analytic properties desired for Abel Plana Summation)

Let's define $F(k)= \log(p_{(k+1)})$

Now ,we can use Abel -Plana or Ramanujan Summation as follows :

$$i\int_0^\infty\frac{F(ix)-F(-ix)}{e^{2\pi x}-1}\,dx = \frac{1}{2}F(0)+\sum_{n=1}^\infty F(n) $$

Now ,we can see that $F(0) = 0$

so , now

$$i\int_0^\infty\frac{F(ix)-F(-ix)}{e^{2\pi x}-1}\,dx = \log(4π²)$$

Question : Can we "guess" the function $F(x)$ ( as we desire ) from the above equation ?

Approach :

Consider we have some "suitable" infinite product like $4π² = \prod ^ \infty a_n $ ( lots of infinite product for numbers like $π$ and $\sqrt2$ exist and lots of we can successively construct.)

Hence , we can get an infinite sum by taking $\log$ on both sides .

Now , equation (t):

$$i\int_0^\infty\frac{\log(\frac{p(1+ix)}{p(1-ix)})}{e^{2\pi x}-1} dx = \sum \log(a_n)$$

Now , Assume $$i\log(\frac{p(1+ix)}{p(1-ix)})$$ is continuous and $\infty$ - differentiable ( should be ?).

So ,it has a Taylor series around $0$ : $$\sum \frac{c_n x^n}{\Gamma(n+1)}$$

Now ,we know the integral $$\int_0^\infty \frac{x^{(a-1)}}{e^{b x} - 1} = \frac{\Gamma(a)\zeta(a)}{b^a}$$

So ,the equation $(t)$ becomes $$\sum _{n=1}^\infty \frac{c_n\zeta(n+1)}{(2π)^{(n+1)}}$$

Now only thing remaining is correlation between $\log(a_n)$ and $\frac{c_n\zeta(n+1)}{(2π)^{(n+1)}}$.

Can we guess the better $p_n$ using above analysis (by using some known facts like approximately $p_n$ is $n\ln(n)$ or using bounds on $p_n$) ?

( Motivation : I calculated various regularized products by using this method and it worked. )

(Note : I know this is an $\infty$-ly underdetermined system . So don't consider this a trivial or anything I'm just asking for suitable "guess" using the known facts. )

(Apologies in advance if there are any partial mistakes )

(Edit 1: consider $F(k)=\ln(k\ln(k)+ E(k))$ for initiation. ( Use of calculation of variation /?) .Here $E$ is sort of an error function. You can replace the $k\ln(k)$ by better known bounds on $p_n$ for example by P. Dusart but it'll become mess)

Edit 2: the general solution to the integrand in equation (t) is :

$$p(x) = e^{\frac{f(x)}{2}}E(x)$$ Where, $E(x)$ is any even function

$$f(x) =\sum_{n=0}^\infty \frac{a_n(2π)^{2n+2}(-1)^{n+1}(x-1)^{2n+1}}{\zeta(2n+2)\Gamma(2n+2)}$$

with $a_n$ being $\log$ of terms in the infinite product for $4π²$

Now ,the question is for $E(x)$.

(Can we use calculus of variations to get more insights? I haven't tried it.)

Edit 3:

Had an email exchange with Sir Roger Heath-Brown about the validity of argument . He confirms that argument is valid but also suggested that to uniquely find out the function for nth prime I have to use some additional known unconditional information of $n$th prime like growth condition of $n$th prime and then carefully selecting the coefficients to construct the function explicitly.

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