Steven Clark

Here's a link to my OEIS revisions. Could you please provide a link to your proposed OEIS entry?

It might be worth mentioning conjectured formula (2) which I verified for $1\le n\le 10,000$ in your proposed OEIS entry, but it'd be better if we had a proof as they tend to favor proven formulas over conjectured formulas in OEIS entries.

I believe your sequence $$a(n)=\frac{1}{i \pi} \log(\sec(\pi\, \pi(n)))=\frac{1}{i \pi} \log\left((-1)^{\pi(n)}\right)\tag{1}$$ can be evaluated as $$a(n)=-\sum\limits_{k=1}^n \text{A325699}(k)\, M\left(\left\lfloor\frac{n}{k}\right\rfloor\right)\tag{2}$$ where $$M(n)=\sum\limits_{k=1}^n \mu(k)\tag{3}$$ is the Mertens function given by OEIS Entry A002321 and $\mu(n)$ is the Möbius function given by OEIS Entry A008683.

I found the seemingly related OEIS entry A325699 which gives the number of distinct even prime indices of $n$ minus the number of distinct odd prime indices of $n$.

I suppose if you want, but there's really no need.

I suppose if you defined it as $$a(n)=\frac{1}{i \pi} \log(\sec(\pi\, \pi(n)))=\frac{1}{i \pi} \log\left((-1)^{\pi(n)}\right)$$ it'd qualify as an integer sequence.

You plot is basically a discrete plot of $$\log(\sec(\pi \,\pi(n)))=\log\left((-1)^{\pi(n)}\right)$$ which takes on the values $i \pi$ when the prime-counting function $\pi(n)$ is odd and $0$ when the prime-counting function $\pi(n)$ is even which explains its alternating behavior at $n\in\mathbb{P}$.

Riemann derived the explicit formula for $\Pi(x)=\sum\limits_{n=2}^x \frac{\Lambda(n)}{\log(n)}$ via a Fourier inversion, but I don't believe Fourier analysis is limited to primes or the Riemann hypothesis. I believe Fourier analysis can be applied to any function of the form $f(x)=\sum\limits_{n=1}^x a(n)$ where the corresponding Dirichlet series $\sum\limits_{n=1}^\infty \frac{a(n)}{n^s}$ converges at least for $\Re(s)\ge2$ (see this answer I posted to a question on an entire function interpolating $\mu(n)$).

This second formula is related to an analytic representation of the prime-power counting function $K(x)=\sum\limits_{p^j\le x} 1$ and its corresponding first-order derivative $K'(x)$.

Or try evaluating $$\lim\limits_{N,f\to\infty} \left(-\frac{1}{f} \sum\limits_{n=1}^N\frac{\mu(n)\, \nu(n)}{n} \sum\limits_{k=1}^{f n}\cos\left(\frac{2 \pi k x}{n}\right)\right)$$ where $\mu(n)$ is the Möbius function, $\nu(n)$ is the number of distinct primes dividing $n$ (see OEIS Entry A001221), and the evaluation frequency $f$ is once again assumed to be a positive integer and you'll see it converges exactly to $1$ at prime-powers $x=p^j$ and to zero at other integer values of $x$ when $0<x\le N$.

Try evaluating $$\lim\limits_{N,f\to\infty} \left(-\frac{1}{f} \sum\limits_{n=1}^N\frac{\mu(n)\, \log(n)}{n} \sum\limits_{k=1}^{f n}\cos\left(\frac{2 \pi k x}{n}\right)\right)$$ where $\mu(n)$ is the Möbius function and the evaluation frequency $f$ is assumed to be a positive integer and you'll see it converges exactly to $\log(p)$ at prime-powers $x=p^j$ and to zero at other integer values of $x$ when $0<x\le N$. This formula is related to another analytic representation of $\psi(x)$ and its corresponding first-order derivative $\psi'(x)$.

Your MatLab code seems to take positive steps at odd-indexed primes $a(n)=p_{2 n-1}$ (see OEIS Entry A031368) and negative steps at even-indexed primes $a(n)=p_{2 n}$ (see OEIS entry A031215), but what does this have to do with your question here?

I was thinking of a power series for $f(t)$ in the context of deriving an alternative representation of $F(s)=\mathcal{M}_t[f(t)](s)$ to compare against your derived formula for $F(s)=\mathcal{M}_t[f(t)](s)$ to serve as a double-check on the correctness of your derivation. But I believe the answer to the question you linked is no for the reason I pointed out.

I posted the results of my investigation below which perhaps provides some insight with respect to a closed-form representation for $F(s)=\mathcal{M}_t[f(t)](s)$.

I've been thinking about deriving a power series for $f(t)$ via reversion of a power series for $t\, \text{csch}(t)$, but since $t\, \text{csch}(t)$ has poles at $2 \pi i c\,,\ c\in\mathbb{Z}\land c\ne 0$, there is no power series for $t\, \text{csch}(t)$ that converges for $0<t<\infty$. Therefore a power series for $f(t)$ derived via reversion of a power series for $t\, \text{csch}(t)$ would not converge for $0<t<1$. For example, the MacClaurin series $$t\, \text{csch}(t)=1-\sum\limits_{n=1}^{\infty} \frac{\left(2^{2 n}-2\right) B_{2 n}}{(2 n)!} t^{2 n}$$ has a radius of convergence of $\pi$.

So I believe you have sign errors in your first two formulas for $M_t(f(t))$ which you seem to have compensated for in the linked Mathematica code. Your formula for $\operatorname{sinhc}^{-1}(x)$ initially indicates $\pm$, but then once again corrects for the sign error. While its true $t\, \text{csch}(t)$ is an even function of $t$, the inverse Mellin transform of $\mathcal{M}_x[f(\tau)](s)$ should return the positive solution, not the negative solution.

For the Mellin transform $$F(s)=\mathcal{M}_{\tau}[f(\tau)](s)=\int\limits_0^1 f(\tau)\, \tau^{s-1} \, d\tau,$$ I believe the substitution $\tau\to g(t)=t\, \text{csch}(t)$ leads to $$F(s)=\int_{\infty}^0 f(g(t))\, g(t)^{s-1} \, dg(t)=\int_{\infty}^0 t\, g(t)^{s-1}\, g'(t) \, dt.$$ I mentioned you had the limits reversed in an earlier comment as $\tau=0$ and $\tau=1$ before the substitution correspond to $t=\infty$ and $t=0$ respectively after the substitution.

After thinking about it some more, I don't believe Ramanujan's master theorem will be useful here at its based on a power series around $x_o=0$ and $f(x)$ doesn't converge at $x=0$ which seems to imply such a power series doesn't exist.

It seems to work for $$\mathcal{M}_x[x\, \text{csch}(x)](s)=\left(2-2^{-s}\right) \Gamma(s+1)\, \zeta(s+1)=\Gamma(s)\, \phi(-s)$$ because in $$\phi(s)=\frac{\left(2-2^s\right) \Gamma(1-s)\, \zeta(1-s)}{\Gamma(-s)}$$ the poles of $\Gamma(1-s)$ in the numerator cancel the poles of $\Gamma(-s)$ in the denominator, but you have to evaluate it as $$\phi(n)=\underset{s\to n}{\text{lim}}\frac{\left(2-2^s\right) \Gamma(1-s)\, \zeta (1-s)}{\Gamma (-s)}.$$

I don't understand your notion of applying Ramanujan's master theorem which is used to derive the Mellin transform from a series representation of a function. Is your notion to derive $\mathcal{M}_x[f(x)](s)=\int\limits_0^{\infty} f(s)\, x^{s-1} \, dx$ by applying Ramanujan's master theorem to a series representation of $f(t)$ derived via series reversion (Mathematica InverseSeries function) of a series representation for $t\, csch(t)$?

I don't understand the early part of your derivation. You define $f(x)$ as the inverse of $x\, \text{csch}(x)$ and then you indicate the substitution $t\to f(t)$, but you actually use the substitution $t\to t\, \text{csch}(t)$. It appears you have the limits reversed after this substitution as $t=0$ and $t=1$ before the substitution corresponds to $t=\infty$ and $t=0$ respectively after the substitution.

I believe the $(-1)^{n-s}$ term in your binomial sum should just be $(-1)^n$ as in $$\left(\frac{t}{e^t-e^{-t}}\right)^s=\sum\limits_{n=0}^{\infty} \binom{-s}{n}\, (-1)^n\, t^s\, e^{-(2 n+s) t}$$ and since $$-\frac{2^s}{s} \int_0^{\infty } t^s\, e^{-(2 n+s) t} \, dt=-2^s\, \Gamma(s)\, (2 n+s)^{-s-1}$$ I believe the Mellin transform result should actually be $$-2^s\, \Gamma(s)\, \sum\limits_{n=0}^{\infty} \binom{-s}{n}\, (-1)^n\, (2 n+s)^{-s-1}$$ and consequently it would appear your subsequent inverse Mellin transform is wrong.

The question here seems to be related to this earlier question. It's easy to show there's at least one other case other than the zeros of $\zeta(s)$. If $\Im(\zeta(\alpha+i \beta))=0$, then $\zeta (\alpha+i \beta)=\zeta (\alpha-i \beta)$ independent of whether or not $\Re(\zeta(\alpha+i \beta))=0$.

For $$f(s)=1-\sum\limits_{n=2}^{\infty}\, (-1)^n\, n^{-\Re(s)} \left(\sin\left(\frac{\pi}{2}-\Im(s) \log(n)+\varphi_1\right)+i\, \sin(-\Im(s) \log(n)+\varphi_2)\right)$$ can you define examples of the values of the pair $\left(\varphi_1,\varphi_2\right)$ that you think both preserve and don't preserve the locations of the zeta-zeros? The only thing obvious to me is $\left(\varphi_1,\varphi_2\right)=\left(2 k_1 \pi,2 k_2 \pi\right)$ where $k_1,k_2\in\mathbb{Z}$ preserves the locations of the zeta-zeros since in this case $f(s)$ evaluates to $\eta(s)$.

You need to account for the $n=1$ term which can be accomplished as follows $$\eta (s)=1-\sum\limits_{n=2}^{\infty}\, (-1)^n\, n^{-\Re(s)} \left(\sin\left(\frac{\pi}{2}-\Im(s) \log(n)\right)+i\, \sin(-\Im(s) \log(n))\right)$$ but I don't understand what you're doing with the phase shifting.

I don't understand your motivation or what you're doing. You can't shift the sine functions phases by an integer multiple $k$ of $\frac{\pi}{\ln(n)}$ because for $n=1$ you end up with the infinite expression $\frac{k\, \pi}{0}$.

The zeros of $\zeta(s)$ are located where the contours of $\Re(\zeta(s))=0$ intersect the contours of $\Im(\zeta(s))=0$ (see Contour Plots of the Zeta Function).

It doesn't really matter if you can evaluate $\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-s}$ first and then separate the real and imaginary parts at the end, or evaluate the real and imaginary parts separately along the way. You can also apply trigonometric identities to make $\eta(s)=\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-\Re(s)}\, (\cos(-\Im(s)\, \log(n))+i\, \sin(-\Im(s) \log (n)))$ look like something different, but I don't really see the point.

I'm not sure I understand what you're trying to do, but its true the Dirichlet eta function $$\eta(s)=\left(1-2^{1-s}\right) \zeta (s)=\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-s}\,,\quad\Re(s)>0$$ has the same zeros as the Riemann zeta function in the critical strip $0<\Re(s)<1$, and it's also true that $$n^{-s}=n^{-\Re(s)}\, (\cos(-\Im(s)\, \log(n))+i\, \sin(-\Im(s)\, \log(n)))$$ so the Dirichlet eta function can also be evaluated as $$\eta(s)=\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-\Re(s)}\, (\cos(-\Im(s)\, \log(n))+i\, \sin(-\Im(s) \log (n))),\quad\Re(s)>0.$$

Do you mean is it correct to argue the following assuming the Riemann hypothesis?

Does it really have to be a "text file" as suggested by the title, or can it be a binary file which would provide the ability to encode more primes?

Assuming $f(t)=\left(\left\lceil\frac{t!}{t^2}\right\rceil-\left\lfloor\frac{t!}{t^2}\right\rfloor\right)$ and $t\in\mathbb{Z}$, there is no correct result for $\underset{t\to\infty}{\text{lim}}f(t)$ since $f(t)$ takes on the values $0$ and $1$ both infinitely often. The only thing that can be said is $f(t)$ evaluates to $1$ much less frequently than it evaluates to $0$ as $t\to\infty$ which is predicted by the prime number theorem $\pi(t)\approx\frac{t}{\log t}$.

Mathematica simplifies $\left(\left\lceil\frac{t!}{t^2}\right\rceil-\left\lfloor\frac{t!}{t^2}\right\rfloor\right)$ to $\left(\left\lceil\frac{\Gamma(t)}{t}\right\rceil-\left\lfloor\frac{\Gamma(t)}{t}\right\rfloor\right)$, and note that $\left(\left\lceil\frac{\Gamma(t)}{t}\right\rceil-\left\lfloor\frac{\Gamma(t)}{t}\right\rfloor\right)$ is only zero when $t$ is composite, so for continuous $t$ it evaluates to one the vast majority of the time since the reals are uncountably infinite.

Could you please give a link to your Wolfram Alpha result? I'm curious about the exact expression you evaluated. Did you just evaluate $\underset{t\to \infty }{\text{lim}}\left(\left\lceil \frac{t!}{t^2}\right\rceil -\left\lfloor \frac{t!}{t^2}\right\rfloor \right) \left(\left\lceil \frac{(t+2)!}{(t+2)^2}\right\rceil -\left\lfloor \frac{(t+2)!}{(t+2)^2}\right\rfloor \right)$?

@TymaGaidash It's actually a finite sum. I've now verified the Mathematica results for $t_{1}$ to $t_{14}$ are consistent with the OEIS entry and the formula I gave above. Each successive $t_n$ value takes longer and longer for Mathematica to derive the result for the limit of the derivative, so I'm going to abort the evaluations.

@TymaGaidash The OEIS sequence starts with $n=0$ instead of $n=1$. Starting at $n=1$, Mathematica gives the first $12$ values of $t_n=\underset{w\to 1}{\text{lim}}\left(\frac{\partial^{n-1}}{\partial w^{n-1}}\left(\frac{w-1}{e^{\left(\frac{1}{w}-1\right) e^{\frac{1}{w}-1}}-1}\right)^n\right)$ as {-1,5,-41,468,-6854,122582,-2589978,63129392,-1743732192,53827681152,-1836453542472,68620052332752} which matches the corresponding values of oeis.org/A305981 except in sign, and which exactly matches the formula $t_n=(-1)^n \sum\limits_{k=1}^n \left| S_n^{(k)}\right| k^k$.

@TymaGaidash I said it's related, but it's not an exact match, which is why the $(-1)^n$ precedes the sum in the formula I gave where the sum is based on oeis.org/A305981.

@TymaGaidash Based on the first 10 values of $t_n$, I believe $t_n$ is actually related to oeis.org/A305981 and is given by $t_n=(-1)^n \sum\limits_{k=1}^n \left| S_n^{(k)}\right| k^k$ where $S_n^{(k)}$ is the Stirling number of the first kind.

The discrepancy between the evaluation of the sum and the numerical evaluation of the integral made me suspicious of a possible error in the derivation based on the Abel-Planar formula. I reviewed the derivation and couldn't find an error, but perhaps I missed something. I have a series representation of the integral which evaluates similar to the numerical evaluation which I'll post when I find a bit more time.

With respect to derivation of a closed form result for the integral, I was investigating integration by parts and I thought I had a result but I discovered I accidentally wrote the denominator as E^(2 Pi x - 1) instead of E^(2 Pi x) - 1.

I just subtracted the $\frac{1}{2}$ term from your earlier formula where the $\frac{1}{2}$ term was based on your earlier comment and observational convergence of $\int_0^y\left(C(x)+S(x) -\sqrt{\frac \pi 2}\right) dx\to -\frac{1}{2}$ as $y\to\infty$. I see you updated the formula in your question in a similar manner. I thought I was close to a result for $2\int_0^\infty\frac{S(x)-C(x)}{e^{2 \pi x}-1}\,dx$ but I found an error in my derivation so back to the drawing board.

With respect to the discrepancy between the evaluation of the sum and the numerical evaluation of the integral, I tried increasing the precision of the evaluations and also evaluating the sum out to $100,000$ terms but it didn't seem to resolve the discrepancy.

I obtained $2\int_0^\infty\frac{S(x)}{e^{2 \pi x}-1}\,dx-2\int_0^\infty\frac{C(x)}{e^{2 \pi x}-1}\,dx-\frac{1}{2}-\frac{\sqrt{\pi }}{2 \sqrt{2}}$. I'm not sure it's equivalent but I noticed you deleted the $2$ preceding the two integrals, changed the sign of one of the two integrals, and added $\frac{1}{2}$ instead of subtracting $\frac{1}{2}$.

I don't understand how you derived the final result $-\frac{\sqrt\pi}{2\sqrt2}-2\int_0^\infty\frac{C(x)}{e^{2\pi x}-1}dx+2\int_0^\infty \frac{S(x)}{e^{2\pi x}-1}dx$ in the derivation based on the Abel-Plana formula in the question above. Also I believe there's one spot where $\frac{1}{2}$ should be $\frac{\sqrt\pi}{2\sqrt2}$ in the derivation.

It seems to me your final result is missing the two integrals $\int_0^\infty\left(C(x)-\frac{\sqrt{\pi }}{2 \sqrt{2}}\right)\,dx+\int_0^\infty\left(S(x)-\frac{\sqrt{\pi }}{2 \sqrt{2}}\right)\,dx$. Are you saying these two integrals sum to zero?

If you're using the simpler definitions it seems to me the $\frac{1}{2}$ terms in your title and related formulas should really be $\frac{1}{2}\sqrt{\frac{\pi}{2}}=\sqrt{\frac{\pi}{8}}$.

There are several representations at functions.wolfram.com/GammaBetaErf/FresnelC and functions.wolfram.com/GammaBetaErf/FresnelS but I haven't had time to look at them. The simple definitions correspond to $\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} x\right)$ and $\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} x\right)$ when using the Wolfram definitions.

Actually I believe the formulas for erfc(y) and erf(y) at math.stackexchange.com/q/2380164 are valid for something like Re(y)>0 & |Re(y)|>|Im(y)| which explains why the formulas for erfc(sqrt(z)) and erf(sqrt(z)) are valid for Re(z)>0.

See wolframalpha.com/input/… and wolframalpha.com/input/….