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?

it's related because the summation he is using over the $\cos(x)$ function can be replaced with an integral and $\tan(x)$ can be used instead which includes the cosine function

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 $\psi'(x)$.

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 $0<x\le N$.

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 $K'(x)$.

Well, trigonometry clearly has a lot to do with RH, and that's what I claim in my two preprints . here a snippet from the first preprint "$G(n) = \text{Imaginary}(f(n))/\pi$, where $f(n) = \ln(\sec(\pi \cdot n\log(n)))$. This expansion involves sine and cosine functions. After substitution and rearrangement, we obtain: $$G(n) = \ln(\sin(\pi \cdot n\log(n))) - \ln(\sec(\pi \cdot n\log(n)))$$ From the above analysis, we conclude that $a(n) \equiv G(n)$, which can be expressed as: $$a(n) \equiv G(n)$$"

in here a(n) includes the Möbius function as in this equation expressed as $\mu(k)$ $$a(n) = 1 + \sum_{k=1}^{\infty} \left(\frac{\mu(k)}{k}\right) \sum_{j=1}^{\infty} \left(\frac{n^{\rho_j/k}}{\rho_j}\right) + O(\log n)$$

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 $\mu(n)$).

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 $n\in\mathbb{P}$.

It could be submitted to OEIS, right?

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.

Can I mention your name as a collaborator?

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

Alright, am waiting for my account approval

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 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.

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.

Done, waiting for approval of the sequence edit.

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

Here is the link