Are there infinite prime of this form ? $\lfloor n\sin^{2}\left(n\right)+1\rfloor$ where $n$ is a prime .

Question

If $n$ is a prime let's define (we are in radians):

$$\lfloor n\sin^{2}\left(n\right)+1\rfloor$$

Do we have infinitly primes of the form $$\lfloor n\sin^{2}\left(n\right)+1\rfloor$$

For small primes we have :

$(11,11),(13,3),(61,57),...$

As a random trial, we have an example: $n=100043$. It gives us $72467$, which is a prime. In this case we have: $$100043=2+3 \cdot 33347$$ $33347$ is a prime, so we have a new pair: $(33347,17837)$.

Perhaps we can use a famous result due to Euler:

$$x\sin^{2}\left(x\right)=x\left(x\prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)\right)^{2}$$

Following the advice @Peter if $1<y<2$:

$$y=x\sin^{2}\left(x\right)+1=\exp\left(-\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(x\sin^{2}\left(x\right)\right)^{n}}{n}\right)$$

If We have :

$$f_{mi}\left(x\right)=\exp\left(-\sum_{n=1}^{i}\frac{\left(-1\right)^{n}\left(x\left(x\prod_{k=1}^{m}\left(1-\frac{x^{2}}{k^{2}\pi^{2}}\right)\right)^{2}\right)^{n}}{n}\right)$$

Then $C$ a positive integer :

$$f_{mi}(C\pi)=f_m(C\pi)f_i(C\pi)=1$$

Perhaps it's a multiplicative function https://en.wikipedia.org/wiki/Multiplicative_function .But I'm not sure...

Nota bene :

For $p$ an integer sufficiently large ,$i,m\in[p,\infty]$ $\exists \varepsilon \in[0,1],x\in[C\pi-\varepsilon ,C\pi+\varepsilon]$ such that :

$$x\sin^2(x)+1\leq \exp\left(-\sum_{n=1}^{i}\frac{\left(-1\right)^{n}\left(x\left(x\prod_{k=1}^{m}\left(1-\frac{x^{2}}{k^{2}\pi^{2}}\right)\right)^{2}\right)^{n}}{n}\right)$$

Last attempt :

We need to find see Willans formula https://en.wikipedia.org/wiki/Formula_for_primes :

$$p_n-1=\sum_{i=1}^{2^{n}}\operatorname{floor}\left(\left(\frac{n}{\sum_{j=1}^{i}\operatorname{floor}\left(\left(\cos\left(\frac{\left(\left(j-1\right)!+1\right)}{j}\pi\right)\right)^{2}\right)}\right)^{\frac{1}{n}}\right)=\lfloor k\sin^2(k)\rfloor$$

Do we have infinite primes of this kind ? If yes, how do we show it? As in my trial example, what is the maximum number of pairs we can make if we subtract by two and divide by three?

PS: Can you propose a name for this kind of pair of primes?

Answer 1

Not and answer, but a possible source of inspiration ...

Let's modify the expression a little bit $$a_n=\lfloor n\cdot\sin^2{n}+1\rfloor= \lfloor n +1-n\cdot\cos^2{n}\rfloor$$ If we enforce $0<1-n\cdot\cos^2{n}<1$ then $a_n=n$.

This means $$1 > n\cos^2{n}>0 \Rightarrow \sin^2{n}>(n-1)\cdot\cos^2{n} \Rightarrow\\ \tan^2{n}> n-1$$ Possible solutions are scarce:

but are grouped around $n \approx\frac{k\pi}{2}$, leading to the numerators of the convergents of $\frac{\pi}{2}$ or A096456. Question is (thus, not an answer): how many of those numerators are prime?

I found a larger one 4846147.

And another one 1169809367327212570704813632106852886389036911

I also checked the 1st 100 numerators.

Answer 2

Partial answer.

Conjecture in radian :

Let $\exists p$ be a prime then we have :

$$\sin^{2}\left(p\right)<\frac{Cp}{p+1}$$

If i'm not wrong we have :

$$\sum_{i=1}^{n}\operatorname{floor}\left(\left(\frac{n}{\sum_{j=1}^{i}\operatorname{floor}\left(\left(\cos\left(\frac{\left(\left(j-1\right)!+1\right)}{j}\pi\right)\right)^{2}\right)}\right)^{\frac{1}{n}}\right)=n$$

Remains to show that for $n$ sufficiently large :

$$A=p_{n-1}-1=\sum_{i=1}^{2^{{n-1}}}\operatorname{floor}\left(\left(\frac{n}{\sum_{j=1}^{i}\operatorname{floor}\left(\left(\cos\left(\frac{\left(\left(j-1\right)!+1\right)}{j}\pi\right)\right)^{2}\right)}\right)^{\frac{1}{n}}\right)=Ct\left(\operatorname{floor}\left(\sum_{i=1}^{p_n}\operatorname{floor}\left(\left(\frac{n}{\sum_{j=1}^{i}\operatorname{floor}\left(\left(\cos\left(\frac{\left(\left(j-1\right)!+1\right)}{j}\pi\right)\right)^{2}\right)}\right)^{\frac{1}{n}}\right)\right)\right)=B=\frac{Cp_n}{p_n+1}$$

Where $t(x)=x^2/(x+1)$

But the inverse function of $t(x)$ is :

$$T(x)=\frac{1}{2}\left(x+\sqrt{x}\sqrt{x+4}\right)$$

So we need to show :

$$B=\frac{1}{2C}\left(A+\sqrt{A}\sqrt{A+4C}\right)$$

Or :$$(2C-1)B=\sqrt{A}\sqrt{A+4C}$$

Or :

$$(2C-1)(p_n-1)=\sqrt{p_{n-1}-1}\sqrt{p_{n-1}+4C}$$

Or :

$$(2C-1)^2(p_n-1)^2=(p_{n-1}-1)(p_{n-1}+4C-1)$$

But with Legendre's conjecture :

$$p_n<p_{n-1}+4\sqrt{p_{n-1}}+3$$

Remains to show that :

$$(2C-1)^2(x+4\sqrt{x}+3)^2\leq (x-1)(x+4C-1)$$

Which is true for example for $C=3/4$ and $\exists x,x>M> 1$.

So the equality is false in this case .It show that it cannot be the precedent prime under these assumptions .