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