# 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:

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

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?

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Feb 14 at 15:56

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.

I also checked the 1st 100 numerators.

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

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 .

• So, this isn't actually an answer, is it? Feb 11 at 10:18
• @GerryMyerson It's a partial answer . Feb 11 at 10:38
• @Peter Do you think it's realistic ? Can we go further ? Feb 11 at 15:07