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 :


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:


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


If We have :


Then $C$ a positive integer :


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?


2 Answers 2


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:

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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.


Partial answer.

Conjecture in radian :

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


If i'm not wrong we have :


Remains to show that for $n$ sufficiently large :


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

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


So we need to show :


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

Or :


Or :


But with Legendre's conjecture :


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 .

  • 4
    $\begingroup$ So, this isn't actually an answer, is it? $\endgroup$ Feb 11 at 10:18
  • $\begingroup$ @GerryMyerson It's a partial answer . $\endgroup$ Feb 11 at 10:38
  • $\begingroup$ @Peter Do you think it's realistic ? Can we go further ? $\endgroup$ Feb 11 at 15:07

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