Linked Questions

19
votes
2answers
2k views

The n-th prime is less than $n^2$? [duplicate]

Let $p_n$ be the n-th prime number, e.g. $p_1=2,p_2=3,p_3=5$. How do I show that for all $n>1$, $p_n<n^2$?
1
vote
0answers
54 views

$n$-th prime is less than a polynomial of $n$ [duplicate]

Let $p_n$ be the $n$-th prime. Is there an elementary way to prove the statement that there exists a polynomial function $P$ such that $$p_n \leq P(n),$$ for all $n \in \mathbb{N}$? (Or, alternatively,...
6
votes
5answers
1k views

n-th prime number is less than $4^n$

Prove that $\forall n \in \mathbb{N}^*, P_n \lt 4^n$ where $P_n$ is the $n$th prime number. I'm searching for a proof that doesn't use induction and uses only the elementary concepts of number theory. ...
2
votes
2answers
439 views

some properties of consecutive primes [closed]

If $p_1$ and $p_2$ are two odd consecutive primes and n is their midpoint, then $p_1p_2$ is the largest odd multiple of $p_1$ not exceeding $n^2$. This sounds obvious, but I still have problem to ...
2
votes
2answers
188 views

Show that $n^{\pi\left(2n\right)-\pi\left(n\right)}<2^{2n}$ and $2^n\le\left(2n\right)^{\pi(2n)}$ for all $n>2$

I may not use the Prime Number Theorem (and even if I could, it would only show the desired claim for sufficiently-large $n$). I have tried induction on $n^{\pi\left(2n\right)-\pi\left(n\right)}<2^...
2
votes
3answers
144 views

Prime number logic

It is conjectured that for every intever $n\geq1$ there is a prime $p$ with $n^2<p<(n+1)^2$. Show that if this conjecture is true then $\pi(x)\geq\lfloor\sqrt{x}\rfloor$ for all $x\geq2$. I ...
2
votes
3answers
197 views

Proving $p_{n+1}<p_n^2$ without Bertrand's postulate

How can we prove that $$p_{n+1}<p_n^2$$ Where $p_n$ is the nth prime number. Using Bertrand's Postulate it becomes easy. But how can we prove it without using this deep result?
1
vote
1answer
143 views

Prove that for every $ϵ>0$ there exist infinitely many natural numbers $n$, such that $τ(n)\geq 2^{(1-ϵ)(\log(n)/\log(\log(n))}$.

Prove that for every $ϵ>0$ there exist infinitely many natural numbers $n$, such that $$\tau(n)\geq 2^{(1-ϵ)\frac{\log(n)}{\log\log(n)}}$$ It is obvious to look at squarefree naturals of the form $...
4
votes
1answer
136 views

Show that $p_n^{1-\epsilon}\le n$ using PNT

Assuming PNT $$\pi(x)\sim \frac{x}{\log{x}}$$ How can we show that given any $\epsilon>0$ $$p_n^{1-\epsilon}< n,$$ for all sufficiently large $n$ ($p_n$ denotes the $n^{th}$ prime.) My work: ...
2
votes
1answer
127 views

question regarding nth prime related to Bertrands postulate.

Let $p_n$ denote the $n$-th prime. Let $k>0$ be an integer. Assume that there exists $N_0$ s.t. $p^{k}_{N_{0}+1} < p_1 p_2 ... p_{N_{0}}$. Show that $$p^{k}_{n+1}< p_1 p_2 ... p_{n}, \space \...
2
votes
0answers
129 views

There are more primes than square

Will someone please help me in understanding the claims of corollary $2$ and $3$? I understand that corollary $2$ means for large $N$ there are more primes than squares in the interval $[1,N]$ which ...
2
votes
1answer
73 views

PNT can't be applied Legendre's Conjecture as it can be to Bertrand's Postulate?

1. Background Bertrand's Postulate says that there is at least one prime between an integer $n$ and its double $2n$. $$n < p < 2n$$ Although not a proof, the Prime Number Theorem (PNT), that the ...
1
vote
1answer
49 views

On primes of the form $2\pi(n)p_n+1$, for some $n\geq 1$ being $\pi(x)$ the prime-counting function and $p_k$ the $k$th prime number

I was inspired in the sequence A088348 from the OEIS to explore if there are primes of the form $$2\pi(n)p_n+1,\tag{1}$$ when $1\leq n$ runs over integers, where $\...
1
vote
0answers
52 views

How to express a coprime relation as a Congruence relation

When $(n,k) \in \mathbb P^2 $, the following coprime relations appear to hold: $$\gcd\Bigl(n^k,{\Bigl\lfloor \frac{p_n}{n} \Bigr\rfloor}^{k}\Bigr)=1$$ $$\gcd\Bigl(n^k,\Bigl\lfloor \frac{p_n^k}{n^k} \...