New answers tagged prime-numbers
0
votes
On the Second Hardy-Littlewood conjecture
$$2\frac {x+y}{\log (x+y) -1.1}\geq\frac {2x}{\log (2x)-1}+\frac {2y}{\log (2y)-1}+\log_2{x}+\log_2{y}$$
Let
$$f(x,y):=\frac {2x+2y}{\log (x+y) -1.1}-\frac {2x}{\log (2x)-1}-\frac {2y}{\log (2y)-1}-\...
- 130k
0
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Is a sieve that can be used to generate prime numbers for the formula produce by 6n±1, a trait of prime numbers greater than 3??
It's hard to understand just what you are asking.
You should be aware that numbers of the form $6k\pm 1$ (i.e. numbers that lack $2$ and $3$ as factors) form a semigroup which is closed under ...
- 6,812
1
vote
Accepted
Chebyshev’s theorem and Bertrand's postulate : An exercise .
We have
$\pi(x)\le{bx\over\log x}$,
$\pi(2x)\ge{2ax\over\log(2x)}$, and
$a/b>3/5$.
So,
$$
\pi(2x)-\pi(x)\ge{6bx\over5\log(2x)}-{bx\over\log x}=\dotsb={bx\log(x/32)\over5\log x\log(2x)}>0
$$
...
- 173k
4
votes
How would one show that any given prime p_i must be a factor of some (p_j - 1)? Is that a true property of primes even?
As stated in the comments, Dirichlet's theorem on arithmetic progressions implies that, for any prime $p,$ then since $p$ and $(p+1)$ are coprime, there is a prime of the form $p_j= (p+1) + np;\ $ in ...
- 16.4k
1
vote
What is the fastest way to write an even number as the sum of two primes?
You could apply Sieve of Eratosthenes that gives primes in a range. You can sieve the range $(2, n/2)$. The sieving takes $O(n \log \log n)$ time and $O(n)$ space.
You could then test pairs of primes ...
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3
votes
Accepted
The set of all sums of 3 primes covers almost all of $2\Bbb{Z}+1$ (result of Helfgott), what can one say about $\Bbb{P} - \Bbb{P} = 2\Bbb{Z}$ problem?
You define a binary operation '$\star$' on $\Bbb{P}_o^2$ and claim that this defines a group operation. But it is not clear that this binary operation is even well-defined; in writing
$$[a,b]\star[c,d]...
- 60.2k
1
vote
Accepted
$\omega$ of a highly composite number.
Your conjecture is not true. OEIS sequence A002182 is the sequence of highly composite numbers, and it gives
$$\begin{align*}h_{25} &= 27720 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \\
h_{26} &...
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1
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Does Euler product formula give any hints about asymptotics of primes?
Intuitively, the limit $s \to \infty$ shouldn't be expected to give asymptotic information, because the contributions from the tail of the series $\sum_{n} n^{-s}$ and $\prod_{p} \frac{1}{1 - p^{-s}}$ ...
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0
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Convergence for infinite sum of all primorials reciprocals.
This answer summarizes my comments. Let $\{p_1, p_2, p_3, ...\}=\{2, 3, 5, ...\}$ be the sequence of primes. We are interested in the number
$$ x = \sum_{n=1}^{\infty} \prod_{k=1}^n\frac{1}{p_k}$$
Fix ...
- 21.5k
0
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Is this a valid proof by contradiction for why there are infinitely many primes?
... use this reasoning, which says for every prime number P, we can construct a greater prime number.
It does not say that. It proves that we can construct a greater prime number if we assume that ...
2
votes
Accepted
Is this a valid proof by contradiction for why there are infinitely many primes?
This is because I am unsure whether it is correct for us to identify a greatest prime number P and then construct an even greater prime number Q right afterwards- this feels like it was not legitimate ...
- 36.1k
0
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Generating residues with $ a^n + b^n \mod p $
Here are some results:
Claim One: If $2$ is a primitive root modulo $p$, then there do not exist $a$ and $b \ne a$ such that $a^n + b^n$ represents all non-zero residue classes modulo $p$. More ...
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1
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Accepted
Find the least prime $p$, such that for some positive integers $a$, $b$, $c$, $d$, $(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$
Here is a possible approach, which
collects as many ideas as possible to sieve.
Some order is brought into the mess by using a tabular form,
when the information can be systematically presented.
It ...
- 29.5k
0
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Generating residues with $ a^n + b^n \mod p $
Just a comment based upon user1172706 answer :
It SEEMS in general if $(p-1)/2$ factors alot, then the pigeonhole principle together with fermat's little principle forbids us to generate every residue ...
- 14.2k
2
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Generating residues with $ a^n + b^n \mod p $
General conditions for a solution
Let $b=ac$ where $c$ has order $q$ and so
$$a^n+b^n=a^n(1+c^n).$$
It is convenient to consider the necessary conditions in group theoretic terms, where $G$ is the ...
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-1
votes
Generating residues with $ a^n + b^n \mod p $
Of course, a choice of $a,b$ exist for every prime. Simply choose $a$ to be a primitive root, and $b$ to be 1.
Let $a$ have order $\alpha$ and $b$ have order $\beta$ in the multiplicative group $\...
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2
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How to find the poles of $\frac{\zeta'(s)}{\zeta(s)}$?
When $f$ is a non constant meromorphic function defined in an open domain $U \subset \mathbb{C}$, you can write near each point $a \in U$ such that $f(a) = 0$ or $\infty$, $f(x) = \alpha(x - a)^n + \...
- 1,467
5
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How to find the poles of $\frac{\zeta'(s)}{\zeta(s)}$?
That is false. The zeta function is analytic with a nonzero value there: $\zeta(0) = -1/2$. And $\zeta'(s)/\zeta(s)$ at $s=1$ has residue $-1$, not $1$.
Quite generally, when $f(s)$ is a meromorphic ...
- 39.6k
0
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Show $\sum_p\frac{1}{p^{1+i}}$ converges
Using the fact that
$$
\int_2^x{\mathrm dt\over\log t}={x\over\log x}+O\left(x\over\log^2x\right),
$$
it follows from the PNT in OP's question that for all $x\ge y\ge2$,
$$
\pi(x)-\pi(y)=\sum_{y<p\...
- 5,308
1
vote
Primes $p$ such that $5$ is primitive root $\bmod{p}$ , where $p$ is a $321$ prime
joriki's answer gives a great analysis.
Calculations with PARI/GP let me find a counterexample (maybe the first one).
The prime $p = 3\cdot 2^{18123}-1$ apparently found by Ballinger in April 1998 is ...
- 5,015
3
votes
Accepted
How well does $\mathrm{Zi}(x)=\frac1e\sum_{k=1}^\infty\frac{(\ln x)^k}{kk!\phi(k)}$, with $\phi(k)=\sum_{n=1}^\infty e^{-n^k}$, approximate $\pi(x)$?
It appears that
$$Zi(x)=\frac{1}{e} \sum\limits_{k=1}^\infty \frac{\log^k(x)}{k k! \phi(k)}\tag{1}$$
is an approximation to the Gram series
$$G(x)=1+\sum\limits_{k=1}^\infty \frac{\log^k(x)}{k k! \...
- 5,917
1
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Finding the (smallest) next number with the same distinct prime factors as a previous number
The updated problem is more general, so let's consider that one.
Note that $n_{2} = n_{1} + d$ implies that $gcd(n1, n2) \vert d$. Since $n_{1}=p_{1}^{e_{1}^{(1)}}p_{2}^{e_{2}^{(1)}}...p_{k}^{e_{k}^{(...
- 31
2
votes
Are there infinite elements in the super prime set?
Claim
This is not a proof, but I have some heuristic 'evidence' that your Super Prime Set is most likely infinite.
Argument
Let $I_1 = [2,3,5,7]$ be the initial set.
For the next set, $I_2$, we ...
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2
votes
A problem on von Mangoldt function.
Hint: let $\lambda(m) = \log m$ if $m$ is prime and $\lambda(m)=0$ otherwise. Then
\begin{align*}
r(n) - r'(n) &= \sum_{n_1+n_2+n_3=n} \bigl( \Lambda(n_1)\Lambda(n_2)\Lambda(n_3) - \lambda(n_1)\...
- 70.7k
0
votes
How to prove that $\sum_p\frac{\ln p}{p^s(p^s-1)}$ converges at $\Re\{s\} >1/2$?
By the prime number theorem, the $n$th prime $p_n$ is asymptotically $n\log n$. Then, with $\sigma =\operatorname{Re}(s)$,
$$
\left| {\frac{{\log p_n }}{{p_n^s (p_n^s - 1)}}} \right| \sim \frac{1}{{n^...
- 27.2k
1
vote
Find the least prime $p$, such that for some positive integers $a$, $b$, $c$, $d$, $(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$
This answer is a work in progress. It shows that, without loss of generality, any integral solution must have $c$ and $d$ even and $a$ and $b$ odd, and either $a=1$ or $b=1$ or both. It is quite a lot ...
- 60.2k
4
votes
Show $\sum_p\frac{1}{p^{1+i}}$ converges
$$
\sum_{p\leqslant N}p^{-1-i}=\sum_{n=2}^N\big(\pi(n)-\pi(n-1)\big)n^{-1-i}
\\=\pi(N)N^{-1-i}+\sum_{n=2}^{N-1}\pi(n)\big(n^{-1-i}-(n+1)^{-1-i}\big)
$$
and, since $n^{-1-i}-(n+1)^{-1-i}=(1+i)n^{-2-i}+...
- 37.3k
1
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Find the least prime $p$, such that for some positive integers $a$, $b$, $c$, $d$, $(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$
If $a=b=c=1$ then the equation reduces to $2d-1=2023^2p$ and there are solutions for all odd $p$. So we can concentrate on $p=2$.
$$(ab+a-b)(bc+b-c)(cd+c-d)(da+d-a)=2.7^2.17^4$$
For the LHS to be ...
- 1,310
1
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Find the least prime $p$, such that for some positive integers $a$, $b$, $c$, $d$, $(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$
(Hm, this doesnt' work out as easily as I thought it would, so ignore for now.)
$ab+a-b) = (a-1)(b+1) + 1$.
So if $ab+a-b = 1$, this just requires $a = 1$ (with no restrictions on $b$).
Show that $p$ ...
- 64.3k
5
votes
Accepted
Find all two-digit prime number pairs $p$ and $q$, for which $pq+1$ is a perfect square.
We can assign $pq+1$ to equal to another variable, say $k^2$.
in other words, let $$pq+1=k^2$$ where $k$ is a positive integer.
Rearranging this equation yields $$pq=k^2 - 1$$ $$\implies pq = (k-1)(k+...
- 121
8
votes
Find all two-digit prime number pairs $p$ and $q$, for which $pq+1$ is a perfect square.
Any pair of twin primes will work, because $(n+1)(n-1)=n^2-1$. Those are the only possible solutions because the problem is equivalent to saying $n^2-1=(n+1)(n-1)$ is a product of two primes, so ...
- 20.5k
1
vote
Accepted
Recalling a theorem from vague memory: A monoid, in some sense, cannot "describe" the language (over one letter?) of words of prime length.
I guess you refer to the syntactic monoid of a language. See also the French entry of Wikipedia, which gives more details. A key property is that a language is regular if and only if its syntactic ...
- 38.2k
1
vote
Accepted
Odd numbers form pairs infinitely often in this sequence?
Consider the list of numbers which are double a prime.
$$4,6,10,14,22,26,34,38,46, ...$$
Then $I(k)$ is odd if and only if $k$ is an odd number between a $2i-1$th and $2i$th element in the above list. ...
- 1,310
2
votes
Question on Brun's theorem on twin primes
Warning: This is only a partial answer due to using a yet unproven (nor refuted) conjecture. Quite a bit too long to leave as a comment anyway.
If the first Hardy-Littlewood conjecture for prime $k$-...
- 2,062
3
votes
Accepted
Question on Brun's theorem on twin primes
Let $N$ be some even integer. Then it is known that there exists some constant $A>0$ such that
$$
\#\{p\le x:p,p+N\text{ prime}\}\le{Ax\over\log^2x}\prod_{\substack{p|N\\p>2}}{p-1\over p-2}.
$$
...
- 5,308
2
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Is $12$ the only number that has the same digits as its prime factorization?
This is not a complete answer, but too long to fit in a comment. This shows that if $k\geq100$, then $k$ must be divisible by the cube of a prime. There is plenty of room for improvement of the ...
- 60.2k
1
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How to prove that $N!+p(N)$ can't be a perfect power, where $p(N)$ is the $N^{th}$ prime?
For $y$ even, we can easily prove a stronger statement: $$p(N)\text{ is a quadratic non-residue modulo }N!$$
Proof:
For $N <= 9$, check computationally and confirm the statement. Hence we can ...
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0
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Lucas-Lehmer test for Mersenne and Wagstaff numbers?
If you look at the sequence in OEIS A029858 changing the index is equal to the sequence you used then you get
$$S_{p-2}=\frac{3^{2^p-1}-3}{2}$$
for Mersenne numbers
but also
$$S_{p-2}=\frac{3^{2^p-1}-...
- 607
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Why are there always two primes between $2^n$ and $2^{n+1}$?
We have certain properties, that have been proven, and other properties, that are totally sure, but not proven. It has been proven that between $n$ and $2 \times n$, we have at least 1 prime. It has ...
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Why are there always two primes between $2^n$ and $2^{n+1}$?
By the prime number theorem, the prime counting function $\pi(n)$ is quite close to
$$\pi(n) \approx \frac{n}{\log n}$$
so
$$\pi(2^{n+1}) - \pi(2^n) \\= \frac{2^{n+1}}{\log 2^{n+1}} - \frac{2^{n}}{\...
- 688
4
votes
Accepted
Why are there always two primes between $2^n$ and $2^{n+1}$?
This follows from a strengthening of Bertrand's postulate. In fact it can be shown that for every $\varepsilon>0$, for all sufficiently large $n$ there exists a prime between $n$ and $(1+\...
- 60.2k
0
votes
Finding number with prime number of divisors?
For prime factorization $n = p_1^{e_1} \cdots p_r^{e_r} $, the number of divisors is $\sigma(n) = (e_1 + 1) (e_2 + 1)\cdots (e_r + 1)$. For that to be prime, $n$ must be a prime power $p^e$ where $e + ...
- 10.3k
1
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Is $F_n+F_{n^2+1}$ composite for all odd positive integers $n>1$?
This is not an answer to your question, just a comment that would be too long.
I have sieved $n < 10000$ (odd and even) up to $p = 10^{11}$ and the candidates
...
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5
votes
Accepted
Does for every integer $k$ exist a positive integer $n$ such that $F_n+k$ is prime?
No, $f(k)$ is not total.
Reference https://oeis.org/A361902 for $k\ge 0$, where it states that if $k\equiv 14475 \pmod {1647030}$ then there is no solution.
The OEIS page links to a proof by Robert ...
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1
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Find all relative prime positive integers $p$ and $q$ such that $p+q=(p-q)^3$.
Let $p$ and $q$ be two positive integers satisfying
$$p+q=(p-q)^3.$$
Let $r:=p-q$ so that $q=p-r$ and so
\begin{eqnarray}
p&=&\frac{r^3+r}{2}=\frac12r(r^2+1),\\
q&=&\frac{r^3-r}{2}=\...
- 60.2k
2
votes
Find all relative prime positive integers $p$ and $q$ such that $p+q=(p-q)^3$.
Let $n=p-q. $ Then
$$p=\frac{n^3+n}2,\quad q=\frac{n^3-n}2.$$
If $n$ is odd then it divides $p,q$ (because $n^2\pm1$ is even). This is impossible because $q$ would be equal to $\frac{1^3-1}2=0.$
If $n=...
- 22k
2
votes
Accepted
Find all relative prime positive integers $p$ and $q$ such that $p+q=(p-q)^3$.
Since $p-q\mid p+q$, we have $p-q\mid (p+q)+(p-q) = 2p$ and $p-q\mid (p+q)-(p-q) = 2q$, hence $p-q\mid 2\gcd(p, q) = 2$. We divide cases.
(i) $p-q = 1$: then $p = q+1$, so the equation reduces to $2q+...
- 1,023
1
vote
Accepted
Finding an $x$ that satisfies the Jacobsthal function for the primorial of the $n$th prime.
A list, up to $24$, is given at https://oeis.org/A049300/b049300.txt
Here it is:
$$
\matrix
{1& 2\hfill\cr
2 &2\hfill\cr
3& 2\hfill\cr
4& 2\hfill\cr
5 &114\hfill\cr
6 &9440\...
- 173k
0
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If $30$ divides $p_1^4 + p_2^4 + \ldots + p_{31}^4$. Prove that $p_1=2$, $p_2=3$ and $p_3=5$.
Let $a=1$ if $2$ is among the $p_i$'s, and $a=0$ else. Define similarly $b$ for $3$ and $c$ for $5.$
For any prime number $p$ coprime to $30,$ we have $p^4 ≡ 1\bmod2,3,5$ hence $\bmod{30}.$ Therefore, ...
- 22k
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A pattern in primes numbers?
If you want to analyze the numerator and denominator of the series $$H_n=1+\cdots +\frac{1}{n},$$ you can do so by noticing that $$H_{n+1}-H_{n}=\frac{1}{n+1}$$ and put $$H_n=\frac{N_n}{n!}$$ yielding ...
- 1,473
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