## New answers tagged prime-numbers

0

If $n=3q+r$ then $q \leq \frac{n}{3}$ is a divisor of $\binom{n}{3}.$ So the least prime divisor of $\binom{n}{3}$ is at most $\frac{n}{3}.$
The slight flaw in that analysis is that perhaps $q=1.$ And indeed $\binom{4}{3}$ and $\binom{7}{3}$ are exceptions, but there are no others.
However, for $p$ prime and $n=3p$, $\frac{n}{3}=p$ will be the least prime ...

1

The conjecture as written is false:
Let $N=194+(2*3*5*7*11*13)*2n$, $k=N-2$, where $n$ is a natural number.
Then $C(N,k)=C(N,2)=(97+2*3*5*7*11*13*n)(193+2*3*5*7*11*13*2n)$, having no prime factors $\leq 13$.

4

I think this is true. Let $b = a^{\frac{N-1}{2p}} = a^{2^{m-1}p^{n-1}}$, and note that we
have $\frac{b^{p}+1}{b+1} \equiv 0$ (mod $N$).
Now $a$ and $N$ must be coprime, so that $b$ and $N$ are coprime. We have $b^{2p} \equiv 1$ (mod $N$).
Now $b^{p}-1$ and $b^{p} +1$ have gcd dividing $2$. However $\frac{b^{p}+1}{b+1}$ is always odd, so that $\frac{b^{p}+1}...

-6

My answer will focus on the last question, "what's the best known result of $x$"? Unfortunately there is no valid reponse until now, but there is an article by me not published, titled "There Exist Infinitely Many Couples of Primes $(P,P+2n)$, with $2n \ge2$ is a Fixed Distance Between $P$ and $P+2n$". In it, I use linear Diophantine ...

13

Yes. Obviously this $c$ and $N$ are coprime. We get $c^{(N-1)/2}+1=(c^{(N-1)/6}+1)(c^{(N-1)/3}-c^{(N-1)/6}+1)$ is divisible by $N$. Therefore $c^{N-1}-1$ is divisible by $N$, and $N-1$ is divisible by $k:={\rm {ord}}(c)$, where ${\rm ord}(x)$ denotes the multiplicative order of $x$ modulo $N$. But $(N-1)/2$ is not divisible by $k$, since $c^{(N-1)/2}\equiv -...

0

I add this as my contribution, just as a footnote of the answer and comments that were posted. My post isn't an answer for your question, just additional remarks that maybe are interesting in my view, thus you or the professors of this site MathOverflow feel free to comment if this isn't suitable as a contribution.
The online encyclopedia Wikipedia has an ...

5

Corollary 2 of https://arxiv.org/abs/2007.11062 states that every sufficiently large odd integer is the sum of a prime and a practical number.

1

So, for Sierpiński numbers of the Izotov type $(*)$, was
a bigger covering set found between 1995 and 2015?
No, it was not. And it is conjectured that none exists. In the Math Stack Exchange thread linked by Gerry Myerson in a comment to the question, I give other examples of Sierpiński numbers for which it is not likely that they should possess such a ...

4

If $S$ is congruentially equidistributed and contains enough elements .... is it true that $S+S$ contains all the positive integers except a finite number of them?
Let $S=\bigcup_{n=1}^\infty \{2^{2n},2^{2n}+1,\dots, 2^{2n+1}-1\}.$ It is easy to show that $S$ is congruentially equidistributed and $S+S\not\ni 2^{2n}$ for each positive integer $n$.

0

Here I provide some insights about conjecture B. First, it is still a conjecture, and just like the paradox that I discussed here, it defies empirical evidence: the error term in the approximation involves $\log$ and $\log \log$ functions (see here) so you would need to use insanely large numbers to see convergence to uniform distribution in residue classes, ...

1

Erdos has a nice proof in his paper on the Sylvester Schur theorem from 1934. I will sketch part of it here.
Look at the prime factors of the binomial coefficient $B = \binom{m+n}{n}$. An elementary argument says that the prime $p$ divides $B$ to a power that is less than $m+n$. Therefore $B$ has all its prime factors less than $n$ multiply to a number $C$ ...

1

I'll take a stab at this.
As pointed out at the comments, one can obtain the characteristic vector $\chi_{S+T}$ of the multiset
$$S+T = \{s+t \mid s \in S, t \in T\},$$
by defining $\chi_S$ to be the characteristic vector of a set $S$ in $\mathbb{Z}_p$. Then
$$\chi_{S+T} = \chi_S \ast \chi_T,$$
where $\ast$ is the convolution operator. Now using the Fourier ...

15

$2k=1+p_N$ works for $N>1$, but $2k\le 0.56 \, p_N$ will fail if
$p_{N+2}=p_{N+1}+2$.
With $q=p_{N+1}$, we have
$$
\frac{1}{1-q^{-2k}} < \frac{1}{1-x^{-2k}} = \frac{1}{1-q^{-2k}} \prod_{p>q} \frac{1}{1-p^{-2k}} .
$$
It follows that
$$
q^{-2k} < x^{-2k} < q^{-2k} + \sum_{j\ge 2} (q+j)^{-2k} < q^{-2k} +\frac{1}{(q+1)^{2k-1}(2k-1)}.
$$
Taking ...

2

As pointed out in the question, we have that
$$\prod_{p<Q}\left(\frac{x-1}{p}+1\right)=\sum_{k=0}^{\pi(Q)}\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]x^k$$
which can be derived by showing that on both the RHS and the LHS the coefficient of $x^k$ is equal to
$$\sum_{\substack{S\subseteq \{p<Q\} \\ |S|=k}} \left(\prod_{p\in S}\frac{1}{p}\right)\left(\prod_{p\not\...

1

I wrote:
My test power set (see definition in the example section) consists of integers that are even far more rare than pseudo-super-super-primes, yet for them conjecture A works, as expected. Perhaps this is caused by the fact that these integers are far more abundant than pseudo-super-super-primes among the first million integers, but asymptotically they ...

2

For a number $n$ that has $b=1+\log_2 n=O(\ln n)$ bits
NFS has complexity of the order
$$
\exp\left\{(c+o(1)) (\ln n)^{1/3} (\ln \ln n)^{2/3}\right\}
$$
which is subexponential when compared to the input size in bits. An exponential algorithm would have
complexity of the order
$$
\exp(c b)=\exp(c \ln n).
$$
If the unusual term factorial time means an ...

10

Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every $2^n-1$, with $n\ne1$ or $6$, has a primitive prime factor that divides it to an odd power.

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