# Tag Info

6

The authors of the paper you mention, Erdos, Lacampagne, and Selfridge, define $p(m)$ to be the least prime divisor of $m$ and concern themselves what can be said about $p(\binom{n}{k}).$ I suspect that Selfridge wrote the article. It has his style of saying a lot in a succinct way which is puzzling but solvable with some thought on the part of the reader. ...

3

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 $\leq 13$.

2

Smooth random functions, random ODEs, and Gaussian processes (2018) describes an approach that takes a finite Fourier series on the interval $(0,1)$ with randomly chosen coefficients. The integral of this function approaches Brownian motion in the limit that the number $M$ of Fourier coefficients tends to infinity. The plot shows three such functions, for $M=... 4 I think this is true. Let$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}...

9

The group $J_0(35)(\mathbb Q)$ (where $J_0(35)$ is the Jacobian of $X_0(35)$) has rank 0 (as shown for example by a 2-descent computation in Magma); it is isomorphic to ${\mathbb Z}/24{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$, with generators the difference of the two points at infinity on $X_0(35)$ and the 2-torsion point corresponding to the ...

-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 ...

0

I think there is some mistake in the post. (1) The quantity $$\zeta'(\Delta,0)$$ for a surface is not the Euler characteristic of the surface. Note that there is actually a small issue, as there are many Laplacians on a surface ($d\delta+\delta d, \Delta, \Delta_{\partial}, \Delta_{\partial^{*}}, \Delta_{n}^{+}, \Delta_{n}^{-}$, etc) and the sign usually ...

0

To find an answer to your question I suggest to take moment of your time to read this link: https://vixra.org/abs/2007.0090 Here you find a deep connection between the Goldbach conjecture and the binary version of the Mobius sum, using linear Diophantine equations and sieve methods.

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 $c^{(N-1)/2}\equiv -... 0 I add this as my contribution, just as a footnote of the answer and comments that were posted. My post isn't an answer for your question, just additional remarks that maybe are interesting in my view, thus you or the professors of this site MathOverflow feel free to comment if this isn't suitable as a contribution. The online encyclopedia Wikipedia has an ... 5 Corollary 2 of https://arxiv.org/abs/2007.11062 states that every sufficiently large odd integer is the sum of a prime and a practical number. 3 No, there cannot be 17 such numbers in arithmetic progression (and there cannot be 5 such numbers with the corresponding property for triples). Suppose we have such an arithmetic progression of length$k$, say$x,x+d,\ldots,x+(k-1)d$. I claim that if a prime$p$divides any two of them then either it divides all of them (which cannot be the case), or else$p&...

10

0 = P(7) + P(10) + P(-11) = P(3250) + P(2293) + P(-3593) = P(6266) + P(13243) + P(-13695) = P(11700) + P(13277) + P(-15797) = P(37555) + P(131381) + P(-132396) = P(747511) + P(1059490) + P(-1171307) = P(5529835) + P(22681597) + P(-22790636) = P(8042677) + P(13682243) + ...

7

$a_n$ is composite for $4 \le n \le 2016$. $a_{2017}$ appears to be prime (it passes a strong pseudoprime test). I have not tried to certify that it is prime (this would take a while as the number has 5789 digits).

5

Let $P(x):=x^3-2x$. Then \begin{gather} 70=P(2714)+P(1367)+P(-2825),\\ 75=P(16333)+P(14200)+P(-19328),\\ 83=P(6714)+P(-6682)+P(-1627),\\ 86=P(6413)+P(3721)+P(-6806). \end{gather}

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 ...

0

The smallest interesting case of $k=2$ reduces to a family of Pell equations paramaterized by $b$: $$(2c-1)^2 - b^3(2a)^2 = 1.$$ This gives infinitely many solutions. For example, for $b=2$, we have a series of solutions indexed by $n$: $$c_n + a_n\sqrt{8} = \frac{(17+6\sqrt{8})^n+1}2.$$ Numerical values of $c_n$ are listed in OEIS A055792.

1

You might be interested in extensions to the Sylvester Schur theorem, which by your constraints shows that c is bigger than k^2 as the set of consecutive integers in the product must have a single multiple of q^2 for some prime q bigger than k. A paper of Saradha and Shorey from 2003, Almost Squares and Factorizations in Consecutive Integers, shows the ...

3

You may already know this, but numbers of the form $a^2b^3$ are called powerful numbers. A closely related question that might provide information on your question is to ask for binomial coefficients that are powerful. A Google search of "powerful number" and "binomial coefficient" brought up the following paper of Granville: On the ...

1

In practice, sieves have a hard time producing useful lower bounds for prime counting problems due to the so-called parity problem, a phenomenon that is very poorly understood. The best general "prime producing sieve" is due to Bombieri (though the formulation below is due to Friedlander and Iwaniec in this paper. The question is phrased as follows:...

2

Please check my post. It indicates that the number of divisors of $x^2+x+41$ is equal to the number of lattice points of $X^2+163Y^2-2(2x+1)Y-1=0$. This formula is transformed in this way. $$163X^2+163^2Y^2-2\cdot163(2x+1)Y=163$$ $$163X^2+\{163Y-(2x+1)\}^2-(2x+1)^2=163$$ $$163X^2+(163Y-2x-1)^2=4x^2+4x+164$$ $X':=163Y-2x-1,\ Y':=X$ and we divide both sides by ...

5

The Riemann xi-function decreases exponentially as $t\to\infty$, so it can't be universal. The decay comes from the fact that $$\xi(s) = \frac12 s (s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s) .$$ If $s = \sigma + i t$ with $\sigma$ bounded and $t\to \infty$, then $\frac12 s (s-1)$ has polynomial growth, $\pi^{-s/2}$ is bounded, $\Gamma(s/2)$ decays exponentially, ...

12

This is a partial case of the classical result. https://en.wikipedia.org/wiki/Niven%27s_theorem

6

Let $\theta= \arcsin(1/4)$. Assume $\theta$ is a rational multiple of $\pi$. Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$. Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$. This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{... 14 Yes, \arcsin(\frac14)/\pi is irrational. Suppose \arcsin(\frac14)/\pi = m/n, where m and n are integers. Then \sin(n \arcsin(\frac14))=\sin(m \pi)=0. We analyze this usng the formulas from Browmich as cited on Mathworld:$$\frac{\sin(n\arcsin(x))}{n}=x-\frac{(n^2-1^2)x^3}{3!} + \frac{(n^2-1^2)(n^2-3^2)x^5}{5!} + \cdots\frac{\sin(n\arcsin(x))}{...

2

You could get it by brute force - the supersingular j-invariants have to lie in $\mathbb{F}_{2^2}$, so you can just check each of them.

1

Apologies for commenting in an answer when there is already an answer in the comments... When I was a PhD-student (in the 90's), we were discussing variations of the van der Waerden theorem on monochromatic arithmetic progressions. A question that came up was whether we can require the monochromatic sequence to consist of consecutive multiples of some number....

2

It depends on what you mean by $|\cdot|$, but probably no. If by $|\cdot|$ you mean the absolute value on $\mathbb C$, and your algebraic integers are elements of $\mathbb C$, then the answer is no. The logarithmic Weil height of $1-\sqrt3\in\mathbb A^1(\overline{\mathbb Q})$ is $\frac12\log(2)$, which is strictly bigger than $\log(\max(1,|1-\sqrt3|))=0$. (...

3

I have two comments for you, putting them as an answer because I don't have enough credits for a comment: First, the notation $2\mid m$ should be very common, and it is definitely very common in German books. As an example, I own a copy of Algebra from Siegfried Bosch, first edition from 1992, and it uses this notation as well. Second, I think it is great ...

1

$ax^k + by^k = au^k + bv^k\tag{1}$ $a,b,x,y,u,v$ are integer. Case $k=3$: If equation $(1)$ has a known solution, then equation $(1)$ has infinitely many integer solutions below. Let $(x0,y0,u0,v0)$ is a known solution. p,q are arbitrary. Substitute $x=pt+x0, y=qt+y0, u=pt+u0, v=qt+v0$ to equation $(1)$, then we get $$t=\frac{-ax0^2p+by0^2q-au0^2p-bv0^2q}{... 4 Rewriting your equation as, for fixed m and k,$$\displaystyle x^k + my^k = u^k + mv^k, x,y,u,v \in \mathbb{Z},$$we see that this is of the form F(x,y) = F(u,v) for a binary form of degree k and defines a surface X_F \subset \mathbb{P}^3. Heath-Brown showed in this paper that if one deletes the rational lines on this surface, necessarily formed by ... 3 To expand on GNiklasch's answer, and analyse what you write as well: we always have (when complex conjugation is central in the Galois group) have \overline{\alpha^{\sigma}} = {\bar \alpha}^{\sigma} when \alpha is an algebraic integer in K and \sigma is an element of the Galois group of K. In general, there is no reason why |\alpha|^{2} should be ... 10 Zero is the only algebraic integer which has all its conjugates strictly inside the complex unit circle. (Look at the norm.) For explicit examples with conjugates on either side of the unit circle, you can start with a real quadratic field with a totally positive unit that isn't already a square in this field, such as \varepsilon = 2+\sqrt{3}. Then take \... 4 By inclusion-exclusion principle, the number of representations of n as the sum of squares of four nonzero integers equals:$$\sum_{k=0}^4 \binom4k (-1)^k r_{4-k}(n).$$Formulae for r_k(n) are given in this article at MathWorld. If one wants to further restrict the representations to positive integers, the above expression needs to be divided by 2^4=16.... 4 If S is congruentially equidistributed and contains enough elements .... is it true that S+S contains all the positive integers except a finite number of them? Let S=\bigcup_{n=1}^\infty \{2^{2n},2^{2n}+1,\dots, 2^{2n+1}-1\}. It is easy to show that S is congruentially equidistributed and S+S\not\ni 2^{2n} for each positive integer n. 8 This number is expected to be transcendental. This answer gives a conceptual framework for studying the algebraicity of such \Gamma ratios, and in fact a completely explicit criterion (which is only conjectural, when it comes to establishing transcendence). Your number is equal to \begin{equation*} \frac{\Gamma(2/5)^3}{\Gamma(1/5)^2 \Gamma(4/5)} \end{... 3 The value of \eta(i\sqrt{6}) and \eta(i\sqrt{3/2}) involves the use of gamma function values on a 24 basis, so we have:$$\eta(i\sqrt{6})=\frac{1}{2^{3/2}3^{1/4}}\big(\sqrt{2}-1\big)^{1/12}\frac{\Big(\Gamma\big(\tfrac{1}{24}\big) \Gamma\big(\tfrac{5}{24}\big) \Gamma\big(\tfrac{7}{24}\big) \Gamma\big(\tfrac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}\eta(i\...

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, ...

4

The answer is yes. As was observed, the condition of being pre-Pell is simply the stipulation that $k$ is a sum of two squares: that is, if $p | k$ and $p \equiv 3 \pmod{4}$ then $p$ must divide $k$ with even multiplicity. If we assume $k$ is square-free, then it is divisible only by $2$ or primes of congruent to $1 \pmod{4}$. We now work on the equivalences....

1

The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the $N$. We proceed by induction on $k$, the number of polynomials. For $k=1$, either $p_1$ has only rational coefficients or it has at least one irrational. If it has an ...

5

I have nothing to add to the Fourier-type approach suggested in the question, but for those curious, thought it useful to outline the combinatorial solution to the problem that I know (I believe this is the same as the IMO official solution, and claim no originality). One thing to add is that, although induction is a crucial part of the proof, we do not use ...

4

Yes, I have not known Deligne, Kazhdan and Vigneras to lie. A sketch of the proof, at least with the key details for GL(2), is given in Lecture V of Steve Gelbart, Lectures on the Arthur--Selberg Trace Formula Added remarks: In that Lecture, Gelbart addresses both two kinds of simple traces formulas. The one you are asking about is essentially Prop 2.1. ...

5

On a slightly cheeky note, it seems that the shortest path to the best available bound, subject to peer review, is outlined in the following paper of Bloom-Sisak! Congratulations.

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 $$\sum_{\substack{S\subseteq \{p<Q\} \\ |S|=k}} \left(\prod_{p\in S}\frac{1}{p}\right)\left(\prod_{p\not\... 3 We have$$\frac n{\varphi(n)}=\prod_{p\mid n}\bigl(1-p^{-1}\bigr)^{-1} \le2\prod_{\substack{p\mid n\\p\ge3}}\frac32 =2\prod_{\substack{p\mid n\\p\ge3}}3^{\log_3(3/2)} \le2\prod_{\substack{p\mid n\\p\ge3}}p^{\log_3(3/2)} \le2n^{\log_3(3/2)}$$(where p runs over primes), hence$$\varphi(n)\le m\implies n\le(2m)^{(1-\log_3(3/2))^{-1}}=(2m)^{\log_23}.$$Using ... 0 This is just a comment on the accepted answer but it is too long for the comment box. It is not meant to detract from the excellent accepted answer but only contains one or two suggestions that might save a little time. (Feel free to delete.) Perhaps it is worth mentioning that saying ^pH^0i^*\mathcal F=0 is the same as saying that f^{-1}(v_0) does not ... 3 A. Sarközy, A note on the arithmetic form of the large sieve, Sudia Sci Math Hungarica 27, 1992, 83--95 covers the literature until then (there are some later results, adapting this as needed.) I. Ruzsa, On the small sieve. II. Sifting by composite numbers Journal of Number Theory 14 (2), 1982, 260--268 2 Since m is a dummy variable (i.e. a bound variable) and n,n' are "real" variables (i.e. they are free) perhaps we should rewrite the problem accordingly as compute the following$$ f(y,z) = \#\left\lbrace (x_1,... , x_y )\mid x_1 + ... + x_y = m,\ x_i \in \mathbb{N},\ m \leq yz ,\ i < j \implies x_i \leq x_j \right\rbrace." We can ...

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