User alex botros - MathOverflow

Chen Primes in Chen's theorem

2012-09-10T04:35:41Z

In H. Halbertsam and H. Richert's book "Sieve Methods" (Academic Press 1974) the authors go about illustrating the proof of Chen's theorem in Chapter 11 (Pages 321-337). They use a function $S_0$ defined on line (2.4) at the bottom of page 323. I simply want to know how this $S_0$ relates to the number of Chen Primes less than a given magnitude $N$. Are they the same?

Problems with the divisor function in a summation

2012-09-03T21:30:56Z

I'm trying to work with the following sum: $$f:=\sum_{d\leq x}\mu(d)\tau(d) \Big[ \frac{x}{d}\Big] $$ Where $\mu$ is the Mobius function, $\tau(n)$ is the number of positive divisors of $n$ and $h(x)=[x]$ is the floor function.

We know that $$\sum_{d\leq x} \mu(d)\Big[ \frac{x}{d}\Big]=1,$$ $$ \sum_{d\leq x} \Big[ \frac{x}{d} \Big] \sim x\log(x),$$ $$\sum_{x\leq d} \frac{x}{d} \sim x\ln(x), $$ but what happens if we throw $\tau$ into the mix? Is it similar to $$g:=\sum_{d\leq x}\mu(d)\tau(d)\frac{x}{d}?$$ Or is it similar to multiple of $g$? I've tried Mobius inversion with a few different declarations of $f,g$ but either I'm simply not seeing it, or it doesn't work. Any ideas?

Special divisor function summation

2012-05-18T21:16:49Z

What is a good upper bound for $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d}$$ where $\tau(d)$ is the divisor function?

A question on Selberg's sieve

2012-04-28T20:42:52Z

If we define $$D=D(z,w)=\sum_{d\vert P_z, d\leq w}\frac{1}{\phi(d)}$$ $$\lambda(k)=\frac{k}{D} \sum_{k\vert d,d\vert P_z , d\leq w}\frac{\mu(\frac{d}{k})\mu(d)}{\phi(d)}$$ $$p(d)=\sum_{lcm(d_1,d_2)=d }\lambda(d_1)\lambda(d_2)$$

In other words, we're dealing with the Selberg Sieve, is it possible to say, for the Möbius function $\mu(d)$, that $$\sum_{d\vert P_z} \vert \mu(d) \vert \leq \sum_{d\vert P_z} \vert p(d)\vert$$

Twin primes and D primes

2012-03-22T16:47:35Z

If we define $$\pi^2(N)=\vert [p: p\leq N, p\in\mathbb{P}, p-2\in\mathbb{P}]\vert$$ where $\mathbb{P}$ is the set of all primes (as the number of twin primes less than $N$), and we define $$\pi^D(N)=\vert [p: p\leq N, p\in\mathbb{P}, Dp-2\in\mathbb{P}]\vert$$

Can anyone think of how to show that for $D$ prime and large enough with respect to $N$ and $N$ large enough, we have $\pi^2(N)\geq \pi^D(N)$. It seems heuristically true, simply because the probability, given a prime $p$, that $Dp-2$ is also prime is less than the probability that $p-2$ (which is smaller than $Dp-2$ and thus within a denser area of primes) is also prime.
Sieve methods allow us to conclude that the upper bounds of both sets are multiples of each other, but not much else. Any ideas?

Sum of Mobius function and omega function

2011-08-30T03:06:33Z

I am trying to find some work done on the following: $$\sum_{d \vert n}\frac{2^{\omega(d)}}{d}\mu(d)$$ where $\omega(d)$ is the number of distinct prime factors of $d$ and $\mu$ is the mobius function. I saw something about $$\sum_{d \vert n}\frac{\mu(d)}{d}=\phi(n)/n$$ (where $\phi$ is the Euler phi function) on planetmath, but I'm not entirely certain how to use it. Does anyone know of any work done first sum?

Little problem of divisibility

2011-08-06T18:49:33Z

Suppose we have $D$ prime and $d$ square free. Is there a way to express the smallest integer $n$ that makes $d \vert (Dn-2)$? If such an $n$ exists for a given $D$, $d$ it must exist less than $d$, but is there some function of $D$ and $d$ that gives it explicitly?

Answer by Alex Botros for Little problem of divisibility

2011-08-08T03:49:57Z

Hi guys, sorry I caused a fuss. The fact of the matter is (from what my limited understanding tells me) is that is if you solved this question, you could then tackle the weirdly huge error term in Brun's sieve. You're all probably familiar with this sieve, but it frustrates me that we must assume all error terms (typically written) $R_d$ (of fluctuating sign) positive in order to bound its sum as an alternating series. I understand that we expect there to be about $$\frac{N/2}{d}2^{\omega(d)}$$ integers n between $N/2$ and $N$ that will allow $d\vert n(Dn-2)$, and I understand that this may sometimes be an overestimate (resulting in a negative $R_d$) or an underestimate (resulting in a positive $R_d$). Furthermore, because Brun's error is written as an alternating series over these $R_d$, the rectification of signs becomes tragic. I was thinking that if my above question was answerable, we would be able to rectify when Brun's main term is an underestimate, and when it is an underestimate, and thus we would not have to simply take the error as the strictly positive sum over $\vert R_d \vert$ which blows up way quicker than it has to. I have been using fairly standard sieve theory notation without explaining it, and for that I apologize. Furthermore I just graduated high school as Gerhard Paseman correctly pointed out, so I apologize further for the elementary nature of my comment. Thank you sirs for your patience, I will keep looking.

Divisor function inequality

2011-05-24T02:34:18Z

I have been reading a paper on the Goldbach conjecture found at http://people.exeter.ac.uk/pt224/Goldbach.pdf. At one point, the author (Paul Truman), states: Let $z=N^{1/8}$, then $$\sum_{w\leq z}\frac{d(w)}{w}\gg(\log(z))^2\gg(\log N)^2$$ where $d(w)$ counts all the positive divisors of $w$. I am assuming that there's a mistake in the second part of the inequality $(\log(z))^2\gg(\log N)^2$, but this is not the first time I've encountered such a claim: at http://www.m-hikari.com/ijcms-2010/1-4-2010/mollinIJCMS1-4-2010.pdf the author claims (in his proof of the upper bound on the twin prime counting function) that $$\sum_{\substack{d\leq N^{1/3}\ d \ odd}}\frac{f(d)}{d} \geq (\log N)^2$$ where $f(2)=1, f(p)=2$ for all odd primes $p$. This inequality I have seen proved (though I can't recall where) by saying that $f(w)/w\geq d(w)/w$. A lot of people are saying that $$\sum_{w\leq z}\frac{d(w)}{w}\gg(\log(z))^2$$ but others are saying that $$\sum_{w\leq z}\frac{d(w)}{w}=\frac{\log^2(z)}{2}+O(\log x)$$ including http://people.exeter.ac.uk/pt224/Goldbach.pdf. how can both be true?

Work down on the Upper bound of the Twin Primes

2011-03-15T14:25:47Z

It can be shown using the Selberg Sieve method, that the maximum number of Twin primes less than $N$ is $$\frac{CN}{\ln^2(N)}$$ does anyone know if there has been any work done on finding an upper bound for the constant $C$?

Selberg sieve on a certain Set.

2011-03-13T02:04:47Z

I'm new to sieve theory, and I'm trying desperately to understand Selberg's sieve. I would like to apply the sieve to give me a nice upper bound on primes of the set $$A^D(N)= \{ Dq-2 : q\in P, N/2 < q \leq N \} $$ But basically, for a fixed N, I would like $A^D(N)$ to be the set of elements of the form $Dq-2$ for a fixed positive integer $D$ and letting $p$ run through all primes between $N/2$ and $N$. Now, as I said, I'm trying to apply Selberg's sieve, but as I don't really know what I'm doing, I'm a bit confused. Now, if I understand it correctly could I then say that $$S(A^D(N),N/2,N/2) \leq \frac{\pi(N)-\pi(N/2)}{L_p(z)}+O \Big( \frac{z^2}{L_p(x)^2} \Big)$$ Where $S(A^D(N),N/2,N/2)$ is the number of elements of $A^D(N)$ which are prime and $$L_p(z)=\sum_{n\leq z}^{n\vert P} \frac{\mu(n)^2}{\phi(n)}.$$ where $$\frac{1}{\phi(n)}=\frac{1}{n}\prod_{p\vert n} \frac{1}{1-1/p}.$$ I got this from a pater called "Sketch of the Selberg Sieve method" By Sean Prendville (January, 2008) where he describes the Selberg Sieve not on $A^D(N)$ but on the set of all integers between some positive integer $x$, and $x+y$. I'm sure some of it is wrong, or that I totally misinterpreted, but I would like to know if this is right, and if it is, where do I go from here? (especially with dealing with $L_p$). I appreciate any help, but please keep in mind that I'm sixteen years old and live in the Bronx. As dumbed down as possible would be greatly appreciated..this is all new to me. Much appreciated, Alexis D. Botros

Upper bound on Chen primes in an interval?

2011-03-12T00:06:55Z

I'm well aware of the fact that the number of Chen primes between $N/2$ and $N$ for large enough $N$ is at least $$\frac{c_1N}{\ln^2(N)}$$ (Green and Tao). My question is: is there possibly an upper bound for chen primes between $N/2$ and $N$? What I am eventually trying to prove is that there are infinitely many intervals $$\Omega_i=[i^2/2, i^2]$$ such that the number of chen primes in $\Omega_i$ is at least $$\frac{c_1i^2}{13\ln(i^2)}.$$ Does anyone know where to look?

Derivative of Sum over Variable of derivative

2011-03-06T22:52:08Z

I feel stupid for having to ask this, but does anybody have any idea how to handle $$\frac{d}{x}\sum_{n=k}^{g(x)}f(n,x)?$$ Example: $$\frac{d}{dx}\sum_{n=6}^{i^2+2i} \frac{1}{\ln{(i^2)}-\ln{\ln n}}.$$ If we were able to separate the summand into two functions, one with only $i$ as a variable, and one with only $n$ as a variable, this would be super simple. But it is not always the case. Any ideas?

Answer by Alex Botros for Derivative of Sum over Variable of derivative

2011-03-12T00:10:01Z

The easiest way to handle something like this particular example, is simply to expand and derive term by term. I see what you are saying. THank you

Generating Chen primes.

2010-12-02T19:59:51Z

Let

$A_p(n)=\#\{q<n \vert \ qp-2 \ is \ prime\}$

Where p,q are prime, n is an integer. My question is, it seems fairly reasonable to assume that for a fixed $n$, $A_p(n)$ can be bounded by a decreasing function, but can I prove it. Moreover, can it be shown that $\lim_{p \rightarrow \infty} A_p(n)=0$? Does anyone know of any work done on the subject?

Distribution of Chen primes.

2010-12-01T06:26:46Z

In the paper of Green and Tao "Restriction Theory of the Selberg Sieve, with applications," their theorem 6.1 states: Let $N$ be a large integer. Then the number of Chen primes in the interval $(N/2,N)$ is at least $c_1N/\ln^2N$ , for some absolute constant $c_1>0$ .

My question is, what the heck is $c_1$ ? Is it Brun's constant, or is that just wishful thinking?

Zeroes of a tricky function.

2010-11-18T02:02:39Z

I am attempting to show that there does not exist an N past which every open unit interval (k, k+1) -where k is an integer- contains a zero of the following function:

$h(x)=\sum_{n=2}^{[\sqrt(x)]} \frac{\cot(x\pi/n)}{n}+\frac{\cot((x+2)\pi/n)}{n}$

Where the $[\sqrt(x)]$ is the lowest integer of the square root of $x$ . Any thoughts? I had figured that I could consider the interval $(i^2, (i+1)^2)$ in which h(x) is described by the function

$h(x)=h_n(x)=\sum_{n=2}^{i} \frac{\cot(x\pi/n)}{n}+\frac{\cot((x+2)\pi/n)}{n}$

Then try to see on what unit intervals (k, k+1) in here contain a zero of $h_n$ then try to show that $((i+1)^2, (i+2)^2)$ also has such an interval, but I am running out of ideas.

Comment by Alex Botros

2012-05-18T22:48:11Z

well, if we defined the sum over all $d\vert P_z$ we could bound it above nicely by $1/\log(z)^2$ using Euler products...it's the truncation of the sum that kills it. I was hoping that some work had been done on it.

Comment by Alex Botros

2012-05-18T22:46:11Z

yes, the divisor function is the function that counts the divisors

Comment by Alex Botros

2012-05-04T13:47:45Z

Consider that the sign of $\lambda(d_1) \lambda(d_2)$ is exactly that of $\mu(d)$. Thus, if we factor that out, we get $$p(d)=\mu(d) \sum_{lcm(d_1, d_2)}\vert \lambda(d_1) \lambda(d_2) \vert.$$ Now, it looks as though $\vert \lambda(d) \vert$ is an increasing function over squarefree $d$, any ideas of how to show that?

Comment by Alex Botros

2012-04-28T22:04:35Z

Notice, of course, that if $d$ is not squarefree, then $\mu(d)=\p(d)=0.$ If $d$ is squarefree, then $\vert \mu(d) \vert =1$. We could ask if $\vert p(d) \vert$ is usually bigger than $1$. If $d$ is prime, then $$\vert p(d) \vert=\vert \lambda (d) \vert= \frac{k}{D} \sum_{k\vert d, d\vert P_z, d\leq w}\frac{1}{\phi(d)}$$ so, given our definition of $D$, how much is lost when we further restrict $k\vert d$ as it appears in our definition of $\vert \lambda (d) \vert$?

Comment by Alex Botros

2012-04-13T03:14:09Z

The end goal, is that I would like to show $$\pi^D(N/2,N) \leq \frac{N}{log^2(DN)}$$ Not a multiple of that (which is easy to show), but that. This may seem a more reasonable request than $\pi^D(N) < \pi^1(N)$ as it shies away from touchy subjects like the twin prime conjecture. Any ideas?

Comment by Alex Botros

2012-03-23T14:29:09Z

Let me clarify: When I say it may not be the case that $\pi^D(N)$ go to infinity, what I mean is: there may be a way to show a very close relationship between $\pi^D$ and $\pi^2$. THe fact is, we can show that their difference is bounded above by some multiple of $N/log^2(N)$. That's a pretty good start

Comment by Alex Botros

2012-03-23T14:26:47Z

It may not be the case that $\pi^D(N)$ goes to infinity with $N$. Consider Brun's sieve: if $\omega(p)$ is the standard sieving function for which $$A_d=A\frac{\omega(d)}{d}+R_d$$ Then notice that the only difference between $\omega(d)$ in the case of $p,Dp-2$ tuples and $p,p-2$ tuples, is the fact that in $Dp-2$ tuples, $\omega(d)$ is also equal to one if $D\vert d$. That's the only difference between the two. Thus, Brun's "main" terms in his upper bound can be shown to be about equal for large enough $D$. THe only problem lies in coming up with a reasonable way to deal with the remainde

Comment by Alex Botros

2012-03-23T14:19:40Z

I will find it....hold on

Comment by Alex Botros

2012-03-23T00:02:53Z

That's the interesting thing: computationally $3p-2$ is far more often than $p-2$ or any other $Dp-2$, and it was shown that there are infinitely many prime tuples $p,3p-2$. It seems, however, than if $N$ is sufficiently large, and $D$ is bigger than $N^{3/8}$ say, then the above holds. I just don't know why.

Comment by Alex Botros

2011-08-30T04:13:53Z

So, what would then happen if $n$ were even?

Comment by Alex Botros

2011-08-30T04:05:05Z

FANTASTIC!!!! That's true! thank you

Comment by Alex Botros

2011-08-30T03:38:43Z

I'll give it a try. Thanks

Comment by Alex Botros

2011-08-30T03:21:27Z

So, I found the second equality again on Wikiproofs. I think it would suffice if I could prove that the first sum is greater than zero. Does anybody think that's possible without going too deeply into the actual number of prime factors of a random divisor of $n$?

Comment by Alex Botros

2011-08-08T03:50:49Z

The above isn't actually an answer, it was simply to long to write as an additional comment.

Comment by Alex Botros

2011-08-06T19:33:04Z

Yes, I see what you mean, things get all messed up if D=2, or D divides d. Thank you both, I will continue to look into it.