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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?

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For those as ignorant as I am (or was!): A prime $p$ is a Chen prime if $p + 2$ is either a prime or a product of two primes. –  Joseph O'Rourke Mar 12 '11 at 1:00

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Yes. Sieve methods are much better at proving upper bounds than lower bounds. By the Selberg sieve, or alternatively using the combinatorial sieve, you can prove that the number of Chen primes is bounded above by $C_2 N/\log^2 N$ for some particular value of $C_2$. (ed: Please see Terry Tao's important caveat below, which I neglected in my initial answer.)

I don't know the details offhand; it would take some work to determine a particular value of $C_2$, but this is definitely possible. The Selberg sieve is probably easiest, you can read, for example, Halberstam and Richert. (Just read the first chapter on the Selberg sieve -- no need to delve into the more difficult portions of the book.)

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One caveat: in order to get $N/\log^2 N$ type upper bounds, one should place lower bounds on the two primes that are allowed to factor $p+2$ (specifically, they should be at least $p^c$ for some constant $c>0$). Without this restriction, the density of almost primes increases to about $\log \log N / \log N$ (ultimately because of the log-log divergence of $\sum_p \frac{1}{p}$). The "right" way to count almost primes is to weight then by the second von Mangoldt function $\Lambda_2(n) := \sum_{d|n} \mu(d) \log^2 \frac{n}{d}$, which damps out the almost primes with small prime factors. –  Terry Tao Mar 12 '11 at 1:29
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Bombieri's paper ams.org/mathscinet-getitem?mr=396435 has the right asymptotics for this weighting of the twin prime problem, but it is conditional on the Elliot-Halberstam conjecture. However, for crude upper bounds, placing a floor on the prime factors is indeed easier, and the Selberg sieve is probably the simplest way to proceed. –  Terry Tao Mar 12 '11 at 1:32
    
Do any of you think it is possible to prove the second part using the selberg sieve, namely that for infinitely many $\Omega_i$ there are more than $\frac{i^2}{13\ln(i^2)}$ chen primes? –  Alex Botros Mar 12 '11 at 1:59
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Chen's proof might work verbatim for all intervals of the type [N,2N]. In general, the interval [N,2N] "looks like" [1,N] from the standpoint of sieve methods. I don't know the proof well enough to promise this -- in particular the "reversal of roles" trick might interfere with this -- but I believe it is worth a careful reading of Chen's proof. –  Frank Thorne Mar 12 '11 at 3:15

Explicit lower bounds for twin almost primes up to $x$ (for large enough $x$) are known, see e.g., Corollary 25.12 in the book of Friedlander-Iwaniec. (The constant they get for $p$ with $p+2$ having at most two prime factors, each of which is at least $x^{3/11}$, is $1/31$). I don't think the known upper bounds are good enough to deduce the corresponding statement between $x/2$ and $x$. Since moreover the use of primes to start the linear sieve blocks it from being "local" enough, I don't think we know lower bounds for this question for all dyadic intervals (in other words, there does not seem to exist a version of, say, the Bombieri-Vinogradov Theorem for a dyadic interval). But "almost all" such intervals might possibly be accessible differently.

By the way, I'm curious as to why a result of Chen seems to be attributed to Green-Tao in the original question...

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Thanks. I will look that up asap. The lower bounds always seem to require that extra $\ln(N)$ term in the denominator; it's always $\ln^2(N)$, not $\ln(N)$. I don't need the modified lower bound above to be accurate on every $\Omega_i$, I just need it not to be wrong indefinitely. As to why I attributed the original lower bound to Tao and Green, this is simply due to the fact that they talk explicitly about the interval $(N/2,N]$ which is quite useful for me. –  Alex Botros Mar 12 '11 at 9:45
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I had missed that you want $1/log(x)$. Such a lower bound is certainly false for every large enough $x$, since it contradicts the expected asymptotic formula (the gain of $\log\log x$ from allowing small prime factors in the $p_1p_2$ is not sufficient to compensate for a loss of a logarithm), and it can certainly be shown to be false by easy upper bound sieve methods. –  Denis Chaperon de Lauzières Mar 12 '11 at 10:01
    
Yes, this is most definitely not a result of Green and Tao! It is mentioned in one of our papers, but I'm certain that a clear reference to Chen's work is given. –  Ben Green Mar 12 '11 at 10:57
    
Darn. Thought it could work. Sorry if this seems like beating a dead horse, but I feel the need to understand: Denis, you said that the gain of $\log\log x$ from allowing small prime factors in the $p_1p_2$ is not sufficient to compensate for a loss of a logarithm. Can you clarify that, I think it's they key to my foolishness. –  Alex Botros Mar 12 '11 at 17:22
    
The number of $p\leq x$ with $p+2$ prime is too small for what you want, so the only hope would be $p+2=p_1p_2$ with $p$ between $x/2$ and $x$. You can estimate this from above by summing over $p_1\leq x$ the number of primes $p_2\leq x/p_1$ such that $p_1p_2-2$ is prime. Again, for fixed $p_1$, upper-bound sieve will give you an upper bound of type $x/p_1 \times 1/(\log x)^2$, uniformly with respect to $p_1$. The sum over $p_1$ is $\ll x(\log\log x)/(\log x)^2$. –  Denis Chaperon de Lauzières Mar 12 '11 at 18:13

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