Finite sums of prime numbers $\geq x$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:36:20Z http://mathoverflow.net/feeds/question/93002 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93002/finite-sums-of-prime-numbers-geq-x Finite sums of prime numbers $\geq x$ François Brunault 2012-04-03T14:06:30Z 2012-04-05T18:00:47Z <p>Let $S_x$ be the set of finite sums of prime numbers $\geq x$. In other words, let $S_x$ be the submonoid of <code>$(\mathbf{Z}_{\geq 0},+)$</code> generated by the set $\mathcal{P}_{\geq x}$ of prime numbers $\geq x$.</p> <p>It is easy to see that $S_x$ contains every sufficiently large integer. This follows from the classical fact that given two coprime integers $a$ and $b$, every sufficiently large integer, in fact every integer $\geq (a-1)(b-1)$, is of the form $ma+nb$ for some $m, n \in \mathbf{Z}_{\geq 0}$. See for example <a href="http://en.wikipedia.org/wiki/Coin_problem" rel="nofollow">this page</a>.</p> <p>Let $N_x$ be the largest integer which is not in $S_x$.</p> <p><strong>Examples</strong> :</p> <p>If $x=2$ then <code>$S_2 = \{0\} \cup \mathbf{N}_{\geq 2}$</code> so that $N_2=1$.</p> <p>If $x=3$ then <code>$S_3 = \{0,3\} \cup \mathbf{N}_{\geq 5}$</code> so that $N_3=4$.</p> <p>By definition, we have $N_x \geq x-1$ (in fact parity considerations imply that $N_x \geq 2x-2$ for $x \geq 3$).</p> <p>On the other hand, given that there are at least two primes in the interval $[x,2x]$, the above classical fact implies that $N_x \ll x^2$.</p> <blockquote> <p>What is the asymptotic behaviour of $N_x$ as $x \to \infty$ ?</p> </blockquote> http://mathoverflow.net/questions/93002/finite-sums-of-prime-numbers-geq-x/93013#93013 Answer by Woett for Finite sums of prime numbers $\geq x$ Woett 2012-04-03T16:11:21Z 2012-04-03T18:06:29Z <p>(Edited according to the comments below)</p> <p>In his article titled 'Sums of Distinct Primes', Kløve conjectured, on the basis of computational evidence, that $\displaystyle \lim_{x \rightarrow \infty} \dfrac{N_x}{x} = 3$. This would imply the binary Goldbach conjecture (for large enough $x$) in the following way: if every integer larger than $(3 + \epsilon)x$ can be written as sum of primes, where those primes are all larger than $x$, then, in particular, every even number between $(3 + \epsilon)x$ and $4x$ can be written as a sum of two primes.</p> http://mathoverflow.net/questions/93002/finite-sums-of-prime-numbers-geq-x/93032#93032 Answer by Aaron Meyerowitz for Finite sums of prime numbers $\geq x$ Aaron Meyerowitz 2012-04-03T19:06:40Z 2012-04-03T19:06:40Z <p>Just to spell it out, If Golbach's conjecture is false, then at least once $N(x)>4x$, but that seems extremely unlikely. It is certain that $N(x)>3x$ infinitely often, but there is every reason to believe that $N(x)&lt;(3+\epsilon)x$ with finitely many exceptions. If so, then we could equivalently define $N_x$ as the largest odd integer which is neither prime nor a sum of $3$ primes all at least $x$. </p> <p>it will frequently happen that the next prime after $p$ is at least $p+8.$ In these cases we can write $3p$ as a sum of primes at least $x=p$ however at least one of the odd numbers $3p+2,3p+4,3p+6$ is not itself prime and hence not a sum of primes all greater than $x=p.$ This establishes that $N(x)>3x$ infinitely often. </p> <p>There is no proof that every even integer is a sum of two primes. Were there a counter-example, $E$, it would need to be a sum of at least 4 primes and hence furnish an example of $N(x)>4x$ where $x$ is $\frac{E}{4}.$ However there is every reason to believe the stronger statement that for every $\epsilon \gt 0$ there is an $M(\epsilon)$ such that $2m$ is a sum of two primes, both larger than $m(1-\epsilon)$ for all $m \gt M(\epsilon).$ If so, then $N(x)&lt;(3+\epsilon)x$ provided that $x$ is reasonably larger $\frac{M(\epsilon)}{1-\epsilon}.$</p> http://mathoverflow.net/questions/93002/finite-sums-of-prime-numbers-geq-x/93235#93235 Answer by Anonymous for Finite sums of prime numbers $\geq x$ Anonymous 2012-04-05T18:00:47Z 2012-04-05T18:00:47Z <p>According to "The three primes theorem with almost equal summands" by Baker and Harman, every large odd $N$ is a sum of three primes each of size $\sim N/3$. (In fact, within $N^{4/7}$ of $N/3$-- this is much closer than we need, so we could use weaker results of earlier authors, if preferred.) </p> <p>For odd $N$, this gives $N$ as a sum of three primes, each at least $\sim N/3$. If $N$ is even, take a prime $p$ of size $\sim N/4$ (which exists by the prime number theorem), and write $N-p$ as a sum of three primes of size about $N/4$. So we get $N$ as a sum of four primes at least $\sim N/4$. </p>