1
$\begingroup$

I am curious to know if a lower bound on the number of ways (call this $\beta$ and assume $p_1 + p_2$ distinct from $p_2 + p_1$) in which two primes $p_1, p_2$ that add up to a given even integer $n$, has been conjectured (Obviously, I would think not proven yet). For $n = 2^t$, a really good lower bound that I have arrived at is $$\beta \ge \pi(\pi(n))$$

x-axis is $n=2^t$ and y-axis shows $\beta$

enter image description here

A similar chart for $n = 2 * 3^t$ is shown below.

x-axis is $n=2 * 3^t$ and y-axis shows $\beta$

enter image description here

For all other compositions of n, $\pi(\pi(n))$ while not such a good lower bound, provides a lower bound nevertheless.

Here is the main idea. Let K represent the number of odd primes $<n$, therefore $K = \pi(n) - 1$. Let $P_1, P_2, \ldots, P_r$ represent the $r$ odd primes $< \sqrt{n}$. Now we need to find how many of these K primes add up to a prime and here is where we end up with probability. However, we can take a different approach.

Let $n = 2^t$ and let $n = t_1 \bmod P_1 = t_2 \bmod P_2 = \ldots = t_r \bmod P_r$. We are interested in identifying how many of the $K$ primes leave none of these remainders ($t_i$). If a prime P has a different remainder (different from the $t_i$) modulo each of the $r$ primes , then we are sure that $n-P$ is also prime (since then $n-P \ne 0 \bmod P_i$ for $1 \le i \le r$).

But unlike a sequence of integers, the set of consecutive primes do not leave remainders in a "periodic" manner. For example, there are 135 odd primes less than 770. Of these 135, there are 66 primes that are $1 \bmod 3$ and 68 primes that are $2 \bmod 3$. In some cases, some remainders may not occur. However, we can be sure of one thing. The number of remainders that are 1 mod 3 (or 2 mod 3) will be close to $1/2$. If the number of remainders is $> 1/2$, it will be greater than $1/2$ by a $\delta$. The number of remainders that are $1$ or $2$ or $3$ or $4 \bmod 5$ will be close to $1/4$ or higher or lower than $1/4$ by $\delta$ and so on for all $P_i$.

We assumed n is a power of 2. However if n has a prime factor q, then we simply can ignore this as none of the primes will leave a zero remainder modulo q (except q). We can now pass K through the classic Legendre sieve.

Let Q = product of all odd primes less than $\sqrt{n}$. (There may be a more elegant and concise way to write the below using the Mobius function).

Lower Bound for $$\beta \approx K - (\sum_{i=1}^r \lfloor\frac K {P_i-1}\rfloor )+ (\sum_{i=1 \text{ and } i<j<r}^r \lfloor\frac K {(P_i-1)(P_j-1)}\rfloor) - \ldots $$

Suppose $n$ has $s$ odd prime factors such as $q_1, q_2, \ldots, q_s$. Some of these $q_i$ could be equal to some of the $P_i$. So the modified LB will look like:

Lower Bound for $$\beta \approx K - (\sum_{i=1, (P_i \ne q_i \text{ for all }i)}^r \lfloor\frac K {P_i-1}\rfloor )+ (\sum_{i=1 \text{ and } i<j<r \text { and } (P_i \ne q_i \text{for all i)}}^r \lfloor\frac K {(P_i-1)(P_j-1)}\rfloor) - \ldots $$

for $n$ with odd prime factors. Essentially, in the sieve, we ignore any $P_i$ that is a prime factor of $n$.

For large n, actual $\beta$ will be higher as the remainders of $n \bmod P_i$ are not equally distributed in $0,1,2,\ldots,P_i-1$.

$\endgroup$
3
  • 1
    $\begingroup$ This a quantitative form of the Goldbach conjecture (described as the "extended Goldbach conjecture" here). And yes, it's still unproven. $\endgroup$ May 25, 2016 at 6:24
  • $\begingroup$ @Greg, Thanks for your comment. Why does the Sieve method (as shown above) not work in the proof for GC? Sieves do a perfectly good job for counting primes.. Thanks in advance! $\endgroup$
    – sku
    May 26, 2016 at 5:08
  • 1
    $\begingroup$ They don't, really. There are exponentially many terms in your inclusion-exclusion sum, and removing the floor functions results in an error in each term. This is a standard problem which sieves try to overcome, but (so far) unsuccessfully in the case of Goldbach. $\endgroup$ May 26, 2016 at 8:35

2 Answers 2

0
$\begingroup$

The lower bound is one partition unless the Cartesian plane has an inherent contradiction. It is defined by linear equations for prime intercepts:

$x = 2p_{1} $

$y = -0.5x - p_{2}$

where $x < y < n$

The $x$ intercept is $2p_{1}$ and the $y$ intercept is $p_{2}$ such that the lines intersect at the coordinates of one even $n$. Graphically and logically, there must be a point of intersection - and that intersection must lie in the region defined by $x\geq3$ and $y\leq n-3$. If there were no such intersection, the coordinate plane would offer no solution to the intersection of a slope with a vertical line, which is impossible.

Line and slope in partitions:

Linear line and slope in partitions

Line and slope as coordinates:

Linear line and slope as coordinates

$\endgroup$
0
$\begingroup$

N is the sum of primes p and w. There are n-1 ways to write N = p+q for integers p and q. Random p, q would be both prime with probability 1 / ln p and 1 / ln (N-p). However we know that p, q are both odd unless N <= 4, so we only use half the p’s but the q’s = N-p are twice as likely to be prime. We can look at N modulo 6 which influences the number of solutions. Ignoring this, the expected number is N, pi(N), divided by the average of 2 / ln (N-p).

Now you are looking for a lower bound. The count is a bit lower if N is not divisible by 6 and a bit higher otherwise, same for N multiple of 5, 7, 11 etc. so you calculate the expected number for n not divisible by 3, 3 or 5, 3 or 5 or 7. Then you have many values of N; your actual number will be variable. You can try to calculate variance and standard variation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .