Math Genius: For what percentage of numbers does this proof of Goldbach’s conjecture hold?

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For what percentage of numbers does the below inequality hold?

$$ pi(2m) > frac{phi(2 m) -1}{2} $$

where $m$ is not a prime or $1$, $pi(m)$ is the number of primes less than $m$ and $phi$ is the Euler totient function.

Background + Proof

I was trying to prove Goldbach’s Conjecture are realised I could do it for certain numbers.

Consider the following where $m$ is not a prime:

$$ (2m)! = 2m^2 (m^2 -1)(m^2 -4)(m^2 -9) dots (m^2 -(m-1)^2)$$

Notice, $(2m)!$ will contain primes of the form $p_k =m pm n leq 2m$. The number of such primes are $pi(2m)$. Further, the $m$ and $n$ must be co-prime. Then the number of possible “slots” of the form $m pm n$ which can house primes (disregarding $p_1 = 2$) are:

$$ S(m) = phi(2 m) -1 $$

To derive the above the following are considered:

$1$. If $m$ is even then $(m^2 – (text{even})^2) = text{even}$

$2$. $m$ is not a prime as mentioned before.

$3$. $m – (m-1)$ is cannot be a prime.

Now, if:

$$ pi(2m) > frac{S(m)}{2} $$

Then we can prove Goldbach’s conjecture for that number as that would imply one of the slots $m+n = p_i$ has a corresponding slot housing a prime $m+n =p_k$. And thus, if add both of them:

$$ 2m = p_i + p_k$$


Consider $m = 15 = 3 cdot 5$. After considering $30!$ eliminating “slots”:

$$ S(15) = (1-frac{1}{3})(1-frac{1}{5}) 30 – 1= 15 $$

where the remaining slots are:

$$ (m^2 -1),(m^2 -7^2),(m^2 -11^2),(m^2 -13^2),(m^2 -17^2),(m^2 -19^2),(m^2 -23^2),(m + 29) $$

However, the number of primes are $pi(30) = 10$. Since, there are more primes than “slots” available for them. Then one of the slots of the form $m$.

(Note, I didn’t read the background, just the question.)

One has
$$pi(2x) sim frac{2x}{log x}$$
by the prime number theorem and
$$phi(x) gg frac{e^{-gamma} x}{log log x}$$
by standard estimates (see the wikipedia page, so since the latter grows much faster there can only be finitely many integers satisfying your inequality (and this finite set has density zero, obviously). More precisely, one has:

$$frac{n}{phi(n)} le e^{gamma} left( log log n + frac{2.5}{e^{gamma} log log n} right)$$
for all $n ge 3$ except for $n = 3 times 5 times 7 times 11 times 13 times 17 times 19 times 23$,

(See Lemma 4 here:

$$pi(x) < 1.25506 cdot frac{x}{log x}$$

for $x ge 17$ (see and just from these inequalities (computing $phi(223092870)$ by hand) we already get
$$frac{phi(2x) – 1}{2} > pi(2x), quad x > 10^6$$

But then the smaller examples one can check by computer, and find the inequality you want holds for exactly $649$ integers, the largest one being $45045$ with
$$frac{phi(90090) – 1}{2} = 8639.5 < 8726 = pi(90090).$$

We can rephrase your question as: for which $m$ does
pi(2m) geq frac{phi(2 m)}{2}

hold, or equivalentely
frac{phi(2m)}{pi(2m)} leq 2.

Now for large enough $m$, $pi(m) approx frac{m}{log(m)}$, so this inequality becomes (roughly speaking)
frac{phi(2m)log(m)}{m} leq 2.

One expression for $phi(k)$ is $kprod_{p mid k}left(1-frac1pright)$, where the variable $p$ runs over the primes, so we can rewrite this as
log(m)prod_{p mid 2m}left(1-frac1pright) leq 1.

Taking logarithms on both sides,
log(log(m)) + sum_{p mid 2m}logleft(1-frac1pright) leq 0.

For large $p$, $-frac1p$ is a good approximation for $logleft(1-frac1pright)$, so the question becomes
log(log(m)) – sum_{p mid 2m} frac1p leq 0.

Now a standard approximation tells us that this last sum is approximately equal to $log(log(p))$ where $p$ is the largest prime in the sum. Clearly the largest prime in this sum will be much smaller than $m$, because we took the best case $m$: where $m$ is a product of lots of distinct small primes.

Thus your inequality should hold almost never.

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