Weak Goldbach conjecture with distinct primes for odd integers between $4\times 10^{18}$ and $10^{27}$

Question

This is related to the conjecture that all odd integers greater than $17$ can be written as the sum of 3 distinct primes.

Schinzel showed that the Goldbach conjecture implied this in 1959 and as the Goldbach conjecture has been verified up to $4\times10^{18}$ by Oliveria e Silva, Herzog and Pardi, this conjecture holds up to there as well.

Vinogradov's proof that all sufficiently large odd integers are the sum of three primes implies that as the number of representations of a sufficiently large odd integer as the sum of three primes is large enough, it must be the sum of 3 distinct primes.

Harald Helfgott's proof of the weak Goldbach conjecture also implies that all odd integers greater than $10^{27}$ can be represented as the sum of 3 distinct primes (assuming I've understood it correctly).

So the interval between $4\times 10^{18}$ and $10^{27}$ is the remaining interval on which to verify whether all odd integers greater than $17$ can be written as the sum of 3 distinct primes.

In Helfgott's proof, he uses the fact that the Goldbach conjecture is verified up to $4\times 10^{18}$ along with a prime ladder to show that the odd integers in this interval are the sum of 3 primes. But that prime ladder is a list of primes from 3 to beyond $10^{27}$ such that consecutive primes have difference at least 6 and at most $4\times 10^{18}$. Thus for odd $n$ between $4\times 10^{18}$ and $10^{27}$, there is always a $p$ in the ladder such that $n-p$ is equal to an even number less than $4\times 10^{18}$ and is therefore the sum of 2 primes, $q$ and $r$, so $n = p + q + r$.

As the verification of the Goldbach conjecture up to $4\times 10^{18}$ showed that each even integer was the sum of 2 distinct primes, it would be necessary to create a similar prime ladder such that consecutive primes had difference at least $8$ (as $n-p=6$ could only be written as 3 primes as $n = p + 3 + 3$ which is not a representation with distinct primes) and ensuring that odd integers between $4\times 10^{18}$ and $8\times 10^{18}$ are the sum of 3 distinct primes (as $p$ could potentially equal $q$ or $r$ for $n$ in this interval).

How difficult is it to make such a ladder and verify that primes between $4\times 10^{18}$ and $8\times 10^{18}$ are the sum of three distinct primes? For the second part I imagine finding a prime, $s$, just below $4\times 10^{18} - 8$ and a prime, $t$, just above $4\times 10^{18}$ would solve it as by adding any even number less than $s$ to $s$ you could reach any odd number up to just below $8\times 10^{18}$ and by adding even numbers less than $t$ to $t$ you could reach the other odd numbers less than $8\times 10^{18}$.

Answer 1

The slight issue with Will Sawin's answer above for the second part where he suggests Bertrand's postulate is the case where the prime in between $n/2$ and $n$ equals $n-6$, $n-4$ or $n-2$. Therefore it seems better to use the fact that the largest prime gap less than $15\times10^{18}$ is less than $1526$ so for odd $3068 < n < 15\times10^{18}$ there is definitely a prime, $p$, such that $n-1534\leq p \leq n-8$. As $8\leq n - p\leq 1534 < 4\times10^{18}$, $n - p$ is the sum of 2 distinct primes, neither of which could equal $p$ because they are each at most $1534$ and $p>3068-1534=1534$. This implies that all odd numbers $3068<n<15\times10^{18}$ are the sum of 3 distinct primes. As the largest prime gap less than $3100$ is less than $36$, for odd $88<n<3100$ there is a prime, $q$, such that $n-44\leq q\leq n-8$. As $8\leq n-q\leq 44$, $n-q$ is the sum of 2 distinct primes, neither of which could equal $q$ because they are each at most $44$ while $q>88-44=44$. This implies that all odd numbers $88<n<15\times10^{18}$ are the sum of 3 distinct primes. The odd numbers between $19$ and $87$ can easily be checked.

On pg $305$ of Helfgott's proof 'The Ternary Goldbach Problem', it mentions that in (H. A. Helfgott and David J. Platt. Numerical verification of the ternary Goldbach conjecture up to 8.875 · 1030 . Exp. Math., 22(4):406–409, 2013) there is a prime ladder from $3$ to $8.8\times10^{30}$ with consecutive primes on the ladder having difference less than or equal to $4\times10^{18} - 6$. So if $p$ and $q$ are any consecutive primes on the ladder with $p>q$, then $p-q\leq 4\times10^{18} - 6$. Then for any odd n between $8\times10^{18}$ and $10^{27}$, there exist $p,q$ such that $p$ and $q$ are consecutive primes on the prime ladder, $p>q$ and $q + 8\leq n\leq p + 6$, so $8\leq n - q\leq 4\times10^{18}$. Therefore $n - q$ is the sum of 2 distinct primes $r$ and $s$, and so $n$ is the sum of 3 distinct primes (as $q$ cannot be equal to $r$ or $s$ as $r+s=n - q \leq 4\times10^{18}$ so if $n > 8\times10^{18}$ then $q>4\times10^{18}\geq n-q=r+s$. Remember that if $n \leq 8\times10^{18}$ then the argument at the beginning of this shows that $n$ is the sum of 3 distinct primes). I believe that this shows that all odd numbers in the interval $4\times10^{18}$ and $10^{27}$ are the sum of 3 distinct odd numbers (the result pretty much follows from the work done within Helfgott's proof).