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$n/2$ and n$n$ equals n-6$n-6$, n-4$n-4$ or n-2$n-2$. Therefore it seems better to use the fact that the largest prime gap less than 15x10^18$15\times10^{18}$ is 1510less than $1526$ so for odd 4x10^18 < n < 15x10^18$3068 < n < 15\times10^{18}$ there is definitely a prime, p$p$, such that n - 3024 < p < n - 1510 and as 6 < n$n-1534\leq p \leq n-8$. As - p < 4x10^18$8\leq n - p\leq 1534 < 4\times10^{18}$, n - p$n - p$ is the sum of 2 distinct primes, neither of which could equal q as$p$ because they would have to be much smaller than q as n is greater than 4x10^18are each at most $1534$ and $p>3068-1534=1534$. This implies that all odd numbers less than 15x10^18$3068<n<15\times10^{18}$ are the sum of 3 distinct primes.
For As the first part to do withlargest prime ladders, I misstated the problem. The difference between consecutive primes on the ladder could be between 4 and 4*10^18 and the issue was that if the gaps between consecutive primes in the ladder, p and q, p > qgap less than $3100$ is less than $36$, were 4x10^18 apart then if n equals p+2for odd $88<n<3100$ there is a prime, p+4 or p+6$q$, n-q > 4x10^18 and n-p = 2such that $n-44\leq q\leq n-8$. As $8\leq n-q\leq 44$, 4 or 6 none of which are$n-q$ is the sum of 2 distinct primes. This is all from page 28, neither of Helfgott's arxiv pdf released in 2015which could equal $q$ because they are each at most $44$ while $q>88-44=44$. He showsThis implies that 4x10^18 + 2 isall odd numbers $88<n<15\times10^{18}$ are the sum of two3 distinct primes on pg 28. The odd numbers between $19$ and on$87$ can easily be checked.
On pg 305 states$305$ of Helfgott's proof 'The Ternary Goldbach Problem', it mentions that therein (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$3$ to 8.8x10^30$8.8\times10^{30}$ with consecutive primes on the ladder having difference less than or equal to 4x10^18 - 6$4\times10^{18} - 6$. So if p$p$ and q$q$ are any consecutive primes on the ladder, as they have difference less than or equal to 4x10^18 with - 6$p>q$, then $p-q\leq 4\times10^{18} - 6$. Then for any odd n between q + 8$8\times10^{18}$ and p + 6 inclusive$10^{27}$, nthere exist - q is less than 4x10^18$p,q$ such that $p$ and greater than 6$q$ are consecutive primes on the prime ladder, $p>q$ and $q + 8\leq n\leq p + 6$, so $8\leq n - q\leq 4\times10^{18}$. So nTherefore - q$n - q$ is the sum of 2 distinct primes $r$ and $s$, and so n$n$ is the sum of three3 distinct primes (as q$q$ cannot be equal to the 2 distinct primes that sum to n - q$r$ or $s$ as n - q < 4x10^18$r+s=n - q \leq 4\times10^{18}$ so if n > 8x10^18$n > 8\times10^{18}$ then q > n - q and$q>4\times10^{18}\geq n-q=r+s$. Remember that if n < 8x10^18$n \leq 8\times10^{18}$ then the argument at the beginning of this shows that n$n$ is the sum of 3 distinct primes). I believe that this shows that all odd numbers in the interval 4x10^18$4\times10^{18}$ and 10^27$10^{27}$ are the sum of 3 distinct odd numbers (the result pretty much follows from the work done within Helfgott's proof).
