Introduction
I have observed some heuristic results, and also attempted to formulate a proof, of the conjecture I formulated based on those heuristics.
But first, I want to ask about relevant references.
Then, my main part of this post, is,
I will present my conjecture and my idea of a proof. Can we and how, apply my idea to my conjecture, to prove it, and does all this make sense, or am I missing something?
Summary
In short, it looks to me (based on heuristics), that it is sufficient to observe consecutive sums over $k$-subsets ($k$-combinations) of primes, to prove all numbers greater than some constant, are a sum of exactly $k+1$ distinct primes, for every $k\ge 3$. - But I'm not sure how to approach solving the limit case to verify this on whole $\mathbb N$, or how hard will this be.
That is, are my observations relevant, or is there nothing but heruistics, to conclude from them?
(See the "The Conjecture?" section of this post, below "References?")
References?
Consider the statement:
Every integer greater than $n_0$ is the sum of (at most) $x\in\mathbb N$ (not necessarily distinct) primes.
We have $x\ge 3$, as there are arbitrarily large odd numbers which aren't sums of two primes. (Since there are arbitrarily large sequences of consecutive composite numbers.)
The following results would imply the statement (including conditions in parentheses):
Goldbach's conjecture would imply it for $n_0=1$, and $x=3$.
Goldbach's weak conjecture would imply it for $n_0=1$, and $x=4$.$^{[1]}$
The proof of the weak conjecture is "widely accepted as of 2018." according to wikipedia, and the strong (Goldbach's conjecture) is still a famous unsolved conjecture, in 2019.
References?
My first question, are there papers, works, proofs, conjectures,... other than the Goldbach's, that involve some $x \ge 4$, and the above statement?
What about when we use "exactly" instead of "(at most)" in the statement? Or if we say that primes must be "distinct" instead of "(not necessarily distinct)"? Or if both modification to the statement are used?
The Conjecture?
My main question is; Can we prove the following conjecture, using my ideas presented below? And without using "strong" results like "(weak) Goldbach's conjecture".
That is, can you help me turn my "proof idea" (see it further down), into a proof of the following conjecture? Or perhaps my "proof idea", cannot be used for anything more, other than to establish heuristics for this, making this conjecture as hard as mentioned references?
Conjecture. All natural numbers greater than some constant $n_{x}\in\mathbb N,n_{x}\lt \infty$, can be written as the sum of exactly $x$ distinct primes, for every $x \in\mathbb N,x \ge 4$.
This is based on the following definition and proposition;
Definition 1. Let $T(k, n)$ be the length of the longest consecutive run of sums of $k$-subsets of first $n$ primes. Where $n\ge 0$ and $k=0,\dots,n$ and specially $T(0, n)=T(n, n)=1$. Let $t_{k,n}$ be the smallest (first) sum of this longest consecutive run.
Then, it holds:
Proposition 1. If $N$ is the number such that all prime gaps below it are $\le T(k,n)$, then all numbers in the interval $I_{k,n}=[t_{k,n}+p_{n+1},N]$ are "trivially" a sum of exactly $x=k+1$ distinct primes, where $p_{n+1}$ is the $(n+1)$th prime.
Proposition holds by definition, as we are simply observing the fact that if set of sums $S(k, n)$ of $k$-combinations of first $n$ primes contains $T(k, n)$ consecutive values, then this is enough to cover all prime gaps $\le T(k, n)$ when summing those consecutive values with primes $p_{\gt n}$.
This means we can represent all numbers up to some upper limit $N$, as sums of exactly $x=k+1$ distinct primes, starting at some lower limit $n_{x}\le t_{k,n}+p_{n+1}$, where $t_{k,n}$ is the minimum of $S'(k,n)\subseteq S(k,n)$, where $S'(k,n)$ is the set of those consecutive sums: $T(k,n)=|S'(k,n)|$.
The conjecture considers $x\ge 4$ since for $k\ge 3$, $T(k, n)$ is not a constant (is increasing).
For Example, the $T(9,12)$ was used to solve: "How many natural number between 100 and 1000 exist which can be expressed as sum of 10 different primes."
In essence, I can apply the proposition, to bounded subsets of $\mathbb N$, to prove results on those numbers, with the help of computations, and strengthen the conjecture. But I'm not sure how to approach the limiting case, to extend this "result proving" to whole (an unbounded subset of) $\mathbb N$.
Proof idea - limit case of the proposition?
Now, we should be able to use this proposition, to prove the conjecture, by fixing some $k\ge 3$, and observing consecutive $n=k,k+1,k+2,k+3,\dots$ and observing unions of intervals $I_{k,n}=[t_{k,n}+p_{n+1},N]$, to collect all numbers in $\mathbb N$ greater than $n_{x}\le t_{k,n}+p_{n+1}$, as $n\to\infty$.
That is, what is left to prove is, to prove that $T(k,n)$ "grows fast enough", for fixed $k$, as $n\to\infty$, relative to growth of prime gaps. Since we have $T(k,n)$ grows faster for larger $k$, if we prove $k=3$ grows fast enough, then this implies $k\gt 3$ grows fast enough as well?
This appears to be very true, by a lot. We can start examining results given by proposition, to prove all numbers, up to some $N$, can be represented as sums of $x=k+1=4$ distinct primes.
Take $k=3$. We can see ($t_{3,n}=18,\forall n\ge 5$), and we define $G$ as largest prime gap below $N$.
$$ \begin{array}{cccc} n & G \le T(3,n) & [t_{3,n}+p_{n+1},N]=I_{3,n} & \bigcup\limits_{m\le n} I_{3,m} \\ 7 & 8 \le 10 & [37, 89] & [37, 89] \\ 8 & 14 \le 18 & [41, 113] & [37, 113] \\ 9 & 18 \le 22 & [47, 523] & [37, 523] \\ 10 & 20 \le 22 & [49, 887] & [37, 887] \\ 11 & 22 \le 40 & [55, 1129] & [37, 1129] \\ 12 & 34 \le 42 & [59, 1327] & [37, 1327] \\ 13 & 36 \le 46 & [61, 9551] & [37, 9551] \\ 14 & 44 \le 60 & [65, 15683] & [37, 15683] \\ 15 & 52 \le 66 & [71, 19609] & [37, 19609] \\ 16 & 72 \not\le 70 & - & - \\ 17 & 72 \not\le 70 & - & - \\ 18 & 72 \not\le 70 & - & - \\ 19 & 72 \le 100 & [89, 31397] & [37, 31397] \\ 19 & 86 \le 100 & [89, 155921] & [37, 155921] \\ 19 & 96 \le 100 & [89, 360653] & [37, 360653] \\ \dots&\dots&\dots&\dots \end{array} $$
Notice that sometimes we have $\not\le$ so we have to go over multiple values of $n$ (or simply skip to next $n$ such that $T(k,n)$ will be larger than the gap) , to get the next interval.
And sometimes, we can go over multiple gaps with the same $T(k, n)$ value (same $n$), as observed in the last three entries in the above example table of intervals.
Here, we have so far proven all numbers in $[37, 360653]$ are a sum of exactly $x=4$ distinct primes, by observing just $n\le 19$ for $k=3$, and applying the proposition $1$.
Now, can we "take the limit" of this procedure, to prove that all numbers greater than $37$ are a sum of exactly $x=4$ distinct primes?
How can we prove that as $n\to\infty$, that the union of all of these intervals $\bigcup\limits_{m\le n} I_{3,m}$, will contain all natural numbers greater than $n_{4}\le t_{7,n}+p_{7+1}=18+19=37$?
This proves the conjecture for $x=4$.
Similarly, now we consider larger $k$, and we have a similar sequence of intervals $I_{k,n}$.
Can we then prove this for all $k\ge3$? - This would prove the conjecture.
It seems, lower bounds of the intervals $I_{k,n}$ seem much much smaller than $N$'s we cover on the upper bounds of the intervals. But how do we prove this formally, to prove the conjecture?
In my previous question, I try to find a "closed form" for $T(k, n)$. I find that (observe based on computed data), that $T(k, n)\approx \sum \sum (p_{k,n})$ is approximately represented by sums of sums of prime gap sequences $(p_{k,n})$ whose offset (starting prime gap) depends on $k,n$. (This is for for $k\ge 3$).
Can we prove some asymptotic relation like this, and use it to prove the conjecture?
Or am I missing something that makes this much harder than it looks?