To flesh out a comment: the key insight for speeding up this problem is that by using $O(n)$ space, the intermediate sums can be calculated in constant time each: $S_{m,n} = S_{1,n}-S_{1,(m-1)}$, so storing partial sums from $1$ to $i$ lets all other partial sums be easily computed. This removes an $O(n)$ factor from the running time and makes the problem much more tractable.
First of all we will assume that the Goldbach's conjecture is true, so given an integer $n=kp$ with $k\geq 4$ ($k=3,2$ are easy), we will prove that $n$ is the sum of $p$ primes.
If $p=2,$ it's the Goldbach's conjecture and it's true, so we can assume that $p>2$.
Case 1 If $n$ is even, let's write
$$n=\underbrace{2+\cdots+2}_{p-2}+(n-2(p-2)),$$
and because $(n-2(p-2))=(k-2)p+4\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.
Case 2 If $n$ is odd, let's write
$$n=3+\underbrace{2+\cdots+2}_{p-3}+(n-2(p-3)-3),$$
and because $(n-2(p-3)-3)=(k-2)p+3\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.
So we proved the following statement:
$$\text{Goldbach's conjecture} \implies \text{Extension of Goldbach's conjecture}$$
But Goldbach's conjecture is a particular case of the extension of Goldbach's conjecture (when $p=2$), hence:
$$\text{Goldbach's conjecture} \iff \text{Extension of Goldbach's conjecture}$$
Conclusion The extension of Goldbach's conjecture as you define it is equivalent to the Goldbach's conjecture itself, and hence no one will ever find a counterexample or prove it unless (s)he solves Goldbach's conjecture.
Best Answer
Some thoughts
After the first split of $n$ into two or three unique primes, the problem becomes splitting larger primes into a list of smaller primes which is the problem I discuss below.
The minimum number of unique primes a larger prime can be split into is $2$ smaller primes, where $2$ is one of those unique primes e.g. $31=29+2$. If $2$ is not used the minimum number of primes a larger prime can be split into is $3$ primes e.g.
$$31=(23+5+3)=(19+7+5)=(17+11+3)=(13+11+7)$$
user477343 was playing with this idea here
We can now look at how to generate new primes by making the split into $3$ other primes and so on. The split into $2$ primes using $2$ can only be used once.
e.g. the smaller primes above split as follows
$$29=(19+7+3)=(17+7+5)=(13+11+5)$$ $$23=(13+7+3)=(11+7+5)$$ $$19=(17+2)=(11+5+3)$$ $$13=(11+2)$$ $$7=(5+2)$$
In the example of $31$ I first split this into 2 primes because that gives the largest available prime, that is $31=29+2$
I then use the split $29=(19+7+3)$ giving
$$31=19+7+3+2$$
The question is:
Is this "greedy prime" algorithm (taking the largest primes you can get from any recursive split into 2 or mainly 3 primes) always going to get you to a sum with the largest number of unique primes?
[Which includes the constraint to always decompose the largest prime in the new list first (the new list being generated immediately after a previous split).]