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Sums of primes have been studied by number theorists for many years. Goldbach's conjecture is the most famous unsolved problem in this direction.

Here I'd like to consider weighted sums of primes. For a prime number $p$, I view $\pi(p)$ (the index of $p$) as a weight of $p$. For two different primes $p$ and $q>p$, obviously $\pi(p)p\not=\pi(q)q$ since $q$ does not divide $\pi(p)p$. Let's introduce the weighted set of primes: $$S=\{\pi(p)p:\ p\ \text{is prime}\}=\{kp_k:\ k=1,2,3,\ldots\},\tag{1}$$ where $p_k$ denotes the $k$th prime. By the Prime Number Theorem, $$np_n\sim n^2\log n\ \ \text{as}\ n\to+\infty.$$

Motivated by Question 501356, here I pose the following conjecture.

Conjecture. Let $n$ be any integer with $n\ge3070$. Then $n$ is a weighted sum of four primes. That is, we can write $n$ as $\pi(p)p+\pi(q)q+\pi(r)r+\pi(s)s$ with $p,q,r,s$ all prime. Equivalently, there are positive integers $i,j,k,m$ such that $$n=ip_i+jp_j+kp_k+mp_m.\tag{2}$$

I have verified this conjecture for $n$ up to $10^5$.

It seems hopeless to prove the conjecture by the current number theory. Maybe the conjecture holds but it can never be proved. If this is true, it will be a good example to illustrate Godel's incompleteness theorem.

QUESTION. Is there a counterexample to the conjecture? Can one show that the set $$4S:=\{s_1+s_2+s_3+s_4:\ s_1,s_2,s_3,s_4\in S\}$$ contains a complete system of residues modulo $m$ for every positive integer $m$?

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  • $\begingroup$ The sequence $np_n$ is tabulated at oeis.org/A033286 $\endgroup$ Commented Oct 8 at 8:37
  • $\begingroup$ What happens if you considers only $n$ of the form $n=ip_i+jp_j+kp_k$ ? Since the sequence $S$ grows slightly faster than $n^2$ (indeed as $n^2\log n$), probably three terms suffice to express all sufficiently large integers $n$ as $n=ip_i+jp_j+kp_k$. $\endgroup$ Commented Oct 10 at 20:19
  • $\begingroup$ Yes, I did consider the representation $n=ip_i+jp_j+kp_k$. However, my computation shows that even when $n$ is about $10^5$ there are still counterexamples to the representation. So I gave up using three terms. $\endgroup$ Commented Oct 10 at 21:18

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