Every Odd Number Greater Than 1 is the Sum of at Most 485 Primes

Read-back. Let $n$ be any natural number (a non-negative integer, $n \in \{0, 1, 2, \dots\}$). Assume two things about it:

• $n$ is odd, so $n = 2k + 1$ for some natural number $k$;
• $1 < n$, a strict inequality.

Together these say exactly that $n \in \{3, 5, 7, 9, \dots\}$. No upper bound is placed on $n$, and the hypotheses can be satisfied (for example by $n = 3$), so the statement is not vacuous.

The conclusion is that there is a finite multiset $S$ of natural numbers with all three of the following properties. A multiset is an unordered collection in which an element may appear more than once.

Here $|S|$ is the number of elements of $S$ counted with multiplicity, so a prime that appears several times is counted that many times. The sum also counts each element with its multiplicity. "Prime" has its usual meaning: a natural number $p \ge 2$ whose only divisors are $1$ and $p$.

In words: every odd natural number greater than $1$ can be written as a sum of at most $485$ primes, not necessarily distinct, where the order of the terms does not matter. The bound is the non-strict "$\le 485$", so any number of primes from $1$ to $485$ is allowed. An empty collection would have sum $0$, which cannot equal $n$, so $S$ has at least one element. The statement only asserts that such an $S$ exists. It does not say that $S$ is unique and does not specify how to construct it.