Every Odd Number Greater Than 1 is the Sum of at Most 973 Primes

Read-back of odd_sum_le_973_primes. Let $n$ be any natural number $n \in \mathbb{N} = \{0, 1, 2, \dots\}$ with two hypotheses:

• $n$ is odd, meaning $n = 2k + 1$ for some $k \in \mathbb{N}$;
• $1 < n$, which is strict.

Together these say that $n$ is an odd number with $n \ge 3$. Both hypotheses can be satisfied, for example by $n = 3, 5, 7, \dots$. The cases $n = 0$ and $n = 1$ are excluded, and so are all even $n$.

The statement says that for every such $n$ there exists a finite multiset $s$ of natural numbers satisfying all three of these conditions:

Here are the terms and edge cases:

• A multiset is an unordered finite collection in which an element may appear more than once.
• $|s|$ is the number of elements of $s$ counted with multiplicity. The bound is "at most 973" ($\le$), not "exactly 973".
• The sum also counts each element with its multiplicity.
• "Prime" is the usual notion for natural numbers: $p \ge 2$ and the only divisors of $p$ are $1$ and $p$.

So the conclusion says that $n$ can be written as a sum of at most $973$ primes, where repeats are allowed and order does not matter. Nothing requires the primes to be distinct, to be odd, or to be different from $2$. Nothing requires the number of summands to be any particular value other than being at most $973$; a single summand ($s = \{n\}$ when $n$ is prime) is allowed. The empty multiset has sum $0$, so it can never be a witness, because $n \ge 3$. The statement only asserts that such an $s$ exists. It says nothing about $s$ being unique or how to construct it.