Every Odd Number Greater Than 1 is the Sum of at Most 351 Primes

Read-back: odd_sum_le_351_primes

Let $n$ be any natural number, with $\mathbb{N} = \{0, 1, 2, \dots\}$. Assume two things: $n$ is odd, meaning $n = 2k + 1$ for some natural number $k$, and $n > 1$. Together these mean $n$ ranges over the odd numbers $3, 5, 7, 9, \dots$. The value $n = 1$ is excluded because of the strict inequality, and $n = 0$ is excluded because it is not odd. Both hypotheses can be satisfied, for example by $n = 3$, so the statement is not vacuous.

Under these assumptions, the statement says there is a finite multiset $s$ of natural numbers with the following three properties:

A multiset is an unordered finite collection in which an element may appear more than once. Here $|s|$ is the number of elements counted with multiplicity. So the claim is that $n$ can be written as a sum of at most $351$ primes, where the order of the summands does not matter and the same prime may be used several times. Each summand is a prime in the usual sense: a natural number $p \ge 2$ whose only divisors are $1$ and $p$. The prime $2$ is allowed, and nothing requires the summands to be odd or distinct.

The bound $351$ is an upper bound only, and no minimum number of summands is required. If $n$ is itself prime, the one-element multiset $\{n\}$ already satisfies the statement. The empty multiset is never a witness, because its sum is $0 \ne n$, so at least one prime always appears. The statement asserts only that such a multiset exists. It says nothing about whether the multiset is unique or how it could be constructed.