Number as Sum of Distinct Primes
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Theorem
For $n \ne 1, 4, 6$, $n$ can be expressed as the sum of distinct primes.
Proof
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes.
Then $S = \set {2, 3, 5, 7, 11, 13, \dots}$.
By Bertrand-Chebyshev Theorem:
- $s_{n + 1} \le 2 s_n$ for all $n \in \N$.
We observe that every integer $n$ where $6 < n \le 6 + s_6 = 19$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_5} = \set {2, 3, 5, 7, 11}$.
Hence the result by Richert's Theorem.
$\Box$
Here is a demonstration of our claim:
\(\ds 1\) | \(\text {is}\) | \(\ds \text {less than the smallest prime } 2\) | ||||||||||||
\(\ds 2\) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds 3\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds 4\) | \(\ne\) | \(\ds 2 + 3\) | ||||||||||||
\(\ds 5\) | \(=\) | \(\ds 5\) | ||||||||||||
\(\ds 6\) | \(\ne\) | \(\ds 2 + 3 \text { or } 2 + 5\) | ||||||||||||
\(\ds 7\) | \(=\) | \(\ds 7\) | ||||||||||||
\(\ds 8\) | \(=\) | \(\ds 3 + 5\) | ||||||||||||
\(\ds 9\) | \(=\) | \(\ds 2 + 7\) | ||||||||||||
\(\ds 10\) | \(=\) | \(\ds 2 + 3 + 5\) | ||||||||||||
\(\ds 11\) | \(=\) | \(\ds 11\) | ||||||||||||
\(\ds 12\) | \(=\) | \(\ds 2 + 3 + 7\) | ||||||||||||
\(\ds 13\) | \(=\) | \(\ds 2 + 11\) | ||||||||||||
\(\ds 14\) | \(=\) | \(\ds 3 + 11\) | ||||||||||||
\(\ds 15\) | \(=\) | \(\ds 3 + 5 + 7\) | ||||||||||||
\(\ds 16\) | \(=\) | \(\ds 5 + 11\) | ||||||||||||
\(\ds 17\) | \(=\) | \(\ds 2 + 3 + 5 + 7\) | ||||||||||||
\(\ds 18\) | \(=\) | \(\ds 2 + 5 + 11\) | ||||||||||||
\(\ds 19\) | \(=\) | \(\ds 3 + 5 + 11\) |
$\blacksquare$