Integers whose Number of Representations as Sum of Two Primes is Maximum

Theorem
$210$ is the largest integer which can be represented as the sum of two primes in the maximum number of ways.

The full list of such numbers is as follows:
 * $1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 24, 30, 36, 42, 48, 60, 90, 210$

The list contains:
 * $n \le 8$
 * $n \le 18$ where $2 \divides n$
 * $n \le 48$ where $2 \times 3 \divides n$
 * $n \le 90$ where $2 \times 3 \times 5 \divides n$
 * $210 = 2 \times 3 \times 5 \times 7$

Proof
From Number of Representations as Sum of Two Primes, the number of ways an integer $n$ can be represented as the sum of two primes is no greater than the number of primes in the interval $\closedint {\dfrac n 2} {n - 2}$.

The interval $\closedint {\dfrac {210} 2} {210 - 2}$ is $\closedint {105} {208}$.

Let $p$ be a prime in $\closedint {105} {208}$.

Then $2 \le 210 - p \le 105$.

$p$ must be coprime to $2, 3, 5$ and $7$.

Since $210 = 2 \times 3 \times 5 \times 7$, $210 - p$ must also be coprime to $2, 3, 5$ and $7$.

Hence the smallest prime factor of $210 - p$ must be at least $11$.

If $210 - p$ is composite:
 * $210 - p \ge 11^2 = 121$

which is a contradiction.

Thus $210 - p$ must be prime.

This shows that $210$ is the sum of two primes in the maximum number of ways.

Explicitly, the primes in this interval can be enumerated:


 * $107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$

It can be seen there are exactly $19$ of them.

We have:

and as can be seen, there are $19$ such representations, one for each prime in $\closedint {105} {208}$.