# 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 $\closedint {105} {208}$.

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:

 $\ds 11 + 199$ $=$ $\ds 210$ $\ds 13 + 197$ $=$ $\ds 210$ $\ds 17 + 193$ $=$ $\ds 210$ $\ds 19 + 191$ $=$ $\ds 210$ $\ds 29 + 181$ $=$ $\ds 210$ $\ds 31 + 179$ $=$ $\ds 210$ $\ds 37 + 173$ $=$ $\ds 210$ $\ds 43 + 167$ $=$ $\ds 210$ $\ds 47 + 163$ $=$ $\ds 210$ $\ds 53 + 157$ $=$ $\ds 210$ $\ds 59 + 151$ $=$ $\ds 210$ $\ds 61 + 149$ $=$ $\ds 210$ $\ds 71 + 139$ $=$ $\ds 210$ $\ds 73 + 137$ $=$ $\ds 210$ $\ds 79 + 131$ $=$ $\ds 210$ $\ds 83 + 127$ $=$ $\ds 210$ $\ds 97 + 113$ $=$ $\ds 210$ $\ds 101 + 109$ $=$ $\ds 210$ $\ds 103 + 107$ $=$ $\ds 210$

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