The following is not a mathematically interesting property, but it scores high.

$a^2 < a \times 2^{2^{2^{2^2}}}$

It uses:
$1$ (in)equality
$6$ operators
$2$ is the only the number value used
$6$ numbers
which gives a penalty of $1-6-2-6 = -13$.

The penalty is insignificant, because its solutions are the integers

from $1$ (inclusive) to $2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}$ (exclusive). These have an average of $2^{65536}/2=2^{65535} \approx 10^{19728}$.

This can extended in an obvious way - adding one $2$ and one operator which only gives two extra penalty points - for an even more outrageous score.

The following is not a mathematically interesting property, but it scores high.

$a^2 < a \times 2^{2^{2^{2^2}}}$

It uses:
$1$ (in)equality
$6$ operators
$2$ is the only the number value used
$6$ numbers
which gives a penalty of $1-6-2-6 = -13$.

The penalty is insignificant, because its solutions are the integers

from $1$ (inclusive) to $2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}$ (exclusive). These have an average of $2^{65536}/2=2^{65535} \approx 10^{19728}$.

This can extended in an obvious way - adding one $2$ and one operator which only gives two extra penalty points - for an even more outrageous score.

Of course I could instead simply use this property:

$a=2^{2^{2^{2^2}}}$

but that would be even less interesting.

The following is not a mathematically interesting property, but it scores high.

$a^2 < a \times 2^{2^{2^{2^2}}}$

It uses:
1 (in)equality
6 operators
1 number value (2)
7 numbers
which gives a penalty of 1-6-2-7 = -13.

The penalty is insignificant, because its solutions are the integers

from $1$ (inclusive) to $2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}$ (exclusive). These have an average of $2^{65536}/2=2^{65535} \approx 10^{19728}$.

This can extended in an obvious way - adding one $2$ and one operator which only gives two extra penalty points - for an even more outrageous score.

The following is not a mathematically interesting property, but it scores high.

$a^2 < a \times 2^{2^{2^{2^2}}}$

It uses:
$1$ (in)equality
$6$ operators
$2$ is the only the number value used
$6$ numbers
which gives a penalty of $1-6-2-6 = -13$.

The penalty is insignificant, because its solutions are the integers

from $1$ (inclusive) to $2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}$ (exclusive). These have an average of $2^{65536}/2=2^{65535} \approx 10^{19728}$.

This can extended in an obvious way - adding one $2$ and one operator which only gives two extra penalty points - for an even more outrageous score.