It turns out that
storage of primes can be compressed
by an arbitrary amount,
though the storage needed
is exponential in the compression factor.
Note:
This has been corrected and been made more precise.
Let $p_n$ be the $n$-th prime
(with $p_1 = 2$),
and let
$P_n$ be the product of the first $n$ primes,
so that
$P_1 = 2, P_2 = 6,
P_3 = 30, P_4 = 210$.
If we do a
sieve of Eratosthenes,
sieving out multiples of the
first $n$ primes,
all that are left
are the first $n$ primes
and the numbers
relatively prime to
$P_n$.
For example,
for $n=2$,
the numbers left are $2, 3$,
and the forms
$6m+1$ and $6m+5$.
For $n=3$,
the numbers left
are $2, 3, 5$
and the forms
$30m+1,7,11,13,17,19,23,$ and $29$.
For general $n$,
the numbers remaining
are of the form
$P_nm+q_i$,
where the $q_i$
are the numbers from $1$ to
$P_n-1$ relatively prime to $P_n$.
To illustrate,
I will use the case $n=3$.
Each block of 30 numbers
in a range
$30m+1$ to $30m+29$
can have as most $8$
primes as shown above.
Therefore,
only one bit is needed
for each of these $8$ possibilities,
to indicate whether or not
that value is actually prime.
Therefore,
storage of primes can be compressed
by a factor of
$\frac{8}{30} \approx 0.267$.
Here is what happens for
general $n$.
The $P_n$ values
in the range from
$mP_n$ to
$(m+1)P_n-1$
are compressed to
$\phi(P_n)$ bits,
where $\phi(m)$ is Eu;er's phi function
(though he lets me use it)
that counts the
number of integers
from $1$ to $m$
relatively prime to $m$.
Since $\phi(P_n)
= \prod_{i=1}^n (p_i-1)
$,
the compression factor is
$\dfrac{\phi(P_n)}{P_n}
=\prod_{i=1}^n (1-\dfrac1{p_i})
$.
Since this goes to zero
(because
$\sum_{i=1}^n \dfrac1{p_i}
\sim \ln \ln n
$ - this is Merten's theorem),
the amount of compression
can be arbitrarily large,
though about
$e^n/n$ bits are needed.
To see this,
by one of the corollaries
of the prime number theorem,
$P_n \sim e^n$,
and
$\dfrac{\phi(P_n)}{P_n}
=\prod_{i=1}^n (1-\dfrac1{p_i})
\approx e^{-\ln \ln n}
= 1/\ln n
$.
Therefore,
each block of
$P_n$ values
can be represented by
about $\frac{P_n}{n}$
bits.
Therefore, using a sieve
with the first $n$ primes
and the numbers
relatively prime to $P_n$
can compress the primes
by about a factor of $n$.
Therefore the primes can be compressed by
an arbitrary amount,
though the amount of storage needed
is exponential in $n$.
A number of years ago,
I used this idea
with $n=4$,
so I got a compression of
$\frac{1\cdot 2\cdot 4\cdot 6}{2\cdot 3\cdot 5\cdot 7}
=\frac{8}{35}
$.
