This post discusses the integral,
$$I(k)=\int_0^k\pi(x)\pi(k-x)dx.$$
I've noticed that the odd numbers seem to come in pairs separated by exactly $2$ units, but are otherwise quite randomly distributed amongst the even numbers in the sequence. This can be written like $I(k)$ and $I(k+2).$ Here I should note that we pair the first instance, $I(11)$ and $I(13)$ and proceed like that with further pairings. For example, the second pairing is $I(23)$ and $I(25).$
Does this pattern occur infinitely often in the sequence?
Consider the list of numbers which are double a prime.
$$4,6,10,14,22,26,34,38,46, ...$$
Then $I(k)$ is odd if and only if $k$ is an odd number between a $2i-1$th and $2i$th element in the above list. Hence $k$ is one of: $$5,11,13,23,25,35,37,...$$
If $q$ and $r$ are successive odd primes then $2q+1$ and $2q+3$ lie between $2q$ and $2r$ and therefore there are infinitely many pairs of the type you require.
NOTE
$I(k)= \sum_{m\le k} r(m)(k-m)$ where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then $r(m)$ is odd if and only if $m$ is twice a prime.
For $r(m)(k-m)$ to be odd we then also require $k$ to be odd. Hence the result stated above.