How many partition values are expected to be prime?

Let $$p(n)$$ be the partition function. Let $$P(N)$$ count how many $$1\leq n\leq N$$ are such that $$p(n)$$ is prime.

Are there any heuristics for how $$P(N)$$ should behave?

A crude guess at how this function should grow, based on the prime number theorem, is that we'd expect

$$P(N) \approx \sum_{n=1}^N \frac{1}{\ln(p(n))}.$$

Testing this in Sage up to $$N\approx 5000$$ suggests that right hand side is actually an overcount though - indeed, the ratio of actual prime partition to values to what the crude heuristic suggests is almost always around 75%-90%.

Are there any heuristics as to why the right hand side consistently predicts more prime partition values than there are? Is this just the law of small numbers at play, and eventually this heuristic should be accurate?

The value of this ratio doesn't seem particularly well-behaved, but if I had to squint and guess, it looks like it's tending to about 86%.

Congruences of the partition function, such as Ramanujan's $$p(5k+4)\equiv 0\pmod{5})$$ might explain this, since these suggest events like 'divisible by 5' occur more often than one would expect for a random sequence, skewing the count of primes low. Can what is known about these type of congruences be converted into a precise heuristic prediction, a la predicted density of twin primes?

(I'm not expecting any actual proofs here - as far as I know, it's not even known if $$p(n)$$ is prime infinitely often. I'm just curious if there are any reasonable heuristic predictions/explanations known.)

• Let $f$ be the completely multiplicative function defined by $f(p)=1$ for primes $p$ different from $5, 7, 11$ for which Ramanujan congruences occur, $f(p)=(p-2)/(p-1)$ for them (as in that case there is exactly one residue class mod $p$ out of $p-1$ not leading to primes as Dirichlet's theorem would predict). One can try to replace $(1/\log p(n))$ by $f(p(n))/\log p(n)$ in your RHS. – Sylvain JULIEN Dec 29 '18 at 4:59