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For the number of partitions of n into prime parts $a(n)$ it holds $$a(n)=\frac{1}{n}\sum_{k=1}^n q(k)a(n-k)\tag 1$$ where $q(n)$ the sum of all different prime factors of $n$.
Due to https://oeis.org/A000607.

I have found that $$a(p_n)\approx a(p_{n-2})+a(p_{n-1})\tag 2$$ and conjecture the asymptotic relation.

$$\log a(p_n)\sim \log \big(a(p_{n-2})+a(p_{n-1})\big )\tag 3$$

enter image description here

On x-axis the prime numbers $p_n$ are plotted. The blue lines correspond to $a(p_n)$ and the red lines correspond to $a(p_{n-2})+a(p_{n-1})$.

On the oeis site above there is also a formula $$a(n)\sim e^{2\pi\sqrt{n/\log n}\,/\sqrt{3}}\tag 4$$ but I don't know if this really helps?

Can this (reformulated) conjecture be proved?

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    $\begingroup$ Doesn't the PNT give $p_n/\log p_n\sim n$, so $a(p_n)\sim e^{2\pi\sqrt n/\sqrt 3}$? That would mean you'd have to prove something along the lines of $e^{2\pi\sqrt{n}/\sqrt 3}\sim e^{2\pi\sqrt{n-2}/\sqrt 3}+e^{2\pi\sqrt{n-1}/\sqrt 3}$. Maybe try to sum the Taylor series. $\endgroup$
    – Mastrem
    Commented Mar 4, 2019 at 14:29
  • $\begingroup$ @Mastrem: I have reformulated the conjecture. $\endgroup$
    – Lehs
    Commented Mar 5, 2019 at 6:53
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    $\begingroup$ The asymptotic formula gives $\log a(p_n)\sim \frac{2\pi}{\sqrt{3}}\sqrt n$. $\endgroup$
    – Mastrem
    Commented Mar 7, 2019 at 19:06

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