# Almost a prime number recurrence relation

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.

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

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?

• 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. – Mastrem Mar 4 at 14:29
• @Mastrem: I have reformulated the conjecture. – Lehs Mar 5 at 6:53
• The asymptotic formula gives $\log a(p_n)\sim \frac{2\pi}{\sqrt{3}}\sqrt n$. – Mastrem Mar 7 at 19:06