Given the partition function $P(n)$ and let $q_k=e^{-k\pi/5}$. What is the reason why,
$$\sum_{n=0}^\infty P(n) q_2^{n+1}\approx\frac{1}{\sqrt{5}}\tag{1}$$
$$\sum_{n=0}^\infty P(n) q_4^{n+1}\approx\Big(1-\frac{1}{5^{1/4}}\Big)^2\Big(\frac{1+\sqrt{5}}{4}\Big)\tag{2}$$
where the difference is a mere $10^{-14}$ and $10^{-29}$, respectively? (The form of $(2)$ seems to be more than coincidence.)
P.S. I forgot where I found $(2)$. Does anybody recall the paper where this approximation first appears?
I have not derived the approximation (2) yet, but the approximation (1) is easy:
Set $q(\tau)=\mathrm{e}^{2\pi\mathrm{i}\tau}$. We will use the Dedekind eta function $$\eta(\tau) = q\left(\frac{\tau}{24}\right) \prod_{n=1}^\infty\left(1-q(n\tau)\right) = \frac{q\left(\frac{\tau}{24}\right)}{\sum_{n=0}^\infty P(n) q(n\tau)}$$
We also know the following modular forms property of $\eta(\tau)$: $$\eta(\tau) = \sqrt{\frac{\mathrm{i}}{\tau}}\eta\left(\frac{-1}{\tau}\right)$$
Set $\tau=\frac{\mathrm{i}}{5}$. Then $\frac{-1}{\tau} = 5\mathrm{i} = 25\tau$. Thus $$\begin{align} q(\tau) &= \mathrm{e}^{-2\pi/5} \\ q\left(\frac{-1}{\tau}\right) &= q(25\tau) = \mathrm{e}^{-10\pi} \approx 2\cdot10^{-14} \\ \eta\left(\frac{-1}{\tau}\right) &= \eta(25\tau) \approx q\left(\frac{25\tau}{24}\right) \\ \eta(\tau) &= \sqrt{5}\,\eta(25\tau) \approx \sqrt{5}\,q\left(\frac{25\tau}{24}\right) \end{align}$$ That is, the $\prod_{n=1}^\infty\cdots$ used in $\eta(25\tau)$ well approximates $1$.
Thus $$\begin{align} \sum_{n=0}^\infty P(n) q\left((n+1)\tau\right) &= \frac{q(\tau)\,q\left(\frac{\tau}{24}\right)}{\eta(\tau)} = \frac{q\left(\frac{25\tau}{24}\right)}{\eta(\tau)} \\ &\approx \frac{q\left(\frac{25\tau}{24}\right)} {\sqrt{5}\,q\left(\frac{25\tau}{24}\right)} = \frac{1}{\sqrt{5}} \end{align}$$ Approximation (2) seems to require a little more.