A curious partitions coincidence $\sum_{n=0}^\infty P(n) q^{n+1}$? 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?
 A: 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.
A: (Too long for a comment.) Just to give the exact value of $(2)$, we have the identity,
$$A = \frac{1}{e^{5\pi/6}\,\eta\big(\tfrac{2\,i}{5}\big)} = \frac{2^{19/8}(-1+5^{1/4})\,\pi^{3/4}}{e^{5\pi/6}\,(-1+\sqrt{5})^{3/2}\,\Gamma\big(\tfrac{1}{4}\big)} = 0.0887758\dots$$
which is approximated by,
$$B =\Big(1-\frac{1}{5^{1/4}}\Big)^2\Big(\frac{1+\sqrt{5}}{4}\Big)= 0.0887758\dots$$
and $A\approx B$ with the aforementioned difference of a mere $10^{-29}$. Why that is so seems to be still unexplained.
P.S. Other exact values of the Dedekind eta function $\eta(\tau)$ are in this post.
