<s>[INST] Solutions to $\prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) = \varphi$ The constant $\varphi = \frac{1 + \sqrt5}{2}$ seems to show up everywhere. I am wondering what non-trivial function(s) $f$ satisfy
$$\prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) = \varphi$$
I was suprised when I could not find any solutions. The 'closest' solution I have found is $f(n)=n^2+3n+2$, for which the infinite product is approximately $1.61847$.
 [/INST] This is a little too long for a comment and may not be a complete answer, but here is one such $f(n)$.
We know that $2\cos\left(\frac{\pi}{5}\right)=\phi$, hence $\frac{\sin\left(2\pi/5\right)}{\sin\left(\pi/5\right)}=\phi.$ Then by Euler's product for the sine function
$$\begin{align*}
\frac{\sin\left(\frac{2\pi}{5} \right )}{\sin\left(\frac{\pi}{5} \right )} &=\frac{\frac{2\pi}{5}\prod_{n=1}^\infty\left(1-\frac{(2\pi/5)^2}{n^2\pi^2} \right )}{\frac{\pi}{5}\prod_{n=1}^\infty\left(1-\frac{(\pi/5)^2}{n^2\pi^2} \right )} \\ 
 &=2\prod_{n=1}^\infty\frac{(5n)^2-4}{(5n)^2-1} \\ 
 &= 2\prod_{n=1}^\infty\frac{(5n+2)(5n-2)}{(5n+1)(5n-1)}\\ 
 &= \prod_{n=0}^\infty\frac{(5n+2)(5n+3)}{(5n+1)(5n+4)}.\\
\end{align*}$$
Solving for the $f(n)$ in your product gives
$$\phi=\prod_{n=0}^\infty\left(1+\frac{1}{f(n)}\right)$$
where $f(n)=\frac{25}{2}n^2+\frac{25}{2}n+2.$ The only difference is the index shift, which is no big deal.
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