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 $1.61847$.
 A: 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.
A: You can pick any convergent sequence ($a_n$) without two consecutive equal terms, such that $\forall n \rightarrow a_n \neq a_{n+1}$, and without any zero, so $(\forall n) \, a_n\neq 0$. Then, if the sequence converges to $L$ you can first get a new sequence $b_n = a_n \times (\phi/L)$. It is also necessary that $L\neq0$. So this sequence holds the previous properties and converges to $\phi$.
Then you define:
$$
b_1 = 1+\frac{1}{f(1)} \Rightarrow f(1) = \frac{1}{b_1-1}
$$
And create a recurrence:
$$
\frac{b_n}{b_{n-1}} = 1+\frac{1}{f(n)}  \Rightarrow f(n) = \frac{b_{n-1}}{b_n-b_{n-1}}
$$
And voilà, there is your $f(n)$. 
To be explicit:
$$
b_n = \frac{b_n}{b_{n-1}} \times \frac{b_{n-1}}{b_{n-2}} \times \cdots \frac{b_{2}}{b_{1}} \times b_1
$$
Thus:
$$
b_n = \left( 1+\frac{1}{f(n)}  \right) \times \left( 1+\frac{1}{f(n-1)}  \right)  \times \cdots \left( 1+\frac{1}{f(2)}  \right)  \times \left( 1+\frac{1}{f(1)}  \right) 
$$
Then:
$$
\lim_{n \to \infty} b_n = \phi = \prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) 
$$
By plugging in more sequences satisfying the above conditions, this procedure will get you new functions $f(n)$.
