$2\arctan(\phi^{-n})=\arctan\frac{p}{q}$ or $\arctan\frac{p\sqrt{5}}{q}$, where $\phi$ is the Golden Ratio. Is there a pattern in the $\frac{p}{q}$s? It is very interesting to know that 
$$\arctan\frac{1}{\phi} + \arctan\frac{1}{\phi^3}= \arctan 1 = \frac{\pi}{4}$$
where Golden ratio $\phi = \frac12(\sqrt5 +1)$ is in association with circle constant $\pi$.
More interesting phenomenon is evaluation of inverse tan functions of inverse of $\phi$ in its consecutive powers as follows
$$\begin{align}
2\arctan\frac{1}{\phi} &= \arctan 2 &
2\arctan\frac{1}{\phi^2} &= \arctan\frac{2\sqrt{5}}{5}\\
2\arctan\frac{1}{\phi^3} &= \arctan\frac{1}{2} &
2\arctan\frac{1}{\phi^4} &= \arctan\frac{2\sqrt{5}}{15}\\
2\arctan\frac{1}{\phi^5} &= \arctan\frac{2}{11} &
2\arctan\frac{1}{\phi^6} &= \arctan\frac{\sqrt5}{20} \\
2\arctan\frac{1}{\phi^7} &= \arctan\frac{2}{29} &
2\arctan\frac{1}{\phi^8} &= \arctan\frac{2\sqrt5}{105} \\
2\arctan\frac{1}{\phi^9} &= \arctan\frac{1}{38} &
2\arctan\frac{1}{\phi^{10}} &= \arctan\frac{2\sqrt5}{275} \\
2\arctan\frac{1}{\phi^{11}} &= \arctan\frac{2}{199}
\end{align}$$
Here are the observations


*

*Odd powers of inverse $\phi$ in double arctan functions lead to arctan of well defined fractions

*Even powers of inverse $\phi$ in double arctan functions lead to arctan of fractions involving $\sqrt5$.



My curiosity is to know, is there any pattern in these interesting series?

I will be grateful to understand more, if anyone has come across such evaluations.
 A: The explanation is that $$\tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)}$$
so, for $-1 < x < 1$,
$$2 \arctan(x) = \arctan\left(\frac{2x}{1-x^2}\right)$$
Thus
$$ 2 \arctan(1/\phi^n) = = \arctan \left( \frac{2 \phi^{-n}}{1-\phi^{-2n}}\right)
= \arctan\left(\frac{2}{\phi^{n} - \phi^{-n}}\right)$$
Now the Fibonacci numbers $$F_n = \frac{\phi^n - (-1/\phi)^n}{\sqrt{5}}$$
so if $n$ is even, $$2 \arctan(1/\phi^n) = \arctan\left(\frac{2}{\sqrt{5} F_n}\right)$$
On the other hand, the Lucas numbers $$L_n = \phi^n + (-1/\phi)^n$$
so if $n$ is odd, $$2 \arctan(1/\phi^n) = \arctan\left(\frac{2}{L_n}\right)$$
A: Hint:
$$ \arctan(x)\pm\arctan(y) = \arctan(z) $$
where $z$ is:
$$ z = \frac{x\pm y}{1\mp xy} $$
We will take a look at this:
$$ 2\arctan\phi^{-n} = \arctan \phi^{-n}+\arctan\phi^{-n} = \arctan(\frac{\phi^{-n}+\phi^{-n}}{1-\phi^{-n}\phi^{-n}}) = \arctan(\frac{2\phi^{-n}}{1-\phi^{-2n}}) = $$
$$ = \arctan(\frac{2}{\phi^{n}-\phi^{-n}}) = \arctan(\frac{2}{e^{\ln\phi^{n}}-e^{\ln\phi^{-n}}}) =\arctan(\frac{1}{\sinh(n\ln\phi)})$$
A: Here is the pattern:
Given $$2\arctan\frac1{\phi^n}=\arctan A_n$$
the following recursive reciprocals hold, for both odd and even $n$'s,
$$\frac1{A_{n+2}}+\frac1{A_{n-2}}=\frac3{A_{n}}$$
