On the tribonacci constant $T$ and $4\arctan\left(\frac1{\sqrt{T^3}}\right)+\arctan\Big(\frac1{\sqrt{(2T+1)^4-1}}\Big)=\frac{\pi}2$

Question

We have,

$$4\arctan\left(\frac1{\sqrt{\phi^3}}\right)\color{red}-\arctan\left(\frac1{\sqrt{\phi^6-1}}\right)=\frac{\pi}2$$

$$4\arctan\left(\frac1{\sqrt{T^3}}\right)\color{blue}+\arctan\left(\frac1{\sqrt{(2T+1)^4-1}}\right)=\frac{\pi}2$$

with golden ratio $\phi$ and tribonacci constant $T$.

Note that,

$$\begin{aligned}\left(\sqrt{\phi^3}+i\right)^4\left(\sqrt{\phi^6-1}\color{red}-i\right)&=(\phi^3+1)^2\sqrt{-\phi^6}\\ \left(\sqrt{T^3}+i\right)^4\left(\sqrt{(2T+1)^4-1}\color{blue}+i\right)&=(T^3+1)^2\sqrt{-(2T+1)^4}\end{aligned}$$

The first one is from this post, while the second is from a session with Mathematica. However, I'm having a little trouble with the tetranacci constant.

Q: What would be analogous formulas using the tetranacci constant (and higher)?

Answer 1

This is NOT AN ANSWER, but an identity that might help.

In general, if $\psi\gt 1.8$, we have that $$4\arctan(\psi^{-3/2})+\arctan\bigg(\frac{1}{\sqrt{\beta-1}}\bigg)=\frac{\pi}{2} \tag{i}$$ where $$\beta=\frac{(\psi^3+1)^4}{(\psi^6-6\psi^3+1)^2}=\bigg(1+\frac{8\psi^3}{\psi^6-6\psi^3+1}\bigg)^2 \tag{ii}$$

Notice that (ii) suggests that no "nice simplifications" like in your two examples will be possible for $n$-fibonacci constants with $n\gt 24$, since if the expression $$\frac{(\psi^3+1)^4}{(\psi^6-6\psi^3+1)^2}$$ were rewritten as a rational function of $\psi$ whose numerator and denominator had smaller degree, then that would provide a polynomial equation for $\psi$ of degree under $24$. Thus, your problem has no nice solution for $n\gt 24$, but I suspect that one for $n=4$ or $n=5$ might be obtainable (if you have the force of will to make yourself go through all of the messy algebra, or ask a computer to do it for you).