Towards a formula for the Euler $\phi$ function?

Question

1. $\Phi_n(1)$ and $\Phi_n(-1)$ for the cyclotomic polynomials are well-known.

   I am now looking for $$\Phi_n(i)$$ and/or $$\Phi_n(-i)$$ with $i$ the complex unit.

2. The reason is :

   I suppose it is true that $(1-i)^{\phi(n)}\Phi_n(i)$ is either zero or of the form $$(-1)^{f(n)}2^{g(n)}$$ for some rational valued functions. Just now $f(n)$ and $g(n)$ are unknown to me. But if all calculations would succeed by taking logarithms $\phi(n)$ may perhaps be calculated.

   The ideas behind that rely on the polynomials defined by William E Heierman published on his Web Site.

Answer 1

After a long thought, my conjecture is instead the following. Let $\nu=\nu_2(n)$, $\omega_1$ be the number of distinct prime numbers $\equiv 1\pmod{4}$ that divide $n$, $\omega_3$ be the number of distinct prime numbers $\equiv 3\pmod{4}$ that divide $n$. I claim: $$\begin{array}{rl}\textbf{If...}&\textbf{then...}\\ n=2&\Phi_n(i)=1+i\\ n=4&\Phi_n(i)=0\\ n=2^{k},k\geq 2&\Phi_n(i)=2\\ n=4p&\Phi_n(i)=p\\ \omega_1\geq 1,n\neq 4p & \Phi_n(i)=1\\ \nu\geq 3&\Phi_n(i)=1\\ \omega_3\geq 3&\Phi_n(i)=1\\ \omega_1=0,\omega_3=1,\nu=0&\Phi_n(i)=i\\ \omega_1=0,\omega_3=1,\nu=1&\Phi_n(i)=-i\\ \omega_1=0,\omega_3=2,\nu<2&\Phi_n(i)=-1\\ \omega_1=0,\omega_3=2,\nu=2&\Phi_n(i)=1.\\ \end{array}$$ Please check this conjecture. I am too lazy now to prove that it holds in virtue of: $$\Phi_n(i)=\prod_{d\mid n}\left(i^{n/d}-1\right)^{\mu(d)}$$ by an induction depending on eleven different cases, but I can find somewhere the strength to do my math if this is really useful.

Answer 2

I made a calculation for $n\ge 3$ and get

$$\prod_{E_n}\cos\left(\frac{k\pi}{n}\right)=\frac{\Phi_n(-1)}{2^{\phi(n)}}$$

This formula may be totally wrong at this moment.

I have not the CAS Mathematica available and also not the programming skills to check it by other CAS freely available.

Answer 3

I recalculated the formula and it is indeed

$$\prod_{E_n}\cos\left(\frac{k\pi}{n}\right)=(-1)^{\frac{\phi(n)}{2}}\frac{\Phi_n(-1)}{2^{\phi(n)}}$$

for $n>2$. Taking absolute values on both sides allows one then to calculate $\phi(n)$.