Zeros of a sequence related to roots of unity Consider the sequence
$$
a(n) = \prod_{u^n=1,u \neq 1}( (1+u)^n+1)
$$
Some terms are:
$$
1,1,0,9,121,2704,118336, 4092529,0,97734390625, \ldots
$$
Alonso del Arte asks:
Question: What are the multiples of $3$ such that
$$
a(3k) =0
$$
I tried some factorization of cyclotomic polynomials without success.
May be true for all odd $k$  ???
EDIT: Another simple property of the sequence is
(hope this may please the negative voter  (???))
$$
a(p) \equiv 1 \pmod{p}
$$
for any prime $p>3$
since 
$$
a(n) (2^n+1)
$$
is the determinant of a circulant matrix with first line
$$
3,\binom{n}{1}, \ldots,\binom{n-1}{n}
$$
 A: The complex number $a(n)$ is the resultant of the polynomials $P=(X^n-1)/(X-1)$ and $Q=(X+1)^n+1$;
similarly, $(2^n+1)a(n)$ is the resultant of $X^n-1$ and $(X+1)^n+1$.
Since these polynomials have integer coefficients, their resultant  is a rational integer.
The resultant of two polynomials vanishes whenever they have a common root.
So $a(n)=0$ if and only if there exists a $n$th root of unity $u$ such that
$(u+1)^n+1=0$. This implies that $u$ and $u+1$ are both roots of unity, in particular
they belong to the unit circle, so that necessarily $u=e^{2\pi i/3}$ or $u=e^{-2\pi i/3}$,
and $u+1=e^{\pm i\pi/3}$.
If $u$ is a $n$th root of unity, one gets $3|n$; if $(u+1)^n=-1$, one obtains
that $n/3$ is odd. Conversely, if $n=3m$ with $m$ odd, $u=e^{2\pi i/3}$
satisfies $u^n=1$, $u\neq 1$, and $(u+1)^n=-1$, hence $a(n)=0$.
Since the two polynomials $P$ and $Q$ above are monic, their resultant
vanishes mod $p$ if and only if they have a common root when considered as polynomials 
modulo $p$. If $n=p$ is prime, then $X^n-1=(X-1)^p$, so $1$ is the only root of $P$,
with multiplicity $p-1$; it follows that $$ a(n)\equiv ((1+1)^p+1)^{p-1}\equiv (2^p+1)^{p-1}\equiv 3^{p-1} \pmod p.$$ If, moreover, $p\neq 3$, then $a(n)\equiv 1\pmod p$.
A: Suppose that $n=3m$, where $m$ is odd, and $u=e^{2\pi i/3}$. Then 
  $$ (1+u)^n+1 = ((1+u)^3)^m+1 = (-1)^m+1=0, $$
so $a(n)=0$.
