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
$$
3,\binom{n}{1}, \ldots,\binom{n-1}{n}
$$



Zeros of a sequence related to roots of unity
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 $a(n)=0$.


