<s>[INST] A finite alternate binomial sum with finite value (modulo 6) We use Iverson symbol ( [expr] is $1$ when expr is true else $0$), and $n\%6$ for $n$ modulo $6$. .
For any $n >= 2 $. 
Let $ F(n) := \sum_{j>=1} (-1)^j {\binom {n-j-2} j } $
Prove that $F(n) = [n\%6 \in \{0,5\}] - [n\%6 \in \{2,3\}]$
In other words:
IF   $n$ modulo $6$ is $0$ or $5$    then $F(n) = 1$ 
IF  $n$ modulo $6$ is $2$ or $3$ then $F(n) =-1$
IF  $n$ modulo $6$ is $1$ or $4$ then $F(n) =0$
It is true up to $n=15$ by hand, $40$ would be safe
Yet to what kind of identities does this belong? .And how to prove this?
Can we replace  $-2$ by another constant and still have the same behaviour ( finite set of value with periodic recurrence. 
Note : The motivation is to calculate
$ F(n) = \sum (-1)^k[a_1a_2a_3 \dots a_k=2^n] $. Where any $a_i$ is >= 3. 
 [/INST] Working with the following sum
$$\sum_{q=0}^n (-1)^q {n-q\choose q}$$
we have
$$\sum_{q=0}^n (-1)^q {n-q\choose n-2q}
= \sum_{q=0}^n (-1)^q [z^{n-2q}] (1+z)^{n-q}
\\ = [z^n] (1+z)^n \sum_{q=0}^n (-1)^q z^{2q} (1+z)^{-q}.$$
Now we get zero contribution when $2q\gt n$ hence we may write
$$[z^n] (1+z)^n \sum_{q\ge 0} (-1)^q z^{2q} (1+z)^{-q}
\\ = [z^n] (1+z)^n \frac{1}{1+z^2/(1+z)}
= [z^n] (1+z)^{n+1} \frac{1}{1+z+z^2}.$$
This is
$$\mathrm{Res}_{z=0}
\frac{1}{z^{n+1}} (1+z)^{n+1} \frac{1}{1+z+z^2}.$$
With the substitution $z/(1+z)=w$ or $z=w/(1-w)$ we have
$dz = 1/(1-w)^2 \; dw$ and we obtain
$$\mathrm{Res}_{w=0}
\frac{1}{w^{n+1}}
\frac{1}{1+w/(1-w)+w^2/(1-w)^2}  \frac{1}{(1-w)^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^{n+1}}  \frac{1}{(1-w)^2+w(1-w)+w^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^{n+1}}  \frac{1}{1-w+w^2}
= [w^n]  \frac{1}{1-w+w^2}.$$
Starting from
$$G(w) = \frac{1}{1-w+w^2}$$
we have $$(1-w+w^2) G(w) = 1$$
so that
$$[w^0] (1-w+w^2) G(w) = G_0 = [w^0] 1 = 1$$
and furthermore
$$[w^1] (1-w+w^2) G(w) = G_1-G_0 = [w^1] 1 = 0$$
This establishes the initial values $G_0=1$ and $G_1=1.$
Moreover for $n\ge 2$ we find
$$[w^n] (1-w+w^2) G(w) = G_n - G_{n-1} + G_{n-2}
= [w^n] 1 = 0$$
producing the recurrence
$$G_n = G_{n-1} - G_{n-2}.$$
This means  $G_n$ only depends  on the preceding two  values. Starting
from $1,1$ we obtain by applying the recurrence
$$1,1,0,-1,-1,0,1,1,\ldots$$
We have  reached the  period (intial pair  appears), thus  proving the
claim.
 We may also compute the generating function by snake oil method.
Start with
$$G(w) = \sum_{n\ge 0} w^n \sum_{q=0}^n (-1)^q {n-q\choose q}
= \sum_{q\ge 0} (-1)^q \sum_{n\ge q} w^n {n-q\choose q}
\\ = \sum_{q\ge 0} (-1)^q w^q \sum_{n\ge 0} w^n {n\choose q}
= \sum_{q\ge 0} (-1)^q w^q \sum_{n\ge q} w^n {n\choose q}
\\ = \sum_{q\ge 0} (-1)^q w^{2q} \sum_{n\ge 0} w^n {n+q\choose q}
= \sum_{q\ge 0} (-1)^q w^{2q} \frac{1}{(1-w)^{q+1}}  
\\ = \frac{1}{1-w} \frac{1}{1+w^2/(1-w)}
= \frac{1}{1-w+w^2}.$$
</s>