How come $\ n\ $ always divides at least one of the item of the sequence? Given positive integer$\ \displaystyle n,\ $ the sequence is: 
$\displaystyle 2^n$ 
$\displaystyle 2^n - 2^{n-1}$ 
$\displaystyle 2^n - 2^{n-1} + 2^{n-2}$ 
$\displaystyle 2^n - 2^{n-1} + 2^{n-2} - 2^{n-3}$ 
... 
$\displaystyle 2^n - 2^{n-1} + 2^{n-2} - 2^{n-3} \dots \pm 1$ 
If $\ \displaystyle n\ $ is even, the final term is $\ \displaystyle +1;\ $ if $\ \displaystyle n\ $ is odd, then the final term is $\ \displaystyle -1$. 
My question is : how come $\ n\ $ always divides at least one of the item of the sequence?
Python program:
def f(n):
    l = []
    s = 2**n 
    l.append(s)
    t = -1
    for i in range(n, 0, -1):
        s = s + t * 2**(i - 1)
        t *= -1 
        l.append(s) 
    return l
print(f(5)) #output : [32, 16, 24, 20, 22, 21]
print(f(7)) #output : [128, 64, 96, 80, 88, 84, 86, 85]

Source: http://mymathforum.com/number-theory/100912-conjecture-2-n.html
 A: Among the $n+1$ numbers listed, there must be two of the same remainder modulo $n$. Their difference is a multiple of $n$ and, after multiplying with a suitable power of $2$ (and possibly flipping the sign)occurs in the sequence. More specifically, if we name the terms $a_0=2^n$, $a_1=2^n-2^{n-1}$, $a_2=2^n-2^{n-1}+2^{n-2}$ and so on and if $a_i\equiv a_j\pmod n$ with $0\le i<j\le n$, then $2^{j-i}(-1)^i(a_i-a_j)=a_{j-i-1}$.
A: We need only to check $n$ odd, since the conjecture is true for $n$ being powers of 2, for all terms are even but the last one $S(0)=even\pm 1=odd$.
a) $S(n)=2^n=1\cdot 2^{n}$
b) $S(n-1)=2^n-2^{n-1}=1\cdot 2^{n-1}$
c) $S(n-2)=2^{n-1}+2^{n-2}=3\cdot 2^{n-2}$
d) $S(n-3)=3\cdot 2^{n-2}-2^{n-3}=6\cdot 2^{n-3}-2^{n-3}=5\cdot 2^{n-3}$
e) $S(n-4)=5\cdot 2^{n-3}+2^{n-4}=10\cdot 2^{n-4}+2^{n-4}=11\cdot 2^{n-4}$
So the sequence of coefficients $c$ is actually
$$
1,1,3,5,11,\ldots,c_i,2c_i\pm 1,\ldots,c_{n+1}
$$
and the question can be restated as whether or not a sequence of length $n+1$ has an element $c$ multiple of $n$. Lets call it a $C$ sequence.
Notice that we can check whether or not any $C$ sequence of prime length $p$ has a term divisible by $p$, since in this case a positive answer makes the general question follow immediately.
Let $n=p$, and assume $c_i, 2c_i-1, 2^2c_i-1, 2^3c_i-3, 2^4c_i-5, 2^5c_i-11, \ldots, 2^{\frac{p-1}{2}}c_i-c_j$ are $\frac{p+1}{2}$ consecutive coefficients in the $C$ sequence where none is divisible by $p$. Notice that the negatives are again the sequence up to the $\textstyle\frac{p-1}{2}-th$ value, then
$$
c_i, 2c_i-c_1, 2^2c_i-c_2, 2^3c_i-c_3, 2^4c_i-c_4, 2^5c_i-c_5, \ldots, 2^{\frac{p-1}{2}}c_i-c_{\frac{p-1}{2}}
$$
In particular, 
$$
2^{\frac{p-1}{2}}c_i\not\equiv c_\frac{p-1}{2}\pmod p\Rightarrow 2^{p-1}c_i^2\not\equiv c_\frac{p-1}{2}^2\pmod p\Rightarrow c_i^2\not\equiv c_\frac{p-1}{2}^2\pmod p
$$
Since the length of such list of consecutive coefficients is $\frac{p+1}{2}$, and there are $p+1$ terms in the $C$ sequence, then we have $i\in\{1,2,\ldots,\frac{p-1}{2}\}$.
This means that
\begin{array}{ll}
c_1^2&\not\equiv c_\frac{p-1}{2}^2\pmod p\\
c_2^2&\not\equiv c_\frac{p-1}{2}^2\pmod p\\
&\vdots\\
c_\frac{p-1}{2}^2&\not\equiv c_\frac{p-1}{2}^2\pmod p
\end{array}
The last equivalence is absurd, thus $p$ divides at least one term of the $C$ sequence. $\blacksquare$
