TITLE: How prove this $\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$
QUESTION [12 upvotes]: Given the positive integer numbers $n$,prove that
$$\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$$
where $\sigma(n)$ is defined as $$\sigma{(N)}=\sum_{d|N}d$$
This problem I think is very interesting, and have nice methods?
Thank you everyone.
and I find some of this


or can see:http://www.advancesindifferenceequations.com/content/pdf/1687-1847-2013-84.pdf

REPLY [6 votes]: It is difficult to add anything to such a nice answer. So take this more as a comment, too long for the the comment field:
There is another elementary proof, using only
an arithmetic identity, which generalizes a classical formula of Liouville. For details, see the article "Elementary Evaluation of Certain Convolution Sums Involving Divisor Functions" by Huard et al.On page $21$ there is Theorem $5$, which yields 
$$
\sum_{k\le N/2} \sigma(2k-1)\sigma(N-2k+1)=\frac{1}{24}\left( 5\sigma_3(N)-21\sigma_3(N/2)+16\sigma_3(N/4)\right)+(1 − 6N)(σ(N) − 3σ(N/2) + 2σ(N/4))).
$$
For $N=2^n$ there result follows also (with some manipulations given in theorem $9$). The proof is completely elementary, only using the generalization of Liouville’s Formula, which is Theorem $1$.