Divisor Sum of Power of 2
Jump to navigation
Jump to search
Theorem
Let $n \in \Z_{>0}$ be a power of $2$.
Then:
- $\map {\sigma_1} n = 2 n - 1$
Proof
Let $n = 2^k$.
Then:
\(\ds \map {\sigma_1} n\) | \(=\) | \(\ds \dfrac {2^{k + 1} - 1} {2 - 1}\) | Divisor Sum of Power of Prime | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 2^k - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 n - 1\) |
$\blacksquare$