Vaughan's Identity

Theorem
Let $\Lambda$ be von Mangoldt's function.

Let $\mu$ be the Mobius function.

Then for $y,z \geq 1$ and $n > z$


 * $\displaystyle \Lambda(n) = \sum_{\substack{d \backslash n\\d \leq y}}\mu(d) \log\left( \frac nd \right) - \mathop{\sum\sum}_{\substack{dc \backslash n\\d \leq y,c\leq z}}\mu(d)\Lambda(c) + \mathop{\sum\sum}_{\substack{dc \backslash n\\d > y,c > z}}\mu(d)\Lambda(c)$

Proof
By Sum Over Divisors of von Mangoldt is Logarithm we have:


 * $\displaystyle \log n = \sum_{d \backslash n} \Lambda(d)$

From this we deduce:

Now taking the second sum in this last line we have:

Again taking the second sum in the final line we have:


 * $\displaystyle \mathop{\sum\sum}_{\substack{dc \backslash n \\ d > y,c\leq z} } \mu(d)\Lambda(c)=\mathop{\sum\sum}_{\substack{dc \backslash n \\ c\leq z} } \mu(d)\Lambda(c) - \mathop{\sum\sum}_{\substack{dc \backslash n \\d \leq y, c\leq z} } \mu(d)\Lambda(c)$

Putting this together we have:


 * $\displaystyle \Lambda(n) = \sum_{\substack{d \backslash n\\ d \leq y} } \mu(d) \log \left( \frac nd \right) + \mathop{\sum\sum}_{\substack{dc \backslash n \\ d > y,c>z} } \mu(d)\Lambda(c) + \mathop{\sum\sum}_{\substack{dc \backslash n \\ c\leq z} } \mu(d)\Lambda(c) - \mathop{\sum\sum}_{\substack{dc \backslash n \\d \leq y, c\leq z} } \mu(d)\Lambda(c)$

So we are done if we can show that:


 * $\displaystyle \mathop{\sum\sum}_{\substack{dc \backslash n \\ c\leq z} } \mu(d)\Lambda(c) = 0$

We write the sum as:


 * $\displaystyle \mathop{\sum\sum}_{\substack{dc \backslash n \\ c\leq z} } \mu(d)\Lambda(c) = \sum_{\substack{c \leq z\\c \backslash n}}\Lambda(c)\sum_{d \backslash \frac nc} \mu(d)$

Now we have $c \leq z < n$, so $\displaystyle \frac nc > 1$.

Therefore, by Sum of Möbius Function over Divisors, for each $c$ the inner sum vanishes.

This shows that :


 * $\displaystyle \mathop{\sum\sum}_{\substack{dc \backslash n \\ c\leq z} } \mu(d)\Lambda(c) = 0$

as required.