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Heath-Brown's identity states: Let $K \geq 1, z \geq 1.$ Then for any $n < 2 z^K$ we have $$ \Lambda(n) = - \sum_{1 \leq k \leq K} (-1)^k {{K}\choose{k}} \sum_{ \substack{ m_1 \cdots m_k n_1 \cdots n_k = n \\ m_1, \ldots, m_k \leq z }} \mu(m_1) \cdots \mu(m_k) \log n_k. $$

I am trying to better understand this well used identity, and I have a few questions regarding this which I list below. I would greatly appreciate any comments on any of them. Thank you.

1) When is it more advantageous to use Heath-Brown's identity over Vaughan's identity?

2) When applying Heath-Brown's identity, does the terms with $1 \leq k \leq K-1$ are they all type I estimates and the term $k=K$ become type II estimate? (I would appreciate comments on how the identity is used normally...)

3) I have seen in few places that states Heath-Brown's identity is 'more flexible' than Vaughan's identity. What is meant by this?

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Heath-Brown's identity is more flexible (than Vaughan's) in the sense that the participating convolutions have several factors (not just two). This allows greater freedom in choosing the supports of the factors, in particular, the factors can be supported on much shorter intervals.

I am not sure what you mean by "normal use" of Heath-Brown's identity, but you can have a look at the Polymath8a paper, where the best result is achieved via Heath-Brown's identity and Deligne's work on the Riemann hypothesis for varieties over finite fields, while a weaker (but still very good) result is achieved via Vaughan's identity and Weil's work on the Riemann hypothesis for curves over finite fields.

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