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A small lattice decomposition for Dirichlet's hyperbola method

Example

Take $x = 12$ , $U = 3$ , and $V = 4$ , so $U V = 12$ . For $τ = 1 * 1$ , Dirichlet's hyperbola identity becomes

$\sum_{n \leq 12} τ (n) = \sum_{a \leq 3} ⌊ \frac{12}{a} ⌋ + \sum_{b \leq 4} ⌊ \frac{12}{b} ⌋ - 3 \cdot 4.$

Numerically this is

$35 = (12 + 6 + 4) + (12 + 6 + 4 + 3) - 12.$

Facts & Assumptions

Given: The value $x = 12$ and the split $U = 3$ , $V = 4$ .

Verification

technique · direct

1.1given

By The divisor functions arise by Dirichlet convolution, $τ = 1 * 1$ . Therefore Dirichlet's hyperbola method for summatory convolutions applies with $F (y) = G (y) = ⌊ y ⌋$ .

2.1step 1.1givenalgebra

The two arm sums are $\sum_{a \leq 3} ⌊ \frac{12}{a} ⌋ = 12 + 6 + 4 = 22, \sum_{b \leq 4} ⌊ \frac{12}{b} ⌋ = 12 + 6 + 4 + 3 = 25,$ and the overlap is $F (3) G (4) = 3 \cdot 4 = 12$ .

3.1step 2.1givenalgebra∎

Hence the hyperbola identity gives $\sum_{n \leq 12} τ (n) = 22 + 25 - 12 = 35$ . This matches the direct total $1 + 2 + 2 + 3 + 2 + 4 + 2 + 4 + 3 + 4 + 2 + 6 = 35$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Tom Sanders, Topics in Analytic Number Theory, Chapter 1 (standard reference, not scraped)