 Asymmetric Dirichlet hyperbola split

A balanced Dirichlet hyperbola method split takes real thresholds $A=B=\sqrt{N}$. An asymmetric split takes $1\le A\le N$ and $B=N/A$. Every lattice point with $ab\le N$ then has $a\le A$ or $b\le B$, and every point in their overlap rectangle is allowed because $AB=N$. For arithmetic functions $f,g$, define their partial sums by $F(t)={\sum }_{a\le t}f(a)$ and $G(t)={\sum }_{b\le t}g(b)$, always over positive integers. The same inclusion-exclusion principle, now with weight $f(a)g(b)$, yieldsNo multiplicativity assumption is needed: this is a finite rearrangement of Dirichlet convolution. The constant-one arithmetic function in both positions recovers the divisor summatory function. For $N=12$, $A=2$, $B=6$ givesIt uses two columns and six rows, with a twelve-point overlap.

For general positive real thresholds, $AB\ge N$ is sufficient for coverage, since $a>A$ and $b>B$ would imply $ab>AB\ge N$. If $AB>N$, however, the overlap can include points above the hyperbola. The correct subtraction is then ${\sum }_{a\le A,b\le B,ab\le N}f(a)g(b)$; the product $F(A)G(B)$ is justified only when the whole integer rectangle lies below it. For example, $N=12$, $A=B=4$ covers every allowed pair, but its rectangle contains the forbidden pair $(4,4)$, so subtracting all sixteen points would give $34$ instead of $35$.

Conversely, the balanced integer thresholds $r,r$ have $r^2\le N$ and still cover every allowed pair because $(r+1{)}^2>N$. More generally, positive integer thresholds $U,V$ give coverage exactly when $(U+1)(V+1)>N$, and their overlap is a full rectangle exactly when $UV\le N$. These discrete conditions explain why the square-root identity works even when $r^2<N$.