= Asymmetric Dirichlet hyperbola split
{title2=$AB=N$}
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 $f(a)g(b)$, yields
$$
\sum_{n\le N}(f*g)(n)
=\sum_{a\le A}f(a)G(N/a)
+\sum_{b\le B}g(b)F(N/b)-F(A)G(B),
\qquad AB=N.
$$
No <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 $N=12$, $A=2$, $B=6$ gives
$$
D(12)=(12+6)+(12+6+4+3+2+2)-2\cdot6=18+29-12=35.
$$
It uses two columns and six rows, with a twelve-point overlap.
For general positive real thresholds, \b[$AB\ge N$ 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 $r^2<N$.
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