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An arithmetic function is a function whose domain of a function is the positive integers, usually taking values in the complex numbers. Thus $f(1),f(2),\dots$ are its values; $f(0)$ is outside this domain of a function. Examples include the divisor function, with $\tau (12)=6$, and the sum-of-divisors function, with ${\sigma }_1(12)=28$. Studying an arithmetic function packages a question about each positive integer into one object whose values can be combined using Dirichlet convolution. ^[ref]

The direct divisor summatory function formula evaluates $N$ floor divisions. Its balanced Dirichlet hyperbola method formula evaluates $r=\lfloor \sqrt{N}\rfloor$ floor divisions and reuses their sum for the other strip, followed by doubling and square-overlap subtraction. Thus it uses $O(\sqrt{N})$ arithmetic operations, with big O notation, in a model that counts each division, addition and multiplication as one operation. At $N=10^6$, the division counts for these two formulas are $10^6$ and $10^3$.

For a general Dirichlet convolution, the asymmetric formula has $\lfloor A\rfloor +\lfloor B\rfloor$ outer terms; evaluating $F$ and $G$ has an additional cost. Under $AB=N$, equal thresholds minimize $A+B$, but an asymmetric split can be useful if the two partial sums have different evaluation costs. The balanced divisor summatory function gains the further reuse because its two arithmetic functions are identical.

These are arithmetic operation counts, rather than counts of elementary operations on binary digits: the integers grow with $N$, and division of longer integers costs more in that model of computational complexity theory. Computing $r$ also requires an exact integer square root, or checking and correcting a numerical square root until $r^2\le N<(r+1{)}^2$ holds.

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 $r^2<N$.

The constant-one arithmetic function has $1(n)=1$ for every positive integer $n$, including $1(1)=1(12)=1$. It is a completely multiplicative function. Its Dirichlet convolution with an arithmetic function $f$ is the divisor sum $(1\ast f)(n)={\sum }_{d\mid n}f(d)$. Thus $1\ast 1=\tau$, the divisor function. Its pointwise product of arithmetic functions with $f$ is $f$ itself; its Dirichlet convolution with $f$ generally is not. The Identity for Dirichlet convolution is instead $\epsilon$.

The Dirichlet convolution of arithmetic functions $f,g$ is the arithmetic functionIn the first sum, $d$ runs over the positive integer divisors of $n$; the second uses ordered pairs of positive integers whose product is $n$. Both sums are finite. Each pair contributes one weight from $f$ and one from $g$, so Dirichlet convolution combines information attached to complementary factors. It turns a divisor sum into an operation on arithmetic functions. ^[ref]

For $n=12$ the six ordered pairs are $(1,12),(2,6),(3,4),(4,3),(6,2),(12,1)$. The constant-one arithmetic function gives one per pair, while the identity arithmetic function weights the first factor by its size:Here $\tau$ is the divisor function and ${\sigma }_1$ is the sum-of-divisors function. By comparison, the pointwise product of arithmetic functions gives $(1\cdot 1)(12)=1$ and $(\mathrm{id}\cdot 1)(12)=12$.

Commutativity follows by swapping the two entries of every ordered pair. Associativity follows by reindexing either grouping over the same ordered tuples of three positive integers:For example, $((1\ast 1)\ast 1)(12)$ counts ordered tuples of three positive integers with product $12$. Distribute the two occurrences of the prime factor $2$ among three entries in $\left(\genfrac{}{}{0ex}{}{2+2}{2}\right)=6$ ways, and the one occurrence of the prime factor $3$ in $\left(\genfrac{}{}{0ex}{}{1+2}{2}\right)=3$ ways. These binomial coefficients are the stars and bars counts of exponent choices in the nonnegative integers; prime factorization therefore gives $6\cdot 3=18$ such tuples. Directly, the same Dirichlet convolution is ${\sum }_{d\mid 12}\tau (d)=1+2+2+3+4+6=18$.

The Identity for Dirichlet convolution is $\epsilon$, and the Möbius function reverses a divisor sum because $\mu \ast 1=\epsilon$. This makes Dirichlet convolution useful both for counting factorizations and for recovering an arithmetic function from its divisor sum.

The Dirichlet hyperbola method turns a long summation over a Dirichlet convolution into two short ones by covering the lattice points below a hyperbola with two overlapping strips. Its simplest application is an exact count: for a positive integer $N$, every positive integer divisor $a$ of $n\le N$ corresponds to the ordered pair $(a,n/a)$. Thus the divisor summatory function counts precisely the ordered pairs of positive integers with $ab\le N$:Here $\tau$ is the divisor function, and the floor function gives the number of allowed values of $b$ for each $a$. The axes are excluded, while points on $ab=N$ are included.

Set $r=\lfloor \sqrt{N}\rfloor$, using the floor function and the positive square root. Every counted pair has $a\le r$ or $b\le r$. Otherwise $ab\ge (r+1{)}^2>N$. The first strip contains ${\sum }_{a=1}^r\lfloor N/a\rfloor$ points; swapping the two coordinates gives the same count for the second. Their overlap contains all $r^2$ pairs with $1\le a,b\le r$, since $r^2\le N$. The inclusion-exclusion principle therefore gives the exact identityThe subtraction is for the whole overlap square, rather than only the diagonal. The same Dirichlet hyperbola method identity holds for real $x\ge 1$, with $D(x)={\sum }_{n\le x}\tau (n)$ and $r=\lfloor \sqrt{x}\rfloor$. ^[ref]

A Dirichlet series is a series of the formMultiplication of absolutely convergent Dirichlet series corresponds to Dirichlet convolution of their coefficients.

The divisor function $\tau (n)={\sum }_{d\mid n}1$ counts the positive integer divisors of a positive integer $n$. In particular $\tau (1)=1$, because $1$ has just one positive integer divisor. More generally the notation ${\sigma }_k(n)={\sum }_{d\mid n}{d}^k$ gives $\tau ={\sigma }_0$ and the sum-of-divisors function ${\sigma }_1$. Thus the divisor function counts the positive integer divisors, while ${\sigma }_1$ adds their sizes. ^[ref]

If the prime factorization is $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, thenThe short proof is a count: each positive integer divisor has a unique form ${\prod }_{k=1}^r{p}_k^{b_k}$, where $0\le b_k\le a_k$. There are $a_k+1$ choices at each prime number, and choosing them gives every positive integer divisor exactly once. For $n=1$ there are no prime factors and the product is $1$.

For $12=2^2\cdot 3$, choose $i\in \{0,1,2\}$ and $j\in \{0,1\}$ to obtain $2^i{3}^j$. The six values are $1,2,4,3,6,12$; sorted, the positive integer divisors are $1,2,3,4,6,12$. Therefore $\tau (12)=(2+1)(1+1)=6$. The diagram makes the independent exponent choices visible.

This count is useful because a known prime factorization gives $\tau (n)$ without testing every smaller positive integer for divisibility. It also proves the divisor function is a multiplicative function: coprime integers have disjoint prime factors, so their exponent choices combine. For example $\tau (6)=4=\tau (2)\tau (3)$, but $\tau (4)=3\ne 4=\tau (2{)}^2$, so it is not a completely multiplicative function. Equivalently, $\tau =1\ast 1$ under Dirichlet convolution, counting ordered pairs of positive integers whose product is $n$.

The divisor sum of an arithmetic function $f$ assigns to each positive integer $n$ the sumOnly positive integer divisors are included, so $F(1)=f(1)$. The weights $f(d)$ can count the integer divisors, measure their sizes, or record other arithmetic information: the constant-one arithmetic function gives the divisor function, whereas the identity arithmetic function gives the sum-of-divisors function.

Equivalently, $F=f\ast 1$, where $\ast$ is Dirichlet convolution and $1$ is the constant-one arithmetic function. If $f$ is a multiplicative function, then $F$ is a multiplicative function by Multiplicativity of Dirichlet convolution. ^[ref]

Here is the mechanism. For positive coprime integers $m,n$, every positive integer divisor of $mn$ has a unique form $ab$ with $a\mid m$ and $b\mid n$; moreover, $a,b$ are coprime integers. Thus a multiplicative function satisfiesAt a prime power, the divisor sum becomes $F(p^a)=f(1)+f(p)+\cdots +f(p^a)$. This turns a calculation over many integer divisors into short local calculations at prime powers.

The identity arithmetic function is $\mathrm{id}(n)=n$ on the positive integers, for example $\mathrm{id}(1)=1$ and $\mathrm{id}(12)=12$. It is a completely multiplicative function and gives the sum-of-divisors function through $\mathrm{id}\ast 1={\sigma }_1$ under Dirichlet convolution. It is distinct from the Identity for Dirichlet convolution $\epsilon$. The identity $\mathrm{id}=1\ast \phi$, where $\phi$ is Euler's totient function, is equivalent to ${\sum }_{d\mid n}\phi (d)=n$.

The Identity for Dirichlet convolution is the arithmetic functionFor every arithmetic function $f$, it satisfies $f\ast \epsilon =\epsilon \ast f=f$. Indeed, in $(f\ast \epsilon )(n)={\sum }_{d\mid n}f(d)\epsilon (n/d)$, only $d=n$ survives. For example,At $n=1$ the sole term is $f(1)\epsilon (1)=f(1)$. This gives a useful way to check Dirichlet convolution calculations: inserting $\epsilon$ leaves every value unchanged. The three arithmetic functions $1$, $\mathrm{id}$, and $\epsilon$ all take value $1$ at $1$, but at $12$ the constant-one arithmetic function, the identity arithmetic function, and the Identity for Dirichlet convolution take values $1$, $12$, and $0$, respectively. ^[ref]

The Möbius function is the inverse of the constant-one arithmetic function under Dirichlet convolution, meaning $\mu \ast 1=1\ast \mu =\epsilon$. The Möbius divisor-sum identity gives a short proof: for $n>1$ the nonzero terms expand ${\prod }_{p\mid n}(1-1)=0$, while at $1$ the sum is $\mu (1)=1$. At $12$, the computation isThe zeros occur because $4$ and $12$ are divisible by the square of the prime number $2$. Consequently, if $F=f\ast 1$ is a divisor sum, then associativity gives $F\ast \mu =f\ast (1\ast \mu )=f\ast \epsilon =f$: this is Möbius inversion.

For $f(n)=3^{\omega (n)}$,If $p^a\Vert n$, its contribution to $(f\ast \Lambda )(n)$ is $f(n)(a-1+1/3)\log p$. Since $3(a-1+1/3)\ge a$, summing over the prime factors proves the bound. Here $\ast$ is Dirichlet convolution and $\Lambda$ is the Von Mangoldt function.

Suppose $\Psi (x)=x+O(xE(x))$ and ${\sum }_{j\ge 0}{\sup }_{2^j\le u<2^{j+1}}|E(u)|<\infty$. The Möbius divisor-sum identity gives ${\sum }_{d\le x}\mu (d)/d=O(1)$. The exact Dirichlet convolution identity $\mu \log =-\mu \ast \Lambda$ then gives ${\sum }_{n\le x}\mu (n)\log n=O(x)$, because its error is a convergent dyadic harmonic sum. Since ${\sum }_{n\le x}\log (x/n)=O(x)$, the displayed Mertens function bound follows. The classical Prime number theorem error $E(x)=e^{-c\sqrt{\log x}}$ meets the summability condition. This is a deduction from the Chebyshev error itself, without a separate estimate for the reciprocal zeta function.

For every positive integer $n$, the divisor sum of the Möbius function isThe sum includes all positive integer divisors, but only squarefree divisors contribute. Equivalently, $\mu \ast 1=\epsilon$ under Dirichlet convolution, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. ^[ref]

Each squarefree divisor chooses a subset of the distinct prime factors of $n$, contributing $(-1{)}^r$ when that subset has $r$ elements. Expanding a product therefore givesFor $n>1$ at least one prime factor occurs, so the product is zero. At $n=1$ it is an empty product, equal to one. This is the same alternating cancellation that drives inclusion-exclusion.

For $30=2\cdot 3\cdot 5$, all eight positive integer divisors are squarefree integers, givingPair each subset not containing $2$ with the subset obtained by inserting $2$: $(1,2),(3,6),(5,10),(15,30)$. Each pair cancels. The same nonzero terms occur for $60$, because its squarefree divisors are still the integer divisors of $30$; its other integer divisors have Möbius function value zero.

The Möbius divisor-sum identity is the cancellation behind Möbius inversion and reciprocal Dirichlet series: aggregation over integer divisors has an exact inverse.