Arithmetic function ($f:{\mathbb{Z}}_{>0}\to \mathbb{C}$)

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 radical of an integer $n>0$ is the product of its distinct prime factors, each included exactly once. If the prime factorization is $n=p_1^{a_1}\cdots p_r^{a_r}$ with distinct prime numbers $p_i$ and positive integer exponents $a_i$, thenThus the radical of an integer remembers which prime factors occur and forgets their prime valuations. At $n=1$, the empty product gives $\mathrm{rad}(1)=1$. The domain of a function here is the positive integers; zero is excluded. ^[ref]

For example, $72=2^3\cdot 3^2$ has distinct prime factors $2,3$, so $\mathrm{rad}(72)=2\cdot 3=6$. Similarly,using the prime factorizations $30=2\cdot 3\cdot 5$, $32=2^5$ and $36=2^2\cdot 3^2$. The prime omega function counts the retained prime factors, while the total number of prime factors counts their occurrences before repetitions are removed.

The radical of an integer is its largest squarefree divisor. Indeed, a squarefree divisor chooses at most one copy of each prime factor, so it divides $\mathrm{rad}(n)$; the radical of an integer itself is a squarefree divisor. Consequently,The last statement follows because every nonzero prime valuation is then one. It includes $n=1$, a squarefree integer. Applying the radical of an integer twice also changes nothing: $\mathrm{rad}(\mathrm{rad}(n))=\mathrm{rad}(n)$.

For any positive integers $a,b$, the exact product rule isTo prove it, a prime factor dividing just one input occurs once in the numerator and not in the denominator. A prime factor dividing both occurs twice in the numerator and once in the denominator, since it divides their greatest common divisor. Each prime factor of $ab$ therefore remains exactly once. For coprime integers, the denominator is $\mathrm{rad}(1)=1$, proving the radical of an integer is a multiplicative function. It is not a completely multiplicative function: $\mathrm{rad}(4)=2\ne 4=\mathrm{rad}(2{)}^2$.

For a worked example with shared prime factors, take $a=12$ and $b=18$. Their greatest common divisor is $6$, and $ab=216=2^3\cdot 3^3$, soThis explains why prime support of an integer is useful: questions about the presence of prime factors can discard their multiplicities. For example, the squarefree divisors of $72$ are exactly the integer divisors of $\mathrm{rad}(72)=6$, namely $1,2,3,6$. Also $\gcd (m,n)=1$ if and only if $\gcd (m,\mathrm{rad}(n))=1$, because coprimality depends only on shared prime factors.

A further characterization uses mutual divisibility of powers: for positive integers $a,b$,If their prime supports of an integer agree, sufficiently large $k,\ell$ make all the required prime valuations large enough. Conversely, these two divisibility relations force each prime factor of either input to divide the other. For $72$ and $36$, one can take $k=2$ and $\ell =1$: $72\mid 36^2$ and $36\mid 72$.

The prime support of an integer $n>0$ is the finite set of its distinct prime factors. For example, $P(72)=P(36)=\{2,3\}$, $P(30)=\{2,3,5\}$ and $P(1)$ is the empty set. Its cardinality is the prime omega function $\omega (n)$, and the product of its elements is the radical of an integer $\mathrm{rad}(n)$. These are two different ways to summarize the same finite set: a count and a product.

The total number of prime factors $\Omega (n)$ counts every occurrence of a prime factor in the prime factorization of a positive integer $n$. Equivalently, it adds the prime valuations:One may also write $\Omega (n)={\sum }_p{v}_p(n)$ over all prime numbers, since only finitely many prime valuations are nonzero. The empty sum gives $\Omega (1)=0$. By contrast, the prime omega function has $\omega (n)=r$ and counts distinct prime factors. ^[ref]

Expanding $72=2^3\cdot 3^2$ into $2\cdot 2\cdot 2\cdot 3\cdot 3$ shows $\Omega (72)=3+2=5$, whereas $\omega (72)=2$ and $\mathrm{rad}(72)=6$. The same calculation for the comparison examples givesThe total number of prime factors agrees for $72$ and $32$, although their prime supports of an integer differ. Meanwhile $72$ and $36$ have the same radical of an integer and prime omega function, but different values of the total number of prime factors. These examples show exactly which information each arithmetic function forgets.

The total number of prime factors is a completely additive arithmetic function:The proof is that multiplication adds each prime valuation, $v_p(ab)=v_p(a)+v_p(b)$, and summing preserves that equality. No coprimality condition is needed. For example, $12=2^2\cdot 3$ and $18=2\cdot 3^2$ both have total number of prime factors three; their product $216=2^3\cdot 3^3$ has six. By comparison, the prime omega function is only an additive arithmetic function: here $\omega (216)=2\ne 4=\omega (12)+\omega (18)$.

There is a useful way to see the total number of prime factors as a count of elementary building blocks: any expression of $n$ as a product of prime numbers has exactly $\Omega (n)$ factors, by the fundamental theorem of arithmetic. Equivalently, every factorization tree for $n>1$ has $\Omega (n)$ leaves. Repeated prime factors increase this count without introducing new kinds of prime number. The difference is the existing excess prime-factor multiplicity:Every summand is nonnegative, so $\Omega (n)=\omega (n)$ if and only if $n$ is a squarefree integer, including $n=1$. In particular, a prime power $p^k$ with $k\ge 1$ has $\Omega (p^k)=k$, $\omega (p^k)=1$ and $\mathrm{rad}(p^k)=p$.

Historically, Hardy and Ramanujan's 1917 paper explicitly separated the prime omega function from the total number of prime factors, writing them as $f(n)$ and $F(n)$, respectively. Their example $2^3\cdot 3^2\cdot 5$ has counts three and six. That distinction let them compare two measures of how an integer is built from prime factors. ^[ref]

The nonnegative arithmetic function $\Omega (n)-\omega (n)$ counts the repeated occurrences beyond the first of each prime factor. Its expectation under uniform sampling from $[N]$ is bounded by ${\sum }_{p}1/(p(p-1))\le 1$, independently of $N$.

An arithmetic function $f$ is additive when $f(mn)=f(m)+f(n)$ for coprime integers $m,n$. Equivalently, $f(n)={\sum }_{p^k\parallel n}f(p^k)$. This concerns multiplication of integers, and differs from an additive function on a vector space.

For an additive arithmetic function, the model mean is $A_f(x)={\sum }_{p^k\le x}f(p^k)(1-1/p)/p^k$. The exact uniform expectation differs by $O(x^{-1}{\sum }_{p^k\le x}|f(p^k)|)$.

A strongly additive arithmetic function satisfies $f(p^k)=f(p)$ for every prime $p$ and $k\ge 1$, so $f(n)={\sum }_{p\mid n}f(p)$. The prime omega function is an example.

A completely additive arithmetic function satisfies $f(mn)=f(m)+f(n)$ for all positive $m,n$, equivalently $f(p^k)=kf(p)$. The total number of prime factors is an example.

An arithmetic function $f$ is a multiplicative function when $f(1)=1$ and $f(mn)=f(m)f(n)$ for all coprime integers $m,n>0$, meaning their greatest common divisor is $1$. The condition $f(1)=1$ excludes the identically zero function. This definition makes no assertion about a pair whose greatest common divisor exceeds $1$. ^[ref]

For the divisor function, the integer divisors of $2$ are $1,2$, those of $3$ are $1,3$, and those of $6$ are $1,2,3,6$. HenceHere $2$ and $3$ are coprime integers. In contrast, $\tau (4)=3\ne 4=\tau (2{)}^2$, because the integer divisors of $4$ are $1,2,4$. Thus the divisor function is a multiplicative function but not a completely multiplicative function.

The practical gain is that prime factorization reduces all values to values at prime powers. For distinct prime numbers $p_1,\dots ,p_r$ and positive exponents $a_1,\dots ,a_r$,For instance, $f(12)=f(4)f(3)$ since $4$ and $3$ are coprime integers; replacing $f(4)$ by $f(2{)}^2$ additionally requires a completely multiplicative function. The short proof repeatedly applies the defining rule to prime powers belonging to different prime numbers, which are coprime integers.

For an arithmetic function, multiplicativity means $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m,n$ are positive coprime integers. It allows values of an arithmetic function to be assembled from its values on prime powers, as explained under multiplicative function.

An arithmetic function $f$ is a completely multiplicative function when $f(1)=1$ and $f(mn)=f(m)f(n)$ for every pair of positive integers $m,n$. In particular $f(p^a)=f(p{)}^a$ for every prime number $p$ and every nonnegative exponent $a$. Both the constant-one arithmetic function and the identity arithmetic function are completely multiplicative functions: $1(mn)=1=1\cdot 1$ and $\mathrm{id}(mn)=mn=\mathrm{id}(m)\mathrm{id}(n)$. The divisor function fails this stronger condition at $m=n=2$.

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 sum-of-divisors function adds all positive integer divisors of a positive integer, including the number itself:It is the divisor sum of the identity arithmetic function $\mathrm{id}(n)=n$, so $\sigma =\mathrm{id}\ast 1$ under Dirichlet convolution. Both factors are multiplicative functions, hence so is the sum-of-divisors function. The notation ${\sigma }_1$ distinguishes it from other weighted divisor sums. ^[ref]

For a prime number $p$ and a nonnegative exponent $a$, the positive integer divisors of the prime power $p^a$ are precisely $1,p,\dots ,p^a$. Their finite geometric series givesFor example, $\sigma (8)=1+2+4+8=15=(2^4-1)/(2-1)$. For a prime factorization $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, multiplicativity givesFor $n=1$, this is an empty product, equal to $1$. ^[ref]

The distributive property makes the formula especially beautiful: choosing one term from each factor produces each positive integer divisor exactly once, by the fundamental theorem of arithmetic. For $12=4\cdot 3=2^2\cdot 3$, the choices form a three-by-two grid with entries $2^i{3}^j$, where $0\le i\le 2$ and $0\le j\le 1$. Adding its rows givesThe divisor function counts six cells; the sum-of-divisors function adds their six values. Once the prime factorization is known, the product formula avoids testing every candidate integer divisor. Shared prime factors require care: $\sigma (4)=7\ne 9=\sigma (2{)}^2$, so the sum-of-divisors function is not a completely multiplicative function.

Subtracting the number itself gives the proper-divisor sum $s(n)=\sigma (n)-n$. Comparing $s(n)$ with $n$ distinguishes perfect numbers, deficient numbers, and abundant numbers. For $6$, $s(6)=1+2+3=6$; for $28$, $s(28)=1+2+4+7+14=28$. Both are perfect numbers. The calculation $s(8)=1+2+4=7<8$ makes $8$ a deficient number, while $s(12)=1+2+3+4+6=16>12$ makes $12$ an abundant number. These comparisons turn the sizes of integer divisors into a classification of positive integers. ^[ref]

This connection is ancient: Euclid's Elements, Book IX, Proposition 36, constructs a perfect number by multiplying the last term of $1,2,\dots ,2^{k-1}$ by their sum, when that sum is a prime number. In modern notation, put $q=2^k-1$ and $N=2^{k-1}q$, with $k\ge 2$ and $q$ a prime number. The sum-of-divisors function proves the construction immediately:Consequently the proper-divisor sum is $s(N)=N$. Taking $k=2,3$ produces the perfect numbers $6,28$. ^[ref]

The proper-divisor sum, or aliquot sum, adds the positive proper divisors:In particular $s(1)=0$, an empty sum. Removing $n$ from the sum-of-divisors function is essential: $\sigma (6)=12$, but its proper-divisor sum is $6$.

For multiplicative functions bounded by one, the pretentious distance up to $x$ is defined byIt measures how similarly $f$ and $g$ behave on primes.

Let $X\ge 3$, and let $f$ be a normalized multiplicative arithmetic function supported on squarefree integers, with $|f(p)|\le 1$. Form the finite measure $\nu ={\sum }_{n}f(n)n^{-1-1/\log X}{\delta }_{\log n}$ using Dirac measures. Under the transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$, its Euler product is ${\prod }_p(1+f(p)p^{-1-1/\log X-it})$. Expanding its logarithmic modulus, and replacing the damped prime weights by $1/p$ up to $X$ with Mertens first theorem, gives the displayed absolute bound, uniformly in real $t$. Hence $|\hat{\nu }(t)|\ge \delta \log X$ forces $D(f,n^{it};X{)}^2\le \log (C/\delta )$. The distance expression still defines a nonnegative pretentious score for prime values bounded by one; its actual metric interpretation uses unit-modulus values.

Because $\mu (p)=-1$, the squared pretentious distance from one to the Möbius function is twice the reciprocal-prime sum. The Mertens second theorem therefore gives $D(1,\mu ;X{)}^2=2\log \log X+2B_1+o(1)$. In particular the distance diverges, illustrating that a multiplicative function can be far from the constant function even though both have unit-modulus prime values.

For unit-modulus prime values, pretentious distance is the Euclidean metric on the finite vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$. On whole arithmetic functions it is only a pseudometric: it sees neither values on prime powers nor primes above the cutoff. Even for every cutoff, the functions one and ${\mu }^2$ have distance zero but differ on nonsquarefree integers. The norm representation proves the triangle inequality and identifies exactly the quotient on which separation holds.

The pretentious distance obeys

An Archimedean character on the positive integers is the completely multiplicative function $n\mapsto n^{it}$ for a real parameter $t$.

Halász's theorem bounds the mean of a bounded multiplicative arithmetic function in terms of its least pretentious distance from an Archimedean character. A large mean can occur only when this distance is small.

The Möbius function stays far in pretentious distance from every Archimedean character. One useful uniform form isThis follows by estimating the prime sum ${\sum }_{p\le x}(1+\cos (t\log p))/p$.

The pointwise product of arithmetic functions multiplies two arithmetic functions at the same input: $(f\cdot g)(n)=f(n)g(n)$. Thus $(1\cdot 1)(12)=1$ and $(\mathrm{id}\cdot 1)(12)=12$, using the constant-one arithmetic function and the identity arithmetic function. These differ from the corresponding Dirichlet convolution values $6$ and $28$, because Dirichlet convolution sums over every positive integer divisor.

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 convolution of two multiplicative functions is a multiplicative function. For positive coprime integers $m,n$, every positive integer divisor of $mn$ has a unique form $de$ with $d\mid m$ and $e\mid n$, by prime factorization. The pairs $d,e$ and $m/d,n/e$ are each coprime integers, so the defining rules for the two multiplicative functions separate the sum:Also $(f\ast g)(1)=f(1)g(1)=1$. This short argument proves multiplicativity of both the divisor function $1\ast 1$ and the sum-of-divisors function $\mathrm{id}\ast 1$. The result need not be a completely multiplicative function, even when both inputs are: $1\ast 1=\tau$ fails at $2\cdot 2$. ^[ref]

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 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.

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]

The divisor summatory function adds the divisor function up to a cutoff $x\ge 1$. It counts all ordered pairs of positive integers whose product is at most $x$. For $N=12$, the twelve individual divisor function values areCounting columns of lattice points instead gives $12+6+4+3+2+2+1+1+1+1+1+1=35$. The Dirichlet hyperbola method needs only the first three column counts:The arithmetic work of the Dirichlet hyperbola method is much smaller even though the answer is identical.

This is useful for elementary averages: individual divisor function values fluctuate, but their summation has the regular behavior given by the divisor summatory asymptotic. Dividing that estimate by a positive integer $N$ giveswhere $\log$ is the natural logarithm, $\gamma$ is the Euler--Mascheroni constant, and $O$ uses big O notation. This describes an average, rather than an estimate for every individual divisor function value. ^[ref]

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 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.

The Dirichlet hyperbola method gives $D(x)=2{\sum }_{a\le r}\lfloor x/a\rfloor -r^2$, where $r=\lfloor \sqrt{x}\rfloor$. The harmonic number estimate $H_r=\log r+\gamma +O(1/r)$, with $\gamma$ the Euler's constant, yields $D(x)=2x(\log r+\gamma )-r^2+O(x/r+r)$. Substitution of $r=\sqrt{x}+O(1)$ gives the displayed asymptotic expansion. Comparison of ${\sum }_{n\le x}\log n$ with its integral consequently gives ${\sum }_{n\le N}(\tau (n)-\log n-2\gamma )=O(\sqrt{N})$ for integer $N$.

Write $B(t)={\sum }_{n\le t}\tau (n)/n$ and $r=\lfloor \sqrt{t}\rfloor$. The weighted Dirichlet hyperbola method givesIndeed, use $H_{\lfloor u\rfloor }=\log u+\gamma +O(1/u)$ and ${\sum }_{a\le r}(\log a)/a=\frac{1}{2}{\log }^{2}r+c\prime +O(\log (2r)/r)$. The latter follows by comparing each summand with the integral on its unit interval: the derivative of $(\log t)/t$ has an integrable tail. Since $r=\sqrt{t}+O(1)$, substitution yields the expansion. Here $H_n$ is the harmonic number and $\gamma$ the Euler--Mascheroni constant.

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 $n$.

For each $\eta >0$, the divisor function satisfies the displayed bound. Write $n=\prod p^{a_p}$, so $\tau (n)=\prod (a_p+1)$. For all sufficiently large primes, $a+1\le 2^a\le p^{\eta a}$ for every $a\ge 0$. For each of the finitely many smaller primes, ${\sup }_{a\ge 0}(a+1)p^{-\eta a}$ is finite. Multiplying these finitely many constants proves the bound. Products of a fixed number of divisor functions obey the same conclusion after reducing $\eta$.

The divisor function satisfies ${\sum }_{n\le X}d(n{)}^2\ll X{\log }^3(2X)$. Partial summation consequently gives ${\sum }_{n\le X}d(n{)}^2/n\ll {\log }^4(2X)$.

The Von Mangoldt function assigns a natural logarithm weight to each positive power of a prime number. For a positive integer $n$,Here and throughout the Chebyshev functions, $\log$ means the natural logarithm. In particular, $\Lambda (1)=0$: the exponent must be positive, even though the convention in prime power allows exponent zero. The support of a function of $\Lambda$ therefore consists precisely of the prime powers greater than one. ^[ref]

For example, $\Lambda (2)=\Lambda (4)=\Lambda (8)=\log 2$ and $\Lambda (3)=\Lambda (9)=\log 3$, whereas $\Lambda (6)=\Lambda (12)=0$. A composite number can have a nonzero weight: $8$ has just one distinct prime factor. But $12=2^2\cdot 3$ has two distinct prime factors, so its weight is zero. At $p^k$ the weight is $\log p$, not $\log (p^k)=k\log p$. This choice lets successive powers record successive copies of the same prime factor in the Von Mangoldt divisor identity.

The Von Mangoldt function is useful because it connects two views of the prime numbers: the prime factorization of an individual integer, through a divisor sum, and the accumulated weights below a cutoff, through the Second Chebyshev function. The latter turns questions about the distribution of prime numbers into questions about a partial sum of an arithmetic function.

The elementary Selberg symmetry formula can be writtenUse the weights ${\lambda }_d=\mu (d)\log (x/d)/\log x$. Their divisor sum is $1_{n=1}+\Lambda (n)/\log x$. Apply them to the integrated Chebyshev sum, then use the Möbius harmonic logarithmic moments. The Chebyshev estimate shows this is equivalent, with an $O(x)$ error, to ${\sum }_{n\le x}\Lambda (n)\log n+{\sum }_{mn\le x}\Lambda (m)\Lambda (n)=2x\log x+O(x)$.