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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]

Two integers $a,b$ are coprime when their greatest common divisor is one:For positive integers, this says that the only positive common divisor is one. For signed integers, use $\gcd (a,b)=\gcd (|a|,|b|)$, with absolute values; changing either sign leaves coprimality unchanged. With the convention $\gcd (0,0)=0$, zero is coprime precisely to $1,-1$, because $\gcd (a,0)=|a|$. Both $1,-1$ are coprime to every integer. ^[ref]

The word concerns shared prime factors, not whether either input is a prime number. For instance $8=2^3$ and $9=3^2$ are coprime composite numbers, even though neither is a square-free integer. Conversely, $6=2\cdot 3$ and $10=2\cdot 5$ are square-free integers but are not coprime. Consecutive integers are always coprime: any common divisor of $n,n+1$ divides their difference one. Thus $14,15$ provide another pair of coprime composite numbers.

For positive integers, let $P(n)$ be the prime support of an integer. The equivalent test isThe set intersection of the two prime supports of an integer must be the empty set. Indeed, any positive common divisor greater than one has a prime factor, and any shared prime factor is a common divisor greater than one. Hence coprimality depends on which prime factors occur, rather than on their prime valuations; replacing either input by its radical of an integer preserves coprimality.

For a list of at least two integers, distinguish pairwise coprimality, requiring every pair to be coprime, from joint coprimality, requiring only the greatest common divisor of the entire list to be one. For example,No prime factor belongs to all three prime supports of an integer, yet each pair shares one. By contrast, $4,9,25$ have pairwise coprimality. The distinction is presented by Poritz under the names pairwise and mutually relatively prime. ^[ref]

One use is simplifying fractions. For a fraction $a/b$ with integer entries and $b>0$, divide its numerator and denominator by $d=\gcd (a,b)$ to obtain a reduced fraction with coprime entries. For example, $42/30=7/5$ since $d=6$, and $-42/30=-7/5$ uses the same $d$. A further common divisor would make $d$ too small, so the reduction is complete. In particular, the reduced fraction for zero is $0/1$.

Another use is reversible modular arithmetic: for an integer modulus $m>1$, an integer $a$ has a modular inverse precisely when $a,m$ are coprime. The proof outline uses Bézout's identity: $ua+vm=1$ makes $u$ a modular inverse, and any inverse relation $ua-km=1$ forces every common divisor to divide one. For example, $3\cdot 5-2\cdot 7=1$ gives $5^{-1}\equiv 3(\mathrm{mod}7)$, whereas $6$ has no modular inverse modulo $10$. Thus coprimality is the exact condition allowing division by a residue in modular arithmetic. ^[ref]

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 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.

Euler's totient function is the arithmetic functionIt counts the congruence classes modulo $n$ represented by integers coprime to $n$. In particular, $\phi (1)=1$, since $\gcd (1,1)=1$. For $12$, the coprime integers among $1,\dots ,12$ are $1,5,7,11$, so $\phi (12)=4$; they form a reduced residue system. ^[ref]

Counting these congruence classes is useful because they are exactly the classes with modular multiplicative inverses. Their number supplies the exponent in Euler theorem. Möbius inversion gives the product formulawhere $p$ runs over distinct prime factors of $n$. Thus $\phi (p^a)=p^{a-1}(p-1)$ for prime powers with $a\ge 1$, and $\phi (12)=12(1-1/2)(1-1/3)=4$. The product for $n=1$ is an empty product, so it gives the same convention $\phi (1)=1$.

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 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.

The Möbius function assigns a cancellation weight to each positive integer:Thus a squarefree integer contributes a sign determined by its number of distinct prime factors; a repeated prime factor makes the weight zero. This packages the alternating signs of inclusion-exclusion into an arithmetic function. ^[ref]

For example, $\mu (2)=\mu (3)=-1$, $\mu (6)=1$, and $\mu (30)=(-1{)}^3=-1$, since $30=2\cdot 3\cdot 5$. In contrast, $\mu (4)=\mu (12)=\mu (60)=0$: all three have the square of the prime number $2$ as a divisor. On prime powers, $\mu (p)=-1$ and $\mu (p^a)=0$ for $a\ge 2$.

The Möbius function is a multiplicative function: for coprime integers, their prime factors are disjoint, so either a squared prime factor forces both sides of $\mu (mn)=\mu (m)\mu (n)$ to vanish, or the counts of distinct prime factors add and their signs multiply. It is not a completely multiplicative function, since $\mu (4)=0\ne 1=\mu (2{)}^2$. Its Möbius divisor-sum identity makes it the inverse of the constant-one arithmetic function under Dirichlet convolution: $\mu \ast 1=\epsilon$, the Identity for Dirichlet convolution. This is why the small set of weights $-1,0,1$ can undo an entire divisor sum.

Möbius inversion recovers an arithmetic function from its divisor sum. For arithmetic functions $f,g$, the following statements are equivalent, with each equality required for every positive integer $n$:Here $d$ ranges over positive integer divisors of $n$ and $\mu$ is the Möbius function. No multiplicativity assumption is needed, and both sums are finite. The right side can also be written ${\sum }_{d\mid n}\mu (n/d)g(d)$ by exchanging complementary integer divisors. ^[ref]

The divisor sum $g(n)$ bundles $f(n)$ together with values at smaller integer divisors. Möbius inversion subtracts aggregates indexed by one distinct prime factor, adds those indexed by two, and continues with the alternating signs of inclusion-exclusion. A divisor containing a squared prime factor receives weight zero. The Möbius divisor-sum identity ensures that every unwanted contribution cancels exactly.

In Dirichlet convolution notation, $g=f\ast 1$ and $\mu \ast 1=\epsilon$, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. Associativity and commutativity give the short proof:Convolving $f=\mu \ast g$ with $1$ proves the reverse implication. At $n=1$, the formula reduces to $f(1)=g(1)$ because $\mu (1)=1$.

For a worked example, take $f(n)=\phi (n)$, Euler's totient function. The totient divisor-sum identity says that $g(n)=n$, the identity arithmetic function:This identity comes from classifying the integers $1,\dots ,n$ by their greatest common divisor with $n$: the group with $\gcd (j,n)=h$ has $\phi (n/h)$ members, since $j/h$ and $n/h$ are coprime integers. Summing the group sizes gives $n$, as detailed under totient divisor-sum identity.

Now undo this aggregation with the Möbius function:The nonzero terms use $d=1,2,3,6$; $d=4,12$ vanish because they contain the square of the prime number $2$. The four surviving integers coprime to $12$ are $1,5,7,11$.

This example also reveals why Möbius inversion is useful: it turns an easy count of all objects into a count satisfying a coprimality restriction. Expanding over subsets of distinct prime factors givesThe product is inclusion-exclusion over multiples of the prime factors. For $12$, it gives $12(1-1/2)(1-1/3)=4$. More generally, whenever a counting problem produces a divisor sum, Möbius inversion lets that aggregate be reversed without solving a new system of equations.

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.

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 prime omega function $\omega (n)$ counts the distinct prime factors of a positive integer $n$, so it is the cardinality of the prime support of an integer $P(n)$. For a prime factorization with distinct prime numbers $p_1,\dots ,p_r$ and positive exponents,At $n=1$ there are no prime factors, so the empty sum gives $\omega (1)=0$. The total number of prime factors instead gives $\Omega (n)={\sum }_i{a}_i$. ^[ref]

For $72=2^3\cdot 3^2$, the prime support of an integer is $\{2,3\}$ and $\omega (72)=2$. The comparison examples give $\omega (30)=3$, $\omega (32)=1$ and $\omega (36)=2$. Raising an existing prime factor to a higher power does not affect the prime omega function: $\omega (p^k)=1$ for a prime number $p$ and $k\ge 1$. Thus the prime omega function is a strongly additive arithmetic function, withThe reason is that coprime integers have disjoint prime supports of an integer. For arbitrary positive integers, their prime support of an integer combines by set union, and the shared prime factors are exactly those of the greatest common divisor. Counting this set union gives the more general identityFor $a=12$, $b=18$, both prime supports of an integer are $\{2,3\}$ and $\gcd (a,b)=6$, so $\omega (216)=2+2-2=2$. In particular $\omega (4)=1\ne 2=\omega (2)+\omega (2)$: the prime omega function is not a completely additive arithmetic function.

The prime omega function is useful for counting choices involving distinct prime factors. Every squarefree divisor of $n$ chooses a subset of its prime support of an integer, so there are $2^{\omega (n)}$ squarefree divisors. For $72$, this gives $2^2=4$, namely $1,2,3,6$. Also $\omega (\mathrm{rad}(n))=\omega (n)$, because the radical of an integer keeps exactly that prime support of an integer. The total number of prime factors records what this count discards, and the two counts agree exactly on squarefree integers.

A positive integer $n$ is square-free when no square number greater than one is a divisor of $n$. It suffices to exclude squares of prime numbers; equivalently, every prime valuation is zero or one:Thus the prime factorization of a square-free integer is a product of distinct prime factors, with no repetitions. One is included: its prime factorization is an empty product and all its prime valuations are zero. These equivalent tests are stated in Grinberg's notes. ^[ref]

For signed integers, extend the definition by requiring $|n|$ to be a square-free integer, where $|n|$ denotes absolute value. Hence $-30$ and both $1,-1$ qualify. Zero does not: every prime number has its square as a divisor of zero. The Möbius function and radical of an integer below use their existing positive integer domains, so apply them to $|n|$ when handling a negative integer.

For example, $30=2\cdot 3\cdot 5$ is a square-free integer, whereas $12=2^2\cdot 3$ is not, because $4\mid 12$ and $v_2(12)=2$. Being a composite number causes no problem: $6,10,15$ are all square-free integers. A square-free integer need not be coprime to another integer: $6$ and $10$ are both square-free integers but have greatest common divisor two. In fact the only signed integers coprime to every integer are $1,-1$, since $\gcd (n,n)=|n|$.

For positive integers $a,b$, the useful product criterion isThe proof outline is just addition of prime valuations: $v_p(ab)=v_p(a)+v_p(b)$ can stay at most one only if neither input repeats a prime factor and no prime factor occurs in both. Thus $6\cdot 5=30$ is a square-free integer, but $6\cdot 10=60$ is not. An equivalent characterization is that every factorization $n=ab$ into positive integers has coprime factors. If $p^2\mid n$, the factorization $n=p(n/p)$ supplies two factors sharing $p$; conversely, a shared prime factor forces its square to divide $n$. For $30$, the unordered factor pairs $(1,30),(2,15),(3,10),(5,6)$ all have greatest common divisor one; for $12$, the pair $(2,6)$ fails.

Two existing arithmetic functions turn the definition into practical tests:The radical of an integer keeps one copy of each prime factor, so it changes precisely the integers with a repeated prime factor. Its definition is given in Evertse's exercise sheet. ^[ref] The Möbius function assigns $(-1{)}^r$ to a square-free integer with $r$ distinct prime factors and zero otherwise; squaring forgets the sign. Consequently ${\sum }_{n=1}^{60}\mu (n{)}^2=37$ counts exactly the green cells. This connects a restriction on repeated prime factors with weighted counting and inclusion-exclusion. ^[ref]

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 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 divisor sum of Euler's totient function is the identity arithmetic function:Equivalently, $\phi \ast 1=\mathrm{id}$ under Dirichlet convolution, with $1$ the constant-one arithmetic function. ^[ref]

To see why, group the integers $j=1,\dots ,n$ by $h=\gcd (j,n)$. Each $h$ is a positive integer divisor of $n$. Within its group, $j=ha$ with $1\le a\le n/h$ and $\gcd (a,n/h)=1$, by the scaling identity for greatest common divisors. Dividing by $h$ gives a bijection with the integers counted by $\phi (n/h)$. The groups form a partition of a set, soFor $n=12$, the groups with greatest common divisor $h=1,2,3,4,6,12$ are respectively $\{1,5,7,11\},\{2,10\},\{3,9\},\{4,8\},\{6\},\{12\}$. Their sizes are $4,2,2,2,1,1$, totaling $12$. Möbius inversion reverses this divisor sum to recover Euler's totient function from $g(n)=n$.