Divisor sum ($F(n)={\sum }_{d\mid n}f(d)$)

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 $6$.