Formulas at prime powers For a
prime number p, :\begin{align} \sigma_0(p) & = 2 \\ \sigma_0(p^n) & = n+1 \\ \sigma_1(p) & = p+1 \end{align} because by definition, the factors of a prime number are 1 and itself. Also, where
pn# denotes the
primorial, : \sigma_0(p_n\#) = 2^n since
n prime factors allow a sequence of binary selection (p_{i} or 1) from
n terms for each proper divisor formed. However, these are not in general the smallest numbers whose number of divisors is a
power of two; instead, the smallest such number may be obtained by multiplying together the first
n Fermi–Dirac primes, prime powers whose exponent is a power of two. Clearly, 1 for all n > 2, and \sigma_x(n) > n for all n > 1, x > 0 . The divisor function is
multiplicative (since each divisor
c of the product
mn with \gcd(m, n) = 1 distinctively correspond to a divisor
a of
m and a divisor
b of
n), but not
completely multiplicative: :\gcd(a, b)=1 \Longrightarrow \sigma_x(ab)=\sigma_x(a)\sigma_x(b). The consequence of this is that, if we write :n = \prod_{i=1}^r p_i^{a_i} where
r =
ω(
n) is the
number of distinct prime factors of
n,
pi is the
ith prime factor, and
ai is the maximum power of
pi by which
n is
divisible, then we have: :\sigma_x(n) = \prod_{i=1}^r \sum_{j=0}^{a_i} p_i^{j x} = \prod_{i=1}^r \left (1 + p_i^x + p_i^{2x} + \cdots + p_i^{a_i x} \right ). which, when
x ≠ 0, is equivalent to the useful formula: :\sigma_x(n) = \prod_{i=1}^{r} \frac{p_{i}^{(a_{i}+1)x}-1}{p_{i}^x-1}. When
x = 0, \sigma_0(n) is: :\sigma_0(n)=\prod_{i=1}^r (a_i+1). This result can be directly deduced from the fact that all divisors of n are uniquely determined by the distinct tuples (x_1, x_2, ..., x_i, ..., x_r) of integers with 0 \le x_i \le a_i (i.e. a_i+1 independent choices for each x_i). For example, if
n is 24, there are two prime factors (
p1 is 2;
p2 is 3); noting that 24 is the product of 23×31,
a1 is 3 and
a2 is 1. Thus we can calculate \sigma_0(24) as so: : \sigma_0(24) = \prod_{i=1}^{2} (a_i+1) = (3 + 1)(1 + 1) = 4 \cdot 2 = 8. The eight divisors counted by this formula are 1, 2, 4, 8, 3, 6, 12, and 24.
Other properties and identities Euler proved the remarkable recurrence: :\begin{align} \sigma_1(n) &= \sigma_1(n-1)+\sigma_1(n-2)-\sigma_1(n-5)-\sigma_1(n-7)+\sigma_1(n-12)+\sigma_1(n-15)+ \cdots \\[12mu] &= \sum_{i\in\N} (-1)^{i+1}\left( \sigma_1 \left( n-\frac{1}{2} \left( 3i^2-i \right) \right) + \sigma_1 \left( n-\frac{1}{2} \left( 3i^2+i \right) \right) \right), \end{align} where \sigma_1(0)=n if it occurs and \sigma_1(x)=0 for x , and \tfrac{1}{2} \left( 3i^2 \mp i \right) are consecutive pairs of generalized
pentagonal numbers (, starting at offset 1). Indeed, Euler proved this by logarithmic differentiation of the identity in his
pentagonal number theorem. For a non-square integer,
n, every divisor,
d, of
n is paired with divisor
n/
d of
n and \sigma_{0}(n) is even; for a square integer, one divisor (namely \sqrt n) is not paired with a distinct divisor and \sigma_{0}(n) is odd. Similarly, the number \sigma_{1}(n) is odd if and only if
n is a square or twice a square. We also note
s(
n) =
σ(
n) −
n. Here
s(
n) denotes the sum of the
proper divisors of
n, that is, the divisors of
n excluding
n itself. This function is used to recognize
perfect numbers, which are the
n such that
s(
n) =
n. If
s(
n) >
n, then
n is an
abundant number, and if
s(
n) n = 2^k, then \sigma(n) = 2 \cdot 2^k - 1 = 2n - 1 and s(n) = n - 1, which makes
n almost-perfect. As an example, for two primes p,q:p, let :n = p\,q. Then :\sigma(n) = (p+1)(q+1) = n + 1 + (p+q), :\varphi(n) = (p-1)(q-1) = n + 1 - (p+q), and :n + 1 = (\sigma(n) + \varphi(n))/2, :p + q = (\sigma(n) - \varphi(n))/2, where \varphi(n) is
Euler's totient function. Then, the roots of :(x-p)(x-q) = x^2 - (p+q)x + n = x^2 - [(\sigma(n) - \varphi(n))/2]x + [(\sigma(n) + \varphi(n))/2 - 1] = 0 express
p and
q in terms of
σ(
n) and
φ(
n) only, requiring no knowledge of
n or p+q, as :p = (\sigma(n) - \varphi(n))/4 - \sqrt{[(\sigma(n) - \varphi(n))/4]^2 - [(\sigma(n) + \varphi(n))/2 - 1]}, :q = (\sigma(n) - \varphi(n))/4 + \sqrt{[(\sigma(n) - \varphi(n))/4]^2 - [(\sigma(n) + \varphi(n))/2 - 1]}. Also, knowing and either \sigma(n) or \varphi(n), or, alternatively, p+q and either \sigma(n) or \varphi(n) allows an easy recovery of
p and
q. In 1984,
Roger Heath-Brown proved that the equality :\sigma_0(n) = \sigma_0(n + 1) is true for infinitely many values of , see .
Dirichlet convolutions By definition:\sigma = \operatorname{Id} * \mathbf 1By
Möbius inversion:\operatorname{Id} = \sigma * \mu ==Series relations==