Editing Divisor function (section)

===Other properties and identities===
[[Euler]] proved the remarkable recurrence:<ref>{{Cite arXiv |eprint = math/0411587|last1 = Euler|first1 = Leonhard|title = An observation on the sums of divisors|last2 = Bell|first2 = Jordan|year = 2004}}</ref><ref>https://scholarlycommons.pacific.edu/euler-works/175/, ''Découverte d'une loi tout extraordinaire des nombres par rapport à la somme de leurs diviseurs''</ref><ref>https://scholarlycommons.pacific.edu/euler-works/542/, ''De mirabilis proprietatibus numerorum pentagonalium''</ref>

:<math>\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}</math>

where <math>\sigma_1(0)=n</math> if it occurs and <math>\sigma_1(x)=0</math> for <math>x < 0</math>, and <math>\tfrac{1}{2} \left( 3i^2 \mp i \right)</math> are consecutive pairs of generalized [[pentagonal numbers]] ({{OEIS2C|A001318}}, 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 <math>\sigma_{0}(n)</math> is even; for a square integer, one divisor (namely <math>\sqrt n</math>) is not paired with a distinct divisor and <math>\sigma_{0}(n)</math> is odd. Similarly, the number <math>\sigma_{1}(n)</math> is odd if and only if ''n'' is a square or twice a square.{{sfnp|Gioia|Vaidya|1967}}

We also note ''s''(''n'') = ''σ''(''n'')&nbsp;−&nbsp;''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 number]]s, which are the ''n'' such that ''s''(''n'') =&nbsp;''n''. If ''s''(''n'') > ''n'', then ''n'' is an [[abundant number]], and if ''s''(''n'') < ''n'', then ''n'' is a [[deficient number]].

If {{mvar|n}} is a power of 2, <math>n = 2^k</math>, then <math>\sigma(n) = 2 \cdot 2^k - 1 = 2n - 1</math> and <math>s(n) = n - 1</math>, which makes ''n'' [[Almost perfect number|almost-perfect]].

As an example, for two primes <math>p,q:p<q</math>, let

:<math>n = p\,q</math>.

Then

:<math>\sigma(n)  = (p+1)(q+1) = n + 1 + (p+q), </math>
:<math>\varphi(n) = (p-1)(q-1) = n + 1 - (p+q), </math>

and

:<math>n + 1 = (\sigma(n) + \varphi(n))/2, </math>
:<math>p + q = (\sigma(n) - \varphi(n))/2, </math>

where <math>\varphi(n)</math> is [[Euler phi|Euler's totient function]].

Then, the roots of

:<math>(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 </math>

express ''p'' and ''q'' in terms of ''σ''(''n'') and ''φ''(''n'') only, requiring no knowledge of ''n'' or <math>p+q</math>, as

:<math>p = (\sigma(n) - \varphi(n))/4 - \sqrt{[(\sigma(n) - \varphi(n))/4]^2 - [(\sigma(n) + \varphi(n))/2 - 1]}, </math>
:<math>q = (\sigma(n) - \varphi(n))/4 + \sqrt{[(\sigma(n) - \varphi(n))/4]^2 - [(\sigma(n) + \varphi(n))/2 - 1]}. </math>

Also, knowing {{mvar|n}} and either <math>\sigma(n)</math> or <math>\varphi(n)</math>, or, alternatively, <math>p+q</math> and either <math>\sigma(n)</math> or <math>\varphi(n)</math> allows an easy recovery of ''p'' and ''q''.

In 1984, [[Roger Heath-Brown]] proved that the equality

:<math>\sigma_0(n) = \sigma_0(n + 1)</math>

is true for infinitely many values of {{mvar|n}}, see {{OEIS2C|A005237}}.