Summation with Number theory

by fungarwai, May 23, 2019, 12:59 PM

Summation of divisor

$\sum_{1\le d\le n\atop d|n} d^x=\prod_{i=1}^m \sum_{j=0}^{k_i} p_i^{jx}=
\begin{cases} \prod_{i=1}^m (1+k_i) & x=0\\
\prod_{i=1}^m \frac{p_i^{(k_i+1)x}-1}{p_i^x-1} & x>0\end{cases}$

Summation of coprime set

Refer Page 25 from Introduction to Analytic Number Theory (A.J. Hildebrand)

$\displaystyle\sum_{1\le m\le n\atop (m,n)=1}f(m)
=\sum_{1\le m\le n}f(m)e((m,n))
=\sum_{1\le m\le n}f(m)\sum_{d|(m,n)}\mu(d)$
$\displaystyle
=\sum_{d|n}\mu(d)\sum_{1\le m\le n\atop d|m}f(m)
=\sum_{d|n}\mu(d)\sum_{1\le k\le \frac{n}{d}}f(dk)$

where $e(n)=\begin{cases}1 & n=1\\0 & \text{otherwise}\end{cases}$

$\varphi(n)=\sum_{1\le m\le n\atop (m,n)=1}1=n\prod_{p|n}\left(1-\frac{1}{p}\right)$

Proof

$\sum_{1\le m\le n\atop (m,n)=1}m=\frac{n}{2}\left(e(n)+\varphi(n)\right)$

Proof

$\displaystyle\sum_{1\le m\le n\atop (m,n)=1}m^2
=\frac{n}{6}\prod_{p|n}(1-p)+\frac{n^2}{2}e(n)+\frac{n^2}{3}\varphi(n)$

Proof

Summation of GCD or LCM

$\sum_{n=1}^N (n,N)=\prod_{i=1}^m p_i^{k_i-1}(p_i+k_i(p_i-1))$

Proof

Example

$\sum_{n=1}^N [n,N]=\frac{N}{2}\left(1+\prod_{i=1}^m\frac{p_i^{2k_i+1}+1}{p_i+1}\right)$

Proof

Example
This post has been edited 4 times. Last edited by fungarwai, Apr 22, 2024, 11:18 AM

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Notable algebra methods with proofs and examples

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