Divisional sum

The divisional sum of a natural number is the sum of all the divisors of this number, including the number itself. ${\displaystyle \sigma }$

Example:

The number 6 has the divisors 1, 2, 3 and 6. So the divisional sum for 6 is . ${\displaystyle 1+2+3+6=12}$

Partial sums play a role in many problems in number theory, e.g. B. with perfect numbers and friendly numbers .

Definitions

Definition 1: sum of all factors

Let all be divisors of the natural number , then we call the divisional sum of . There are 1 and itself divisors, i.e. included in the set of factors. The function is called the divisor sum function and is a number theoretic function . ${\displaystyle t_{1},t_{2},...,t_{k}}$ ${\displaystyle n}$ ${\displaystyle \sigma (n)=t_{1}+t_{2}+\dotsb +t_{k}}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle \sigma }$

The example above can now be written like this:

${\displaystyle \sigma (6)=1+2+3+6=12}$

Definition 2: Sum of the real factors

The sum of the real divisors of the natural number is the sum of the divisors of without the number itself and we denote this sum with . ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle \sigma ^{*}(n)}$

Example:

${\displaystyle \sigma ^{*}(6)=1+2+3=6}$

Obviously the relationship applies:

${\displaystyle \sigma (n)-n=\sigma ^{*}(n)}$

Definition 3: deficient, abundant, perfect

A natural number is called ${\displaystyle n>1}$

deficient or splitter arm when , ${\displaystyle \sigma ^{*}(n)<n}$

abundant or partial , if , ${\displaystyle \sigma ^{*}(n)>n}$

perfect if . ${\displaystyle \sigma ^{*}(n)=n}$

Examples:

${\displaystyle \sigma ^{*}(6)=1+2+3=6}$, d. H. 6 is a perfect number.

${\displaystyle \sigma ^{*}(12)=1+2+3+4+6=16>12}$, d. H. 12 is abundant.

${\displaystyle \sigma ^{*}(10)=1+2+5=8<10}$, d. H. 10 is deficient.

Properties of the divisor sum

Theorem 1: divisional sum of a prime number

Be a prime number. Then: ${\displaystyle n}$

${\displaystyle \sigma (n)=n+1}$

Proof: Since is a prime, 1 and are the only factors. Hence the claim follows. ${\displaystyle n}$ ${\displaystyle n}$

Theorem 2: Divisional sum of the power of a prime number

Be a prime number. Then: ${\displaystyle n}$

${\displaystyle \sigma (n^{k})={\frac {n^{k+1}-1}{n-1}}}$

Proof: Since a prime number, has only the following division: . The sum is a geometric series . The assertion immediately follows from the empirical formula for a geometric series. ${\displaystyle n}$ ${\displaystyle n^{k}}$ ${\displaystyle n^{0},n^{1},\ldots ,n^{k}}$

Example:

${\displaystyle \sigma (2^{3})={{2^{4}-1} \over {2-1}}={{16-1} \over 1}=15}$

${\displaystyle \sigma (8)=1+2+4+8=15}$

Theorem 3: Divisional sum of the product of two prime numbers

Let and be different prime numbers. Then: ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle \sigma (a\cdot b)=\sigma (a)\cdot \sigma (b)}$

Proof: The number has four different divider 1 , and . It follows: ${\displaystyle ab}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle ab}$

${\displaystyle \sigma (a\cdot b)=1+a+b+ab=(a+1)(b+1)=\sigma (a)\cdot \sigma (b)}$

Example:

${\displaystyle \sigma (3\cdot 5)=\sigma (15)=1+3+5+15=24}$

${\displaystyle \sigma (3)\cdot \sigma (5)=(1+3)\cdot (1+5)=4\cdot 6=24}$

Sentence 4: Generalization of sentence 2 and sentence 3

Let be different prime numbers and natural numbers. Further be . Then: ${\displaystyle p_{1},p_{2},...,p_{r}}$ ${\displaystyle k_{1},k_{2},...,k_{r}}$ ${\displaystyle n=p_{1}^{k_{1}}\cdot p_{2}^{k_{2}}\cdot \ldots \cdot p_{r}^{k_{r}}}$

${\displaystyle \sigma (n)={\frac {p_{1}^{k_{1}+1}-1}{p_{1}-1}}\cdot \ldots \cdot {\frac {p_{r}^{k_{r}+1}-1}{p_{r}-1}}}$

Thabit's theorem

With the help of Theorem 4 one can prove Thabit's theorem from the field of friendly numbers. The sentence is:

For a fixed natural number let and . ${\displaystyle n}$ ${\displaystyle x=3\cdot 2^{n}-1,y=3\cdot 2^{n-1}-1}$ ${\displaystyle z=9\cdot 2^{2n-1}-1}$

If , and prime numbers are greater than 2, then the two numbers and are friends; H. and . ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle z}$ ${\displaystyle a=2^{n}xy}$ ${\displaystyle b=2^{n}z}$ ${\displaystyle \sigma ^{*}(a)=b}$ ${\displaystyle \sigma ^{*}(b)=a}$

proof

${\displaystyle {\begin{aligned}\sigma ^{*}(a)&=\sigma (a)-a\\&=\sigma (2^{n}\cdot x\cdot y)-a\\&=(2^{n+1}-1)(x+1)(y+1)-a\qquad \qquad \qquad \qquad \qquad \qquad &&{\text{(Satz 4)}}\\&=(2^{n+1}-1)(3\cdot 2^{n})(3\cdot 2^{n-1})-2^{n}(3\cdot 2^{n}-1)(3\cdot 2^{n-1}-1)\\&=(2^{n+1}-1)\cdot 9\cdot 2^{2n-1}-2^{n}(9\cdot 2^{2n-1}-6\cdot 2^{n-1}-3\cdot 2^{n-1}+1)\\&=2\cdot 2^{n}\cdot 9\cdot 2^{2n-1}-9\cdot 2^{n}\cdot 2^{n-1}-2^{n}(9\cdot 2^{2n-1}-9\cdot 2^{n-1}+1)\\&=2^{n}(18\cdot 2^{2n-1}-9\cdot 2^{n-1}-9\cdot 2^{2n-1}+9\cdot 2^{n-1}-1)\\&=2^{n}(9\cdot 2^{2n-1}-1)\\&=2^{n}\cdot z\\&=b\end{aligned}}}$

One shows analogously . ${\displaystyle \sigma ^{*}(b)=a}$

Divisional sum as a finite series

For every natural number the divisor function can be represented as a series without explicit reference to the divisibility properties of : ${\displaystyle n}$ ${\displaystyle n}$

${\displaystyle \sigma (n)=\sum _{\mu =1}^{n}\sum _{\nu =1}^{\mu }\cos {2\pi {\frac {\nu n}{\mu }}}}$

Proof: The function

${\displaystyle T(n,\mu )={\frac {1}{\mu }}\sum _{\nu =1}^{\mu }\cos 2\pi {\frac {\nu n}{\mu }},\quad n=1,2,\dots ,\quad \mu =1,2,\dots }$

becomes 1 if is a factor of , otherwise it remains zero. First of all, ${\displaystyle \mu }$ ${\displaystyle n}$

${\displaystyle T(n,\mu )={\frac {1}{\mu }}\lim _{x\to n}\sum _{\nu =1}^{\mu }\cos 2\pi {\frac {\nu x}{\mu }}=\lim _{x\to n}{\frac {1}{2\mu }}\left(\sin 2\pi x\cot {\frac {\pi x}{\mu }}-1+\cos 2\pi x\right)=\lim _{x\to n}{\frac {\sin 2\pi x\cos {\frac {\pi x}{\mu }}}{2\mu \sin {\frac {\pi x}{\mu }}}}}$

The counter in the last expression always goes to zero when goes. The denominator can only become zero if is a divisor of . But then ${\displaystyle x\to n}$ ${\displaystyle \mu }$ ${\displaystyle n}$

${\displaystyle \lim _{x\to n}{\frac {\sin 2\pi x\cos {\frac {\pi x}{\mu }}}{2\sin {\frac {\pi x}{\mu }}}}=\cos {\frac {\pi n}{\mu }}\lim _{x\to n}{\frac {\sin 2\pi x}{2\sin {\frac {\pi x}{\mu }}}}\;=\;\cos {\frac {\pi n}{\mu }}\lim _{x\to n}{\frac {2\pi \cos 2\pi x}{2{\frac {\pi }{\mu }}\cos {\frac {\pi x}{\mu }}}}\;=\;\mu }$

Only in this case is it asserted as above. ${\displaystyle T(n,\mu )=1}$

If you now multiply by and add the product over all values to , then a contribution to the sum is only made if is a factor of . But that is exactly the definition of the general divisor function ${\displaystyle T(n,\mu )}$ ${\displaystyle \mu ^{k}}$ ${\displaystyle \mu =1}$ ${\displaystyle \mu =n}$ ${\displaystyle \mu ^{k}}$ ${\displaystyle \mu }$ ${\displaystyle n}$

${\displaystyle \sigma _{k}(n)=\sum _{\mu =1}^{n}\mu ^{k-1}\sum _{\nu =1}^{\mu }\cos {2\pi {\frac {\nu n}{\mu }}},\quad k=0,\pm 1,\dots }$

whose special case is the simple divisional sum . ${\displaystyle k=1}$ ${\displaystyle \sigma (n)}$

See also

• Content chain
• Partial set

literature

• Paul Erdős , János Surányi : Topics in the Theory of Numbers. (=  Undergraduate Texts in Mathematics ). 2nd Edition. Springer Verlag, New York, NY (et al.) 2003, ISBN 0-387-95320-5 ( MR1950084 - translated from the Hungarian by Barry Guiduli ).
• József Sándor, Dragoslav S. Mitrinović, Borislav Crstici: Handbook of Number Theory. I . Springer Verlag, Dordrecht 2006, ISBN 1-4020-4215-9 ( MR2186914 ).
• József Sándor, Borislav Crstici: Handbook of Number Theory. II . Kluwer Academic Publishers, Dordrecht / Boston / London 2004, ISBN 1-4020-2546-7 ( MR2119686 ).
• Wacław Sierpiński : Elementary Theory of Numbers (=  North-Holland Mathematical Library . Volume 31 ). 2nd revised and expanded edition. North-Holland, Amsterdam / New York 1988, ISBN 0-444-86662-0 ( MR0930670 ).