Difference between revisions of "Sum of divisors function"

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If <math>k=p_1^{\alpha_1}\cdot\dots\cdot p_n^{\alpha_n}</math> is the [[prime factorization]] of <math>\displaystyle{k}</math>, then the sum of all divisors of <math>k</math> is given by the formula <math>s=(p_1^0+p_1^1+...+p_1^{\alpha_1})(p_2^0+p_2^1+...+p_2^{\alpha_2})\cdot\dots\cdot (p_n^0+p_n^1+...+p_n^{\alpha_n})</math>.
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If <math>p_1^{\alpha_1}\cdot\dots\cdot p_n^{\alpha_n}</math> is the [[prime factorization]] of <math>\displaystyle{k}</math>, then the sum of all divisors of <math>k</math> is given by the formula <math>s=(p_1^0+p_1^1+...+p_1^{\alpha_1})(p_2^0+p_2^1+...+p_2^{\alpha_2})\cdot\dots\cdot (p_n^0+p_n^1+...+p_n^{\alpha_n})</math>.
  
 
In fact, if you use the formula <math>1+q+q^2+\ldots+q_n = \frac{q^{n+1}-1}{q-1}</math>, then the above formula is equivalent to
 
In fact, if you use the formula <math>1+q+q^2+\ldots+q_n = \frac{q^{n+1}-1}{q-1}</math>, then the above formula is equivalent to

Revision as of 08:06, 22 June 2006

If $p_1^{\alpha_1}\cdot\dots\cdot p_n^{\alpha_n}$ is the prime factorization of $\displaystyle{k}$, then the sum of all divisors of $k$ is given by the formula $s=(p_1^0+p_1^1+...+p_1^{\alpha_1})(p_2^0+p_2^1+...+p_2^{\alpha_2})\cdot\dots\cdot (p_n^0+p_n^1+...+p_n^{\alpha_n})$.

In fact, if you use the formula $1+q+q^2+\ldots+q_n = \frac{q^{n+1}-1}{q-1}$, then the above formula is equivalent to

$s = \displaystyle\left(\frac{p_1^{\alpha_1+1}-1}{p_1-1}\right)\left(\frac{p_2^{\alpha_2+1}-1}{p_2-1}\right)\ldots\left(\frac{p_n^{\alpha_n+1}-1}{p_n-1}\right)$

Derivation

If you expand the monomial into a polynomial you see that it comes to be the addition of all possible combinations of the multiplication of the prime factors, and so all the divisors.

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