# Difference between revisions of "Sum of divisors function"

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− | If <math> | + | 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 09:06, 22 June 2006

If is the prime factorization of , then the sum of all divisors of is given by the formula .

In fact, if you use the formula , then the above formula is equivalent to

## 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.