Difference between revisions of "Sum of divisors function"
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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 | ||
| − | <math> 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)</math> | + | <math> 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)</math>. |
== Derivation == | == 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. | + | 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. |
Revision as of 12:00, 28 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.
