* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download PDF
Survey
Document related concepts
Law of large numbers wikipedia , lookup
List of important publications in mathematics wikipedia , lookup
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Hyperreal number wikipedia , lookup
Non-standard analysis wikipedia , lookup
Elementary mathematics wikipedia , lookup
Factorization wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Large numbers wikipedia , lookup
Fundamental theorem of calculus wikipedia , lookup
Transcript
determination of abundant numbers with specified prime factors∗ rspuzio† 2013-03-21 22:13:34 The formula for sums of factors may be used to find all abundant numbers with a specified set of prime factors or that no such numbers exist. To accomplish this, we first do a little algebraic manipulation to our formula. Qk 1 Theorem 1. A number n whose factorization into prime numbers is i=1 pm i is abundant if and only if k Y 1 − p−mi −1 i > 2. 1 − p−1 i i=1 Proof. By definition n is abundant, if the sum of the proper divisors of n is greater than n. Using our formula, this is equivalent to the condition k mi +1 Y p −1 i pi − 1 i=1 >2 k Y i pm i . i=1 Dividing the k-th term in the product on the left-hand side by the k-th term on the right-hand side, i i +1 i −1 −1 pi−mi −1 pi − p−m 1 − p−m 1 pm i i i , = = i pm pi − 1 pi − 1 p−1 1 − p−1 i i i so the condition becomes k Y 1 − p−mi −1 i i=1 1 − p−1 i >2 ∗ hDeterminationOfAbundantNumbersWithSpecifiedPrimeFactorsi created: h2013-03-21i by: hrspuzioi version: h39028i Privacy setting: h1i hTheoremi h11A05i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Note that each of the terms in the product is bigger than 1. Furthemore, the k-th term is bounded by 1 pi −1 = p − 1 . 1 − pi i This means that it is only possible to have an abundant number whose prime factors are {pi | 1 ≤ i ≤ k} if k Y pi > 2. pi − 1 i=1 As it turns out, the convers also holds, so we have a nice criterion for determining when a set of prime numbers happens to be the set of prime divisors of an abundant number. Theorem 2. A finite set S of prime numbers is the set of prime divisors of an abundant number if and only if Y p > 2. p−1 p∈S Proof. As described above, if S is a set of prime factors of an abundant number, then we may bound each term in the inequality of the previous theorem to obtain the inequality in the current theorem. Assume, then that S is a finite set of prime numbers which satisfies said inequality. Then, by continuity, there must exist a real number > 0 such that Y p −x >2 p−1 p∈S whenver 0 < x < . Since limk→∞ n−k = 0 when n > 1, we can, for every p ∈ S, find an m(p) such that pm(p) < . p−1 Hence, Y Y p pm(p) 1 − p−m(p)−1 − = >2 p−1 p−1 1 − p−1 p∈S so, by the previous theorem, p∈S Q p∈S pm(p) is an abundant number. 2