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... first few even abundant numbers. To keep things more managable, we shall take advantage of the fact that a multiple of an abundant number is abundant and only look for abundant numbers none of whose proper divisors are abundant. Once we know these numbers, it becomes a rather easy matter to find the ...
... first few even abundant numbers. To keep things more managable, we shall take advantage of the fact that a multiple of an abundant number is abundant and only look for abundant numbers none of whose proper divisors are abundant. Once we know these numbers, it becomes a rather easy matter to find the ...
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... Our motivation for this problem arose from counting certain finite topologies as described below. If j is any point in a finite topological space, let N (j) be the intersection of all open sets containing j. Corollary. Let T be the set of topologies τ on n such that the basis {N (j) : j ∈ n} consist ...
... Our motivation for this problem arose from counting certain finite topologies as described below. If j is any point in a finite topological space, let N (j) be the intersection of all open sets containing j. Corollary. Let T be the set of topologies τ on n such that the basis {N (j) : j ∈ n} consist ...
2.1 Use Inductive Reasoning
... Conjecture You can connect five noncollinear points ________, or _____ different ways. ...
... Conjecture You can connect five noncollinear points ________, or _____ different ways. ...
A RIGOROUS TIME BOUND FOR FACTORING INTEGERS For real
... Unfortunately we are not able to prove that the elliptic curve method can recognize all smooth numbers efficiently. For this reason we introduce the notion of a recognizable smooth number. A result from [28] shows that not only do recognizable smooth numbers have a good probability of being recogniz ...
... Unfortunately we are not able to prove that the elliptic curve method can recognize all smooth numbers efficiently. For this reason we introduce the notion of a recognizable smooth number. A result from [28] shows that not only do recognizable smooth numbers have a good probability of being recogniz ...
Orders of Growth - UConn Math
... Here a > 1, b > 1, and r > 0 (not just r > 1!). All sequences here tend to ∞ as n → ∞, but the rates of growth are all different: any sequence which comes to the left of another sequence on this list grows at a substantially smaller rate, in the sense that the ratio tends to 0. For example, can we f ...
... Here a > 1, b > 1, and r > 0 (not just r > 1!). All sequences here tend to ∞ as n → ∞, but the rates of growth are all different: any sequence which comes to the left of another sequence on this list grows at a substantially smaller rate, in the sense that the ratio tends to 0. For example, can we f ...
Some properties of the space of fuzzy
... with the metric d∞ . In Section 3, we investigate some properties of fuzzy-valued continuous functions defined on a compact subset K and prove that the space C(K , E1 ) of fuzzy-valued continuous functions on K is complete in the supremum metric D. In Section 4, we present a characterization of comp ...
... with the metric d∞ . In Section 3, we investigate some properties of fuzzy-valued continuous functions defined on a compact subset K and prove that the space C(K , E1 ) of fuzzy-valued continuous functions on K is complete in the supremum metric D. In Section 4, we present a characterization of comp ...
Overview Background / Context
... We spent time getting a feel for how big symmetric cipher\ keys needed to be ➔ How big do keys in a public key system need to be? ...
... We spent time getting a feel for how big symmetric cipher\ keys needed to be ➔ How big do keys in a public key system need to be? ...
Full text
... Theorem 1: (Brison) Let p ≥ 5 be a prime number. A Φ2 -sequence (an )n is a complete Fibonacci sequence if and only if an = bn for all n, where b is a Fibonacci primitive root. The new results of this paper concern the case κ = 3. Because of the specific recurrence satisfied by Φ3 -sequences (an+3 = ...
... Theorem 1: (Brison) Let p ≥ 5 be a prime number. A Φ2 -sequence (an )n is a complete Fibonacci sequence if and only if an = bn for all n, where b is a Fibonacci primitive root. The new results of this paper concern the case κ = 3. Because of the specific recurrence satisfied by Φ3 -sequences (an+3 = ...
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... positive integers. The second row contains all the fractions with denominator equal to 2. The third row contains all the fractions with denominator equal to 3, etc. ...
... positive integers. The second row contains all the fractions with denominator equal to 2. The third row contains all the fractions with denominator equal to 3, etc. ...
(Vertex) Colorings
... Critical graphs One way to prove that G can not be properly colored with k − 1 colors is to find a subgraph H of G that requires k colors. How small can this subgraph be? Definition: A graph G is called critical if for every proper subgraph H G , then χ(H) < χ(G ). Theorem 2.1.2: Every graph G cont ...
... Critical graphs One way to prove that G can not be properly colored with k − 1 colors is to find a subgraph H of G that requires k colors. How small can this subgraph be? Definition: A graph G is called critical if for every proper subgraph H G , then χ(H) < χ(G ). Theorem 2.1.2: Every graph G cont ...
The Impossibility of Certain Types of Carmichael Numbers
... noted that instead of merely searching for a number which fit the original Korselt’s criterion, one should search for an integer L with a sizeable number of factors. If enough of these factors were of the form p−1 for a prime p, one could show that some product p1 ...pn of these primes was ≡ 1 (mod ...
... noted that instead of merely searching for a number which fit the original Korselt’s criterion, one should search for an integer L with a sizeable number of factors. If enough of these factors were of the form p−1 for a prime p, one could show that some product p1 ...pn of these primes was ≡ 1 (mod ...
The Right-Angled Triangle - Singapore Mathematical Society
... The problem of determining all congruent numbers has a long history. The examples 5 and 6 were given in an Arab manuscript written more than 1000 years ago [1]. The problem is not completely solved even today. In 1983, using very sophisticated methods in number theory Tunnell [4] discovered a charac ...
... The problem of determining all congruent numbers has a long history. The examples 5 and 6 were given in an Arab manuscript written more than 1000 years ago [1]. The problem is not completely solved even today. In 1983, using very sophisticated methods in number theory Tunnell [4] discovered a charac ...
REPRESENTATIONS OF THE REAL NUMBERS
... follow from the general fact 6 <~ 6' ~ r6, ~_ r6 and the characterizations of the final topologies of p, p< and p>. [] For a given representation 6 of a set M we can ask which informations about x = 6 ( p ) • M are finitely (or continuously) accessible (f.a.) from the name p • dom(6) of x. In the ca ...
... follow from the general fact 6 <~ 6' ~ r6, ~_ r6 and the characterizations of the final topologies of p, p< and p>. [] For a given representation 6 of a set M we can ask which informations about x = 6 ( p ) • M are finitely (or continuously) accessible (f.a.) from the name p • dom(6) of x. In the ca ...
n - UOW
... Question16 Let T = {1, 2, 3, 4, 5, 6, 7, 8, 9} . Five integers are chosen from T. Must there be two which add to 10? Question17 A programmer writes 500 lines of computer code in 17 days. Must there be at least one day the programmer wrote 30 or more lines of code? Question18 Show that at a party of ...
... Question16 Let T = {1, 2, 3, 4, 5, 6, 7, 8, 9} . Five integers are chosen from T. Must there be two which add to 10? Question17 A programmer writes 500 lines of computer code in 17 days. Must there be at least one day the programmer wrote 30 or more lines of code? Question18 Show that at a party of ...