On Divisors of Lucas and Lehmer Numbers
... remark that the factor 104 which occurs on the right hand side of (7) has no arithmetical significance. Instead it is determined by the current quality of the estimates for linear forms in p-adic logarithms of algebraic numbers. In fact we could replace 104 by any number strictly larger than 14e2 . ...
... remark that the factor 104 which occurs on the right hand side of (7) has no arithmetical significance. Instead it is determined by the current quality of the estimates for linear forms in p-adic logarithms of algebraic numbers. In fact we could replace 104 by any number strictly larger than 14e2 . ...
THE FIBONACCI SEQUENCE MODULO p2 – AN INVESTIGATION
... 2.1.1. It is quite surprising that the Fibonacci sequence still keeps secrets. But there are at least two of them. Problem 2.1.2. The first open problem is “What is the exact value of κ(p)?”. Equivalently, one should understand precisely the behaviour of the quotient Q p−1 for p ≡ ±1 (mod 5) and Q(p ...
... 2.1.1. It is quite surprising that the Fibonacci sequence still keeps secrets. But there are at least two of them. Problem 2.1.2. The first open problem is “What is the exact value of κ(p)?”. Equivalently, one should understand precisely the behaviour of the quotient Q p−1 for p ≡ ±1 (mod 5) and Q(p ...
100 Statements to Prove for CS 2233 The idea of this
... 100 Statements to Prove for CS 2233 The idea of this document is to provide a list of statements that any student of CS 2233 (Discrete Mathematical Structures) at UTSA ought to be able to prove. The point of being able prove 100 statements is to avoid simple memorization. The proofs cover many, but ...
... 100 Statements to Prove for CS 2233 The idea of this document is to provide a list of statements that any student of CS 2233 (Discrete Mathematical Structures) at UTSA ought to be able to prove. The point of being able prove 100 statements is to avoid simple memorization. The proofs cover many, but ...
... In concluding the Introduction, we want to attract the reader’s attention to the fact that many of the formulae presented here are very combinatorial in nature (see Sections 4, 5, 8). This raises the natural question as to whether it is possible to find combinatorial proofs for them. Indeed, a combin ...
SIMPLE GROUPS ARE SCARCE X)-log log x
... (log A)2 and A1'2. A proof cannot be based on the "very large" primes greater than A1'2. The reason why is contained in Theorem 4 below, which may be of independent number-theoretic interest. In the following, 7r(x) denotes the number of primes less than or equal to x. We use the fact that ^,n£x(l/n ...
... (log A)2 and A1'2. A proof cannot be based on the "very large" primes greater than A1'2. The reason why is contained in Theorem 4 below, which may be of independent number-theoretic interest. In the following, 7r(x) denotes the number of primes less than or equal to x. We use the fact that ^,n£x(l/n ...
DM- 07 MA-217 Discrete Mathematics /Jan
... in Königsberg and Berlin. His first love was the theory of elliptic functions, but he also wrote in other branches of analysis and in geometry and mechanics. Interested in the history of mathematics, Jacobi was a prime mover in the publication of Euler’s collected work. He and Dirichlet were close ...
... in Königsberg and Berlin. His first love was the theory of elliptic functions, but he also wrote in other branches of analysis and in geometry and mechanics. Interested in the history of mathematics, Jacobi was a prime mover in the publication of Euler’s collected work. He and Dirichlet were close ...
Modular Numbers - Department of Computer Sciences
... The Euclidean algorithm is based on the quotient remainder theorem which states: For any integer a ∈ Z and natural number n there exists a unique integer q and a unique natural number 0 ≤ r < n such that a = q · n + r The Euclidean algorithm can be summarized by the recursive function described in t ...
... The Euclidean algorithm is based on the quotient remainder theorem which states: For any integer a ∈ Z and natural number n there exists a unique integer q and a unique natural number 0 ≤ r < n such that a = q · n + r The Euclidean algorithm can be summarized by the recursive function described in t ...
Lecture Notes: Cryptography – Part 2
... To understand how the algorithm was designed, and why it works, we shall need several mathematical ingredients drawn from a branch of mathematics known as Number Theory, the study of whole numbers. This branch of mathematics has been studied from antiquity because it was (and is) found to be profoun ...
... To understand how the algorithm was designed, and why it works, we shall need several mathematical ingredients drawn from a branch of mathematics known as Number Theory, the study of whole numbers. This branch of mathematics has been studied from antiquity because it was (and is) found to be profoun ...
William Stallings, Cryptography and Network Security 3/e
... when she took them seven at a time they came out even. What is the smallest number of eggs she could have had? What other possible number of eggs could she have? [Hint: x = 1 (mod 2,3,4,5,6), x = 0 (mod 7).] ...
... when she took them seven at a time they came out even. What is the smallest number of eggs she could have had? What other possible number of eggs could she have? [Hint: x = 1 (mod 2,3,4,5,6), x = 0 (mod 7).] ...
Decomposition numbers for finite Coxeter groups and generalised
... non-crossing partitions, which we also explain in the same section. This is followed by an intermediate section in which we collect together some auxiliary results that will be needed later on. In Section 4, we recall Goulden and Jackson’s formula [23] for the full rank decomposition numbers of type ...
... non-crossing partitions, which we also explain in the same section. This is followed by an intermediate section in which we collect together some auxiliary results that will be needed later on. In Section 4, we recall Goulden and Jackson’s formula [23] for the full rank decomposition numbers of type ...
Circular Flow and Circular Chromatic Number in the Matroid Context
... of examples which was most illuminating. I am grateful to him for this contribution. I sincerely appreciate the work of the examining committee members who so kindly agreed to read this work. In particular Dr. Marni Mishna who kindly provided useful feedback on an early draft of this thesis. I thank ...
... of examples which was most illuminating. I am grateful to him for this contribution. I sincerely appreciate the work of the examining committee members who so kindly agreed to read this work. In particular Dr. Marni Mishna who kindly provided useful feedback on an early draft of this thesis. I thank ...
Exam 2 Review Problems
... a) 33 (= 27) ≡ −3 mod 15 ⇒ 36 (= (33 )2 ) ≡ (−3)2 ≡ 32 mod 15. We repeatedly use the fact that 36 ≡ 32 mod 15 by taking suitable powers. If we cube both sides, we get 318 ≡ 36 ≡ 32 mod 15. Next, we square both sides of the previous equation to get 336 ≡ 34 mod 15. Multiply 31 4 on both sides to get ...
... a) 33 (= 27) ≡ −3 mod 15 ⇒ 36 (= (33 )2 ) ≡ (−3)2 ≡ 32 mod 15. We repeatedly use the fact that 36 ≡ 32 mod 15 by taking suitable powers. If we cube both sides, we get 318 ≡ 36 ≡ 32 mod 15. Next, we square both sides of the previous equation to get 336 ≡ 34 mod 15. Multiply 31 4 on both sides to get ...
PPT
... We want to show that n is a power of a prime number OR a product of powers of prime numbers. Consider two cases. n is a power of a prime number Case 1: n is a prime number. n is a product of two smaller numbers, namely a and b (where a < n and b < n) Case 2: n is not a prime number. Thus, n = a . b ...
... We want to show that n is a power of a prime number OR a product of powers of prime numbers. Consider two cases. n is a power of a prime number Case 1: n is a prime number. n is a product of two smaller numbers, namely a and b (where a < n and b < n) Case 2: n is not a prime number. Thus, n = a . b ...
Solutions for the 2nd Practice Midterm
... (b) Choose two of the numbers you found in the first part and compute their reciprocals mod 100. The easiest reciprocal is that of 99, because 99 ≡ −1 mod 100 so the reciprocal of 99 is 1/ − 1 = −1 ≡ 99 mod 100. Since 91 ≡ −9 and 97 ≡ −3, the reciprocals of 91 and 97 can both be obtained using the f ...
... (b) Choose two of the numbers you found in the first part and compute their reciprocals mod 100. The easiest reciprocal is that of 99, because 99 ≡ −1 mod 100 so the reciprocal of 99 is 1/ − 1 = −1 ≡ 99 mod 100. Since 91 ≡ −9 and 97 ≡ −3, the reciprocals of 91 and 97 can both be obtained using the f ...
arXiv:math/0604314v2 [math.NT] 7 Sep 2006 On
... Theorem 4 The only squarefull integers not in R are 1, 4, 8, 9, 16 and 36. We recall that an integer n is said to be squarefull if for every prime divisor p of n we have p2 |n. An integer n is called t-free if pt ∤ m for every prime number p. (Thus saying a number is squarefree is the same as saying ...
... Theorem 4 The only squarefull integers not in R are 1, 4, 8, 9, 16 and 36. We recall that an integer n is said to be squarefull if for every prime divisor p of n we have p2 |n. An integer n is called t-free if pt ∤ m for every prime number p. (Thus saying a number is squarefree is the same as saying ...
Full text
... We distinguish several cases on n. First, if n 6= F2k , F2k − 1, F2k+1 , F2k+1 − 1 then by definition of the indicator functions, ε1 (n) = ε1 (n − 1) = ε2 (n) = ε2 (n − 1) = 0. Furthermore, by (2.7) and (2.8), ε1 (bnµc) = ε2 (bnµc) = 0. Then, the equalities (2.9) and (2.10) hold by (1) of Lemma 2.2. ...
... We distinguish several cases on n. First, if n 6= F2k , F2k − 1, F2k+1 , F2k+1 − 1 then by definition of the indicator functions, ε1 (n) = ε1 (n − 1) = ε2 (n) = ε2 (n − 1) = 0. Furthermore, by (2.7) and (2.8), ε1 (bnµc) = ε2 (bnµc) = 0. Then, the equalities (2.9) and (2.10) hold by (1) of Lemma 2.2. ...
2 Sequences: Convergence and Divergence
... can hold only for finitely many n. Similarly, the sequence {2n } diverges. (ii) The sequence defined by {(−1)n } is {−1, 1, −1, 1, . . .}, and this sequence diverges by oscillation because the nth term is always either 1 or −1. Thus an cannot approach any one specific number a as n grows large. Also, w ...
... can hold only for finitely many n. Similarly, the sequence {2n } diverges. (ii) The sequence defined by {(−1)n } is {−1, 1, −1, 1, . . .}, and this sequence diverges by oscillation because the nth term is always either 1 or −1. Thus an cannot approach any one specific number a as n grows large. Also, w ...
lecture12-orig - School of Computer Science
... Every two recursive calls, the input numbers drop by half. ...
... Every two recursive calls, the input numbers drop by half. ...
The Chinese Remainder Theorem
... exploiting the advantage of congruence arithmetic, we can do this fairly efficiently, once we see by long division that 70 ≡ 1 (mod 3). This immediately yield 2 × 70 ≡ 2 × 1 ≡ 2 (mod 3), so that for the second number we can choose 140 = 2 × 70. Now consider the congruence modulo 5. Since the other t ...
... exploiting the advantage of congruence arithmetic, we can do this fairly efficiently, once we see by long division that 70 ≡ 1 (mod 3). This immediately yield 2 × 70 ≡ 2 × 1 ≡ 2 (mod 3), so that for the second number we can choose 140 = 2 × 70. Now consider the congruence modulo 5. Since the other t ...