STRUCTURAL RESULTS ON MAXIMAL k-DEGENERATE - DML-PL
... (2) Iterate the following operation. Let v = v0 be vertex of degree k in G with neighbors {u1 , . . . , uk }. Successively add k − 1 vertices {v1 , . . . , vk−1 } with degree k when added so that the neighbors of vi are all in {u1 , . . . , uk , v0 , . . . , vi−1 }. Then add a new vertex v ′ adjacen ...
... (2) Iterate the following operation. Let v = v0 be vertex of degree k in G with neighbors {u1 , . . . , uk }. Successively add k − 1 vertices {v1 , . . . , vk−1 } with degree k when added so that the neighbors of vi are all in {u1 , . . . , uk , v0 , . . . , vi−1 }. Then add a new vertex v ′ adjacen ...
On perfect numbers which are ratios of two Fibonacci numbers∗
... divisible by q. In particular, they are all congruent to ±1 (mod q); (ii) p ≡ 1 (mod 5) and p ≡ 1 (mod q). Furthermore, N ≡ 1 (mod 5) and N ≡ 1 (mod q); (iii) If qi ≡ 1 (mod q) for some i = 1, . . . , s, then ai > 2q − 2; (iv) We have q ≡ ±1 (mod 20). In particular, Fq ≡ 1 (mod 5); (v) Fq 6= p. Proo ...
... divisible by q. In particular, they are all congruent to ±1 (mod q); (ii) p ≡ 1 (mod 5) and p ≡ 1 (mod q). Furthermore, N ≡ 1 (mod 5) and N ≡ 1 (mod q); (iii) If qi ≡ 1 (mod q) for some i = 1, . . . , s, then ai > 2q − 2; (iv) We have q ≡ ±1 (mod 20). In particular, Fq ≡ 1 (mod 5); (v) Fq 6= p. Proo ...
Fibonacci numbers that are not sums of two prime powers
... We use the same method employed in [2, 3, 9]. However, there are additional difficulties which arise because we want our numbers to belong to the Fibonacci sequence. Let us quickly recall how one can create a residue class of odd integers most of which are not sums of two powers. Well, assume that n ...
... We use the same method employed in [2, 3, 9]. However, there are additional difficulties which arise because we want our numbers to belong to the Fibonacci sequence. Let us quickly recall how one can create a residue class of odd integers most of which are not sums of two powers. Well, assume that n ...
On condition numbers of polynomial eigenvalue problems
... Dedicated to the memory of James H. Wilkinson (1919–1986) Abstract In this paper, we investigate condition numbers of eigenvalue problems of matrix polynomials with nonsingular leading coefficients, generalizing classical results of matrix perturbation theory. We provide a relation between the condi ...
... Dedicated to the memory of James H. Wilkinson (1919–1986) Abstract In this paper, we investigate condition numbers of eigenvalue problems of matrix polynomials with nonsingular leading coefficients, generalizing classical results of matrix perturbation theory. We provide a relation between the condi ...
Full text
... (mod 281), and x2 ≡ 97, 39 (mod 211), all of which are impossible. (b) Assume that n ≡ −1, 3, 7, 8, 9 (mod 14), n ≡ 7, 11 (mod 16), n ≡ 14 (mod 28), and n ≡ −1, −13, 3 (mod 48). In this case, using congruence (12) of [1], we find that un ≡ ±u−1 , ±u3 , ±u7 , ±u8 , ±u9 (mod F7 ), un ≡ u7 , u11 (mod F ...
... (mod 281), and x2 ≡ 97, 39 (mod 211), all of which are impossible. (b) Assume that n ≡ −1, 3, 7, 8, 9 (mod 14), n ≡ 7, 11 (mod 16), n ≡ 14 (mod 28), and n ≡ −1, −13, 3 (mod 48). In this case, using congruence (12) of [1], we find that un ≡ ±u−1 , ±u3 , ±u7 , ±u8 , ±u9 (mod F7 ), un ≡ u7 , u11 (mod F ...
CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS
... REMARKS. The natural number N of the preceding definition surely depends on the positive number . If 0 is a smaller positive number than , then the corresponding N 0 very likely will need to be larger than N. Sometimes we will indicate this dependence by writing N () instead of simply N. It is a ...
... REMARKS. The natural number N of the preceding definition surely depends on the positive number . If 0 is a smaller positive number than , then the corresponding N 0 very likely will need to be larger than N. Sometimes we will indicate this dependence by writing N () instead of simply N. It is a ...
background on constructible angles
... By taking the real part of that sum, we obtain a formula for cos(nθ) using a sum whose terms have products of integers, cos(θ), and sin(θ). All those factors are constructible numbers, so cos(nθ) ∈ F . (Similarly, the imaginary part shows that sin(nθ) ∈ F too.) Proving an Angle is Non-Constructible ...
... By taking the real part of that sum, we obtain a formula for cos(nθ) using a sum whose terms have products of integers, cos(θ), and sin(θ). All those factors are constructible numbers, so cos(nθ) ∈ F . (Similarly, the imaginary part shows that sin(nθ) ∈ F too.) Proving an Angle is Non-Constructible ...
Modular curves, Arakelov theory, algorithmic applications
... general situation. This problem is solved as follows. We take D0 = gO, where O is a rational cusp of X1 (n). With this choice, there may be points of J1 (n)[m] for which the representation in the form [D − D0 ] is not unique. For every x ∈ J1 (n)[m](Q) we therefore consider the least integer dx such ...
... general situation. This problem is solved as follows. We take D0 = gO, where O is a rational cusp of X1 (n). With this choice, there may be points of J1 (n)[m] for which the representation in the form [D − D0 ] is not unique. For every x ∈ J1 (n)[m](Q) we therefore consider the least integer dx such ...
Elementary Number Theory
... Here are some examples of outstanding unsolved problems in number theory. Some of these will be discussed in this course. A solution to any one of these problems would make you quite famous (at least among mathematicians). Many of these problems concern prime numbers. A prime number is an integer gr ...
... Here are some examples of outstanding unsolved problems in number theory. Some of these will be discussed in this course. A solution to any one of these problems would make you quite famous (at least among mathematicians). Many of these problems concern prime numbers. A prime number is an integer gr ...
MATH 25 CLASS 16 NOTES, OCT 26 2011 Contents 1. Fast
... one, but also is one that is beyond the scope of this class. Another interesting question is how many Carmichael numbers there are. As a matter of fact, it is a difficult theorem of Alford, Granville, and Pomerance in 1992 that there are infinitely many Carmichael numbers, and they show that there a ...
... one, but also is one that is beyond the scope of this class. Another interesting question is how many Carmichael numbers there are. As a matter of fact, it is a difficult theorem of Alford, Granville, and Pomerance in 1992 that there are infinitely many Carmichael numbers, and they show that there a ...
arXiv
... When it is, p cannot divide a, and so the Fermat quotient q p (a) is an integer. In fact, (1) shows that a prime p is a Wieferich prime base a if and only if p does not divide a but does divide q p (a). In 1909, while still a graduate student at the University of Münster in Germany, Wieferich creat ...
... When it is, p cannot divide a, and so the Fermat quotient q p (a) is an integer. In fact, (1) shows that a prime p is a Wieferich prime base a if and only if p does not divide a but does divide q p (a). In 1909, while still a graduate student at the University of Münster in Germany, Wieferich creat ...
Elementary Number Theory
... Here are some examples of outstanding unsolved problems in number theory. Some of these will be discussed in this course. A solution to any one of these problems would make you quite famous (at least among mathematicians). Many of these problems concern prime numbers. A prime number is an integer gr ...
... Here are some examples of outstanding unsolved problems in number theory. Some of these will be discussed in this course. A solution to any one of these problems would make you quite famous (at least among mathematicians). Many of these problems concern prime numbers. A prime number is an integer gr ...
1. (a) Find and describe all integers x such that the conditions x ≡ 9
... the book and your notes, but do not ask other people for help, and do not collaborate with anyone else on these exam problems. Do not use computers to solve any of these problems. If you need another copy of this exam you can download the pdf file at: http://faculty.uml.edu/dklain/numtest2home.pdf ...
... the book and your notes, but do not ask other people for help, and do not collaborate with anyone else on these exam problems. Do not use computers to solve any of these problems. If you need another copy of this exam you can download the pdf file at: http://faculty.uml.edu/dklain/numtest2home.pdf ...
LANDAU`S PROBLEMS ON PRIMES 1. Introduction In his invited
... first two are really generalizations of the Twin Prime Conjecture, the third one, (2.16) is obviously trivial if the difference is two. As the cited lines of Hilbert’s lecture also indicate, both Goldbach’s Conjecture and the Twin Prime Conjecture are special cases of linear equations of type (2.17) ...
... first two are really generalizations of the Twin Prime Conjecture, the third one, (2.16) is obviously trivial if the difference is two. As the cited lines of Hilbert’s lecture also indicate, both Goldbach’s Conjecture and the Twin Prime Conjecture are special cases of linear equations of type (2.17) ...
Higher-order Carmichael numbers
... About 2#P(L) squarefree composite n built from p ∈ P(L). “Each such n has 1/ϕ(L) chance of being 1 modulo L.” So we expect #C(L) ≈ 2#P(L) /ϕ(L). Goal: Find L with #P(L) very large. ...
... About 2#P(L) squarefree composite n built from p ∈ P(L). “Each such n has 1/ϕ(L) chance of being 1 modulo L.” So we expect #C(L) ≈ 2#P(L) /ϕ(L). Goal: Find L with #P(L) very large. ...
Modular Arithmetic
... Problem 2. How do we generate a random number? There is nothing random about random() or rand() function. The standard technique is to create a pseudorandom sequence. The most commonly used procedure for generating pseudorandom numbers is the linear congruential method (Lehmer, 1949). We generate a ...
... Problem 2. How do we generate a random number? There is nothing random about random() or rand() function. The standard technique is to create a pseudorandom sequence. The most commonly used procedure for generating pseudorandom numbers is the linear congruential method (Lehmer, 1949). We generate a ...
- ScholarWorks@GVSU
... raining, then Daisy is riding her bike,” that I have not told the truth. So in this case, the statement P ! Q is false. 3. Now suppose that P is false and Q is true or that it is raining and Daisy is riding her bike. Did I make a false statement by stating that if it is not raining, then Daisy is ri ...
... raining, then Daisy is riding her bike,” that I have not told the truth. So in this case, the statement P ! Q is false. 3. Now suppose that P is false and Q is true or that it is raining and Daisy is riding her bike. Did I make a false statement by stating that if it is not raining, then Daisy is ri ...
Ramanujan, Robin, highly composite numbers, and the Riemann
... Nicolas (email to Sondow, 2012): I keep a very strong souvenir of the conference organised in Urbana-Champaign in 1987 for the one hundred anniversary of Ramanujan. It is there that I discovered the hidden part of “Highly Composite Numbers” [first published in [15], later in [16], and again in [2]]. ...
... Nicolas (email to Sondow, 2012): I keep a very strong souvenir of the conference organised in Urbana-Champaign in 1987 for the one hundred anniversary of Ramanujan. It is there that I discovered the hidden part of “Highly Composite Numbers” [first published in [15], later in [16], and again in [2]]. ...
Exercises
... 29. A certain top secret is held in the form a very large integer N . The secret is known to the Airforce commander only. The ...
... 29. A certain top secret is held in the form a very large integer N . The secret is known to the Airforce commander only. The ...