modulo one uniform distribution of the sequence of logarithms of
... and the sufficiency of Weyl's criterion proves the sequence {y.}°° to be uniformly distributed mod 1. Lemma 2. If a is a positive algebraic number not equal to one, then In a is irrational. Proof. Assume, to the contrary, In a = (p/q), where p and q are non-zero integers. Then e p / q = a9 so that e ...
... and the sufficiency of Weyl's criterion proves the sequence {y.}°° to be uniformly distributed mod 1. Lemma 2. If a is a positive algebraic number not equal to one, then In a is irrational. Proof. Assume, to the contrary, In a = (p/q), where p and q are non-zero integers. Then e p / q = a9 so that e ...
On the Number of False Witnesses for a Composite Number
... Lagrange's theorem gives F(«)|<|>(«), whereis Euler's function.
There are composite numbers « for which F(n) = («),such as « = 561. Such
numbers are called Carmichael numbers and probably there are infinitely many of
them, but this has never been proved. It is known that Carmichael numbers ar ...
... Lagrange's theorem gives F(«)|<|>(«), where
A Pascal-Type Triangle Characterizing Twin Primes
... (3) and (8). Of course, these do look rather convoluted compared with the triangular recurrence for the Pascal triangle. Given the denominators on the right-hand sides of (10) and (11), it is by no means clear that the numbers a(k, s) and b(k, s), if they were defined in this way, should all be inte ...
... (3) and (8). Of course, these do look rather convoluted compared with the triangular recurrence for the Pascal triangle. Given the denominators on the right-hand sides of (10) and (11), it is by no means clear that the numbers a(k, s) and b(k, s), if they were defined in this way, should all be inte ...
A formally verified proof of the prime number theorem
... The reason it is useful is that there are two terms in the sum on the left, each sensitive to the presence of primes in different ways. Selberg’s proof involves cleverly balancing the two terms off each other, to show that in the long run, the density of the primes has the appropriate ...
... The reason it is useful is that there are two terms in the sum on the left, each sensitive to the presence of primes in different ways. Selberg’s proof involves cleverly balancing the two terms off each other, to show that in the long run, the density of the primes has the appropriate ...
All Elite Primes Up to 250 Billion
... these numbers, some of their properties and open problems can be found, e.g., in the work of Hardy and Wright [3], in section A3 of Guy’s problem book [2] or in the 17 lectures on this topic by Křížek, Luca and Somer [4]. We call a prime number p elite, if there is an integer index m for which al ...
... these numbers, some of their properties and open problems can be found, e.g., in the work of Hardy and Wright [3], in section A3 of Guy’s problem book [2] or in the 17 lectures on this topic by Křížek, Luca and Somer [4]. We call a prime number p elite, if there is an integer index m for which al ...
Notes on the large sieve
... and Vaughan [MV1] in 1973. The −1 was contributed by Selberg (unpublished), using a different method, but as we show below, it can be established by a minor addition to the proof of Montgomery and Vaughan. We now give (a) a short proof of the theorem with a weaker constant, (b) a longer proof, givin ...
... and Vaughan [MV1] in 1973. The −1 was contributed by Selberg (unpublished), using a different method, but as we show below, it can be established by a minor addition to the proof of Montgomery and Vaughan. We now give (a) a short proof of the theorem with a weaker constant, (b) a longer proof, givin ...
1 Introduction: Historical Background
... Proof. First remember that each member of C is square-free. If a prime p belongs to C , then by Theorem 2.1 we have p , 1 j p, which implies p = 2. Hence, 2 is the only prime belonging to C . If p < q are two primes and pq 2 C , then we get by Theorem 2.3 that p , 1 j q and q , 1 j p. By p , 1 j q w ...
... Proof. First remember that each member of C is square-free. If a prime p belongs to C , then by Theorem 2.1 we have p , 1 j p, which implies p = 2. Hence, 2 is the only prime belonging to C . If p < q are two primes and pq 2 C , then we get by Theorem 2.3 that p , 1 j q and q , 1 j p. By p , 1 j q w ...
2340-001/lectures - NYU
... • For any real number x, the floor of x, written x, is the unique integer n such that n x < n + 1. It is the max of all ints x. • For any real number x, the ceiling of x, written x, is the unique integer n such that n – 1 < x n. What is n? • If k is an integer, what are x and x + 1/2? ...
... • For any real number x, the floor of x, written x, is the unique integer n such that n x < n + 1. It is the max of all ints x. • For any real number x, the ceiling of x, written x, is the unique integer n such that n – 1 < x n. What is n? • If k is an integer, what are x and x + 1/2? ...
Structure and Randomness in the prime numbers
... Bob cannot unlock the box… but he can put his own padlock b on the box. He then sends the doubly locked box gab back to Alice. ...
... Bob cannot unlock the box… but he can put his own padlock b on the box. He then sends the doubly locked box gab back to Alice. ...
On the number of real quadratic fields with class number divisible by 3,
... where c |b. We write b = cd, then f (x) = (x + c)(x2 − cx + d). This implies a = d − c2 . Thus P a is uniquely determined by the number of divisors of b. It is well known that |b|≤B d(b) ≤ 2B log B. Here d(b) represents the number of positive divisors of b. Thus the cardinality of this set is ≤ 2B l ...
... where c |b. We write b = cd, then f (x) = (x + c)(x2 − cx + d). This implies a = d − c2 . Thus P a is uniquely determined by the number of divisors of b. It is well known that |b|≤B d(b) ≤ 2B log B. Here d(b) represents the number of positive divisors of b. Thus the cardinality of this set is ≤ 2B l ...
Full text
... [3]. The three preceding ones were given by Gillies [2].) It is not known whether there are any odd perfect numbers, though many necessary conditions for their existence have been established. The most interesting of recent conditions are that such a number must have at least eight distinct prime fa ...
... [3]. The three preceding ones were given by Gillies [2].) It is not known whether there are any odd perfect numbers, though many necessary conditions for their existence have been established. The most interesting of recent conditions are that such a number must have at least eight distinct prime fa ...
Lecture notes, sections 2.5 to 2.7
... • (52n−1 + 1) is divisible by 6 • (n2 + n + 17) is a prime number • A set with n elements has 2n subsets • n can be written as the sum of at most three prime numbers Consider the first of these examples. In this case, S(2) is a true statement, while S(4) is a false statement. What do you think about ...
... • (52n−1 + 1) is divisible by 6 • (n2 + n + 17) is a prime number • A set with n elements has 2n subsets • n can be written as the sum of at most three prime numbers Consider the first of these examples. In this case, S(2) is a true statement, while S(4) is a false statement. What do you think about ...
On Exhaustion of Domains - Department of Mathematics, Statistics
... therefore for the question of exhaustion one can consider the whole class {G1 } of domains biholomorphically equivalent to G1 . We now introduce the set E(D) of all classes {G}, where G is a bounded domain in Cn that can be exhausted by D. Obviously E(D) is not empty: it always contains {D}. By #E(D ...
... therefore for the question of exhaustion one can consider the whole class {G1 } of domains biholomorphically equivalent to G1 . We now introduce the set E(D) of all classes {G}, where G is a bounded domain in Cn that can be exhausted by D. Obviously E(D) is not empty: it always contains {D}. By #E(D ...
On the Divisibility of an Odd Perfect Number by the Sixth Power of a
... least) the sixth power of aprime, or N > 109118. Proof. Theorem 3 proves that if N is not divisible by the sixth power of a prime then the least prime factor of TVis ^ 101. Karl Norton's paper [5] contains a table which shows that any odd perfect number whose least prime factor is S: 101 has at leas ...
... least) the sixth power of aprime, or N > 109118. Proof. Theorem 3 proves that if N is not divisible by the sixth power of a prime then the least prime factor of TVis ^ 101. Karl Norton's paper [5] contains a table which shows that any odd perfect number whose least prime factor is S: 101 has at leas ...
1. Prove that 3n + 2 and 5 n + 3 are relatively prime for every positive
... b) d bxce = d xe 11. Factor 51, 948 into a product of primes. 12. Prove that n5 − n is divisble by 30 for every integer n. 13. Prove that n, n + 2, n + 4 are all primes if and only if n = 3. n 14. The integer Fn = 22 +1 is called the nth Fermat number. Show that for n = 1, 2, 3, 4 Fn is prime. 15. S ...
... b) d bxce = d xe 11. Factor 51, 948 into a product of primes. 12. Prove that n5 − n is divisble by 30 for every integer n. 13. Prove that n, n + 2, n + 4 are all primes if and only if n = 3. n 14. The integer Fn = 22 +1 is called the nth Fermat number. Show that for n = 1, 2, 3, 4 Fn is prime. 15. S ...
Prime and Composite Terms in Sloane`s Sequence A056542
... One consequence of the theorem is the fact that for a given prime p it is enough to search among the terms ap , ap+1 , . . . , a2p−1 for multiples of p. Numerical computations lead to the results presented in Table 1. Furthermore, we can now use the theorem to sift the prime candidates and retain mo ...
... One consequence of the theorem is the fact that for a given prime p it is enough to search among the terms ap , ap+1 , . . . , a2p−1 for multiples of p. Numerical computations lead to the results presented in Table 1. Furthermore, we can now use the theorem to sift the prime candidates and retain mo ...
A Note on Nested Sums
... which allowed us to write a sum of binomial coefficients as a single binomial coefficient and rewrite a product of two binomial coefficients so that we could pull one term out. When our function is no longer a single binomial coefficient then we might no longer be able to rewrite the product so simp ...
... which allowed us to write a sum of binomial coefficients as a single binomial coefficient and rewrite a product of two binomial coefficients so that we could pull one term out. When our function is no longer a single binomial coefficient then we might no longer be able to rewrite the product so simp ...
(pdf)
... Remark 1.4. Because of the identity αλ = 1, we can express every element of OK in the form a0 + a1 α + a2 α2 + ... + aλ−1 αλ−1 , where a0 , ..., aλ−1 are integers. And because of this form which resembles a polynomial in α, Kummer used the notation f (α), g(α), ... to denote cyclotomic integers. Def ...
... Remark 1.4. Because of the identity αλ = 1, we can express every element of OK in the form a0 + a1 α + a2 α2 + ... + aλ−1 αλ−1 , where a0 , ..., aλ−1 are integers. And because of this form which resembles a polynomial in α, Kummer used the notation f (α), g(α), ... to denote cyclotomic integers. Def ...
M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS 1
... Now suppose we have an infinite sequence a0 , a1 , a2 , . . . , of real numbers, and assume that ai ≥ 1 for i ≥ 1. Define pi and qi by the recurrence given above. We call pqii the ith convergent to the continued fraction: a0 + ...
... Now suppose we have an infinite sequence a0 , a1 , a2 , . . . , of real numbers, and assume that ai ≥ 1 for i ≥ 1. Define pi and qi by the recurrence given above. We call pqii the ith convergent to the continued fraction: a0 + ...
