A PRNG specialized in double precision floating point numbers
... 6 Reducible transition function in affine case Usually, to make sure that the period is maximal, we need to check the primitivity of χf . This is often computationally difficult, since we need the integer factorization of 2deg(χ(t)) − 1, which is hard if the degree is high (say, > 10000). There are two ...
... 6 Reducible transition function in affine case Usually, to make sure that the period is maximal, we need to check the primitivity of χf . This is often computationally difficult, since we need the integer factorization of 2deg(χ(t)) − 1, which is hard if the degree is high (say, > 10000). There are two ...
2 Values of the Riemann zeta function at integers
... the Riemann zeta function is as mysterious as ever. For instance, except for the so-called trivial zeros at −2, −4, −6, . . . , the position of the other zeros is still an open conjecture. It is a subject of the Riemann Hypothesis. No unsolved conjecture is more celebrated nor more desirable than th ...
... the Riemann zeta function is as mysterious as ever. For instance, except for the so-called trivial zeros at −2, −4, −6, . . . , the position of the other zeros is still an open conjecture. It is a subject of the Riemann Hypothesis. No unsolved conjecture is more celebrated nor more desirable than th ...
Notes for Lesson 8-1: Factors and Greatest Common Factors
... Factors that are shared by two or more numbers are called common factors. The greatest of these factors is called the greatest common factor or GCF. You can either list all of the factors to see what factors are in common or more efficiently use the prime factorization of the numbers and take the co ...
... Factors that are shared by two or more numbers are called common factors. The greatest of these factors is called the greatest common factor or GCF. You can either list all of the factors to see what factors are in common or more efficiently use the prime factorization of the numbers and take the co ...
Solutions - CMU Math
... the only way to get a number divisible by 3 by adding three of these is 1 + 1 + 1, so those scores must be entered first. Now we have an odd sum, so we must add 71 in order for the sum to be divisible by 4. That leaves 80 for the last score entered. 6. Find the number of integers n, 1 ≤ n ≤ 25 such ...
... the only way to get a number divisible by 3 by adding three of these is 1 + 1 + 1, so those scores must be entered first. Now we have an odd sum, so we must add 71 in order for the sum to be divisible by 4. That leaves 80 for the last score entered. 6. Find the number of integers n, 1 ≤ n ≤ 25 such ...
Full text
... which implies that m > 2n + 2. By the Primitive Divisor Theorem (see, for example, [1]), we know that for any k > 12, Fk is divisible by a prime number p with p ≡ ±1 (mod k). In particular, p ≥ k − 1. Thus, since m ≥ 2n + 2 and n ≥ 16, it follows that Fm is divisible by a prime number p ≥ m − 1 ≥ 2n ...
... which implies that m > 2n + 2. By the Primitive Divisor Theorem (see, for example, [1]), we know that for any k > 12, Fk is divisible by a prime number p with p ≡ ±1 (mod k). In particular, p ≥ k − 1. Thus, since m ≥ 2n + 2 and n ≥ 16, it follows that Fm is divisible by a prime number p ≥ m − 1 ≥ 2n ...
