Overpseudoprimes, and Mersenne and Fermat numbers as
... all divisors greater than 1 of n. In particular, we show that it contains all composite Mersenne numbers and, at least, squares of all Wieferich primes. In the fourth section, we give a generalization of this concept to arbitrary bases b > 1 as well. In the final section, we put forward some of its ...
... all divisors greater than 1 of n. In particular, we show that it contains all composite Mersenne numbers and, at least, squares of all Wieferich primes. In the fourth section, we give a generalization of this concept to arbitrary bases b > 1 as well. In the final section, we put forward some of its ...
Equidistribution and Primes - Princeton Math
... this statement is slightly weaker since Dirichlet considers one-sided progressions and here and elsewhere we allow negative numbers and call −p a prime if p is a positive prime). (3) Initial Generalizations: There are at least two well known generalizations of Dirichlet’s Theorem that have been inve ...
... this statement is slightly weaker since Dirichlet considers one-sided progressions and here and elsewhere we allow negative numbers and call −p a prime if p is a positive prime). (3) Initial Generalizations: There are at least two well known generalizations of Dirichlet’s Theorem that have been inve ...
ON THE BITS COUNTING FUNCTION OF REAL NUMBERS 1
... In this article, we will first prove related results when m is allowed to be an irrational number (in Theorem 1) via a study of the function Bn (x) = #{j ≤ n : xj = 1} where n ≥ 0 and x = (x−p · · · x−1 x0 . x1 x2 x3 · · · )2 is the binary expansion of x ≥ 0. The case where x is a rational number of ...
... In this article, we will first prove related results when m is allowed to be an irrational number (in Theorem 1) via a study of the function Bn (x) = #{j ≤ n : xj = 1} where n ≥ 0 and x = (x−p · · · x−1 x0 . x1 x2 x3 · · · )2 is the binary expansion of x ≥ 0. The case where x is a rational number of ...
Prime Numbers in Generalized Pascal Triangles
... the reflection across the x-axis of it which is still on the same elliptic curve. Let us observe that if (x2, y2) = (x1, -y1), these two points define a vertical line which has no third intersection point, thus in this case (x, y) + (x, -y) = (x, -y) + (x, y) = O. It can be proved further that if (x ...
... the reflection across the x-axis of it which is still on the same elliptic curve. Let us observe that if (x2, y2) = (x1, -y1), these two points define a vertical line which has no third intersection point, thus in this case (x, y) + (x, -y) = (x, -y) + (x, y) = O. It can be proved further that if (x ...
Xiaosi Zhou
... ( X a) p X p a (mod X r 1, p) • This must hold if p is prime • The problem now is that some composites n may satisfy the equation for a few values of a and r. • n must be a prime power if the equation holds for several a’s and an appropriately chosen r. ...
... ( X a) p X p a (mod X r 1, p) • This must hold if p is prime • The problem now is that some composites n may satisfy the equation for a few values of a and r. • n must be a prime power if the equation holds for several a’s and an appropriately chosen r. ...
Calculation of the Moments and the Moment Generating Function for
... variance, skewness, kurtosis,etc; second, to study analytically and numerically the moment generating function
... variance, skewness, kurtosis,etc; second, to study analytically and numerically the moment generating function
CSci 2011 Discrete Mathematics
... perfect square A perfect square is a square of an integer Rephrased: Show that a non-perfect square exists in the set {2*10500+15, 2*10500+16} Proof: The only two perfect squares that differ by 1 are 0 and 1 Thus, any other numbers that differ by 1 cannot both be perfect squares Thus, a non-per ...
... perfect square A perfect square is a square of an integer Rephrased: Show that a non-perfect square exists in the set {2*10500+15, 2*10500+16} Proof: The only two perfect squares that differ by 1 are 0 and 1 Thus, any other numbers that differ by 1 cannot both be perfect squares Thus, a non-per ...
Math 5330 Spring 2013 Notes: The Chinese Remainder Theorem
... problem: If 5x2 ` 6x ` 7 is divisible by 35, then it must be divisible by both 5 and 7. Rather than a congruence modulo 25, let’s look at two congruences: 5x2 ` 6x ` 7 ” 0 pmod 5q and 5x2 ` 6x ` 7 ” 0 pmod 7q. The first congruence is the same as x ` 2 ” 0 pmod 5q, or x ” 3 pmod 5q. The second is eq ...
... problem: If 5x2 ` 6x ` 7 is divisible by 35, then it must be divisible by both 5 and 7. Rather than a congruence modulo 25, let’s look at two congruences: 5x2 ` 6x ` 7 ” 0 pmod 5q and 5x2 ` 6x ` 7 ” 0 pmod 7q. The first congruence is the same as x ` 2 ” 0 pmod 5q, or x ” 3 pmod 5q. The second is eq ...
The “coefficients H” Technique - PRiSM
... Adv P RF is the best |P1 − P1∗ | that we can get with such φ algorithms. Proof of the “Main Lemma” Evaluation of P1∗ Let f be a fixed function, and let C1 , . . . , Cm be the successive values that the program φ will ask for the values of f (when φ tests the function f ). We will note σ1 = f (C1 ), ...
... Adv P RF is the best |P1 − P1∗ | that we can get with such φ algorithms. Proof of the “Main Lemma” Evaluation of P1∗ Let f be a fixed function, and let C1 , . . . , Cm be the successive values that the program φ will ask for the values of f (when φ tests the function f ). We will note σ1 = f (C1 ), ...
as a PDF
... the formal proof of the generalized version of Fermat Little’s Theorem, also known as The Fermat-Euler Theorem, ( aφ(n) ≡ 1 mod n when gcd(n, a) = 1). Originally, Fermat had made an observation for a special case restricted to prime numbers instead of n. After verifying it for several primes Fermat ...
... the formal proof of the generalized version of Fermat Little’s Theorem, also known as The Fermat-Euler Theorem, ( aφ(n) ≡ 1 mod n when gcd(n, a) = 1). Originally, Fermat had made an observation for a special case restricted to prime numbers instead of n. After verifying it for several primes Fermat ...
ünivalence of continued fractions and stieltjes transforms1
... the theory of univalent functions on the one hand and Stieltjes transforms and certain classes of continued fractions on the other. The only known result of approximately similar character is one, due to G. Szegö, which is mentioned in Corollary 2.1. In §2 are some theorems on univalence of function ...
... the theory of univalent functions on the one hand and Stieltjes transforms and certain classes of continued fractions on the other. The only known result of approximately similar character is one, due to G. Szegö, which is mentioned in Corollary 2.1. In §2 are some theorems on univalence of function ...
Proof of Euler`s φ (Phi) Function Formula - Rose
... the formal proof of the generalized version of Fermat Little’s Theorem, also known as The Fermat-Euler Theorem, ( aφ(n) ≡ 1 mod n when gcd(n, a) = 1). Originally, Fermat had made an observation for a special case restricted to prime numbers instead of n. After verifying it for several primes Fermat ...
... the formal proof of the generalized version of Fermat Little’s Theorem, also known as The Fermat-Euler Theorem, ( aφ(n) ≡ 1 mod n when gcd(n, a) = 1). Originally, Fermat had made an observation for a special case restricted to prime numbers instead of n. After verifying it for several primes Fermat ...
How To Think Like A Computer Scientist
... Theorem: S can’t be put into 1-1 correspondence with P(S) Suppose f:S->P(S) is 1-1 and ONTO. Let CONFUSE =All x in S such that x is not contained in f(x) There is some y such that f(y)=CONFUSE ...
... Theorem: S can’t be put into 1-1 correspondence with P(S) Suppose f:S->P(S) is 1-1 and ONTO. Let CONFUSE =All x in S such that x is not contained in f(x) There is some y such that f(y)=CONFUSE ...
Full text
... of Diophantine quadruple {a, b, a + b + 2r, 4r (r + a) (r + b)}, if ab + 1 = r2 . In January 1999, Gibbs [8] found the first set of six positive rationals with the above property. In the integer case, there is a famous conjecture: there does not exist a Diophantine quintuple. The case n 6= 1 also ha ...
... of Diophantine quadruple {a, b, a + b + 2r, 4r (r + a) (r + b)}, if ab + 1 = r2 . In January 1999, Gibbs [8] found the first set of six positive rationals with the above property. In the integer case, there is a famous conjecture: there does not exist a Diophantine quintuple. The case n 6= 1 also ha ...
Full text
... of the negative coefficients, which is 66488086. This is easily seen to be the case. The second case follows similarly. 4. LARGER VALUES OF k Theorem 1 naturally leads one to wonder about larger values of k. In fact, we have observed interesting behavior for values of k such as 13, 34, 89, and 233, ...
... of the negative coefficients, which is 66488086. This is easily seen to be the case. The second case follows similarly. 4. LARGER VALUES OF k Theorem 1 naturally leads one to wonder about larger values of k. In fact, we have observed interesting behavior for values of k such as 13, 34, 89, and 233, ...
PPT
... both change and hence the odd parity count changes by 2 – and remains even If 2 people of different parities shake, then they both swap parities and the odd parity count is unchanged ...
... both change and hence the odd parity count changes by 2 – and remains even If 2 people of different parities shake, then they both swap parities and the odd parity count is unchanged ...
The HOMER System for Discovery in Number Theory
... Not ready yet for research mathematics – Limited nature of the background concepts • Nothing likely to surprise the user ...
... Not ready yet for research mathematics – Limited nature of the background concepts • Nothing likely to surprise the user ...
Euclid and Number Theory
... Proof: We have by our theorem 1 = ax + by and we further know that bc = sa and so c = c · 1 = cax + cby = cax + say = a(cx + sy) which says a|c. Theorem: If p is a prime and p|a1 a2 . . . an then p|ak for some k. Proof: We may assume that aj 6= 0 for each j, 1 ≤ j ≤ n. We have gcd(p,aj ) is 1 or p. ...
... Proof: We have by our theorem 1 = ax + by and we further know that bc = sa and so c = c · 1 = cax + cby = cax + say = a(cx + sy) which says a|c. Theorem: If p is a prime and p|a1 a2 . . . an then p|ak for some k. Proof: We may assume that aj 6= 0 for each j, 1 ≤ j ≤ n. We have gcd(p,aj ) is 1 or p. ...