a 1
... 2. for all i = 1 .. t, gcd (a(n-1)/qi - 1, n) = 1, 3. if F > n, then n is prime. If n is prime the probability that a randomly selected a which satisfies Pocklington is (1 – 1/qi) Example: n = 2 ( 3. 11 ) + 1 = 67, F = 3 x 11 and R=2. Is 67 a prime? Proof: 1. gcd ( 2 (67-1)/ 3 –1 , 67 ) = gcd ( 2 ...
... 2. for all i = 1 .. t, gcd (a(n-1)/qi - 1, n) = 1, 3. if F > n, then n is prime. If n is prime the probability that a randomly selected a which satisfies Pocklington is (1 – 1/qi) Example: n = 2 ( 3. 11 ) + 1 = 67, F = 3 x 11 and R=2. Is 67 a prime? Proof: 1. gcd ( 2 (67-1)/ 3 –1 , 67 ) = gcd ( 2 ...
Full text
... & < 0 or k > n. This array has been called a Lucas triangle by Feinberg [1], because rising diagonals sum to give the Lucas numbers 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, ..., in contrast to the rising diagonals in the standard Pascal triangle where rising diagonals sum to give the Fibonacci ...
... & < 0 or k > n. This array has been called a Lucas triangle by Feinberg [1], because rising diagonals sum to give the Lucas numbers 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, ..., in contrast to the rising diagonals in the standard Pascal triangle where rising diagonals sum to give the Fibonacci ...
Finding Factors of Factor Rings over the Gaussian Integers
... If the Gaussian integers were no more than a practice area for generalizing concepts from the standard integers, they would still be of great interest to students and researchers alike. Yet these numbers have been (and still are) far more than just a convenient teaching tool. They have played roles ...
... If the Gaussian integers were no more than a practice area for generalizing concepts from the standard integers, they would still be of great interest to students and researchers alike. Yet these numbers have been (and still are) far more than just a convenient teaching tool. They have played roles ...
I. Precisely complete the following definitions: 1. A natural number n
... VII. Only one of the following will be graded. Please circle the number of the statement you want me to consider. 1. Let p be prime and a, b, m, n be naturals. We say pm exactly divides a when pm divides a, but pm+1 does not. Let pm exactly divide a and pn exactly divide b where m ≠ n. What power o ...
... VII. Only one of the following will be graded. Please circle the number of the statement you want me to consider. 1. Let p be prime and a, b, m, n be naturals. We say pm exactly divides a when pm divides a, but pm+1 does not. Let pm exactly divide a and pn exactly divide b where m ≠ n. What power o ...
A Subrecursive Refinement of the Fundamental Theorem of Algebra
... n = 1, 2, . . . , and this leads to the inequality |zn+1 − zn | < 2 of the constructed sequence z1 , z2 , . . . Of course this can be done also through Lemma 3 , and the rate of the convergence is quite good. Unfortunately the exN ponential dependence of 2−n2 on n is an obstacle to realize such a c ...
... n = 1, 2, . . . , and this leads to the inequality |zn+1 − zn | < 2 of the constructed sequence z1 , z2 , . . . Of course this can be done also through Lemma 3 , and the rate of the convergence is quite good. Unfortunately the exN ponential dependence of 2−n2 on n is an obstacle to realize such a c ...
Remainder Theorem
... 3) Concept of negative remainder Remainder can never be negative; its minimum value can only be 0. Consider an example of -30 / 7. Here, remainder is 5. It would not be (-28 – 2 / 7), but [(-35+5)/7] When you divide, you will get remainder of -2. Since remainder can never be negative, we subtract it ...
... 3) Concept of negative remainder Remainder can never be negative; its minimum value can only be 0. Consider an example of -30 / 7. Here, remainder is 5. It would not be (-28 – 2 / 7), but [(-35+5)/7] When you divide, you will get remainder of -2. Since remainder can never be negative, we subtract it ...
Partitions into three triangular numbers
... Our goal in this paper is to study the partitions of n into three triangular numbers rather than the representations of n into three triangular numbers. For instance, the three representations 28 + 1 + 1, 1 + 28 + 1, and 1 + 1 + 28 stem from one partition, 28 + 1 + 1. Thus the integer 30 can be part ...
... Our goal in this paper is to study the partitions of n into three triangular numbers rather than the representations of n into three triangular numbers. For instance, the three representations 28 + 1 + 1, 1 + 28 + 1, and 1 + 1 + 28 stem from one partition, 28 + 1 + 1. Thus the integer 30 can be part ...
Fundamental units and consecutive squarefull numbers,
... approach is elementary. We will prove the above theorems using only rudimentary facts of the BP equation. Recently, Reuss [18] established a better error term in the case d = 1. He showed S1 (N ) = O(N 29/100+ ), for any > 0. This result uses more sophisticated methods and relies on earlier deep ...
... approach is elementary. We will prove the above theorems using only rudimentary facts of the BP equation. Recently, Reuss [18] established a better error term in the case d = 1. He showed S1 (N ) = O(N 29/100+ ), for any > 0. This result uses more sophisticated methods and relies on earlier deep ...
www.XtremePapers.com
... You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of ...
... You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of ...
Full text
... fi/W = x, z2(x) = x +2, zn(x)-= xzn-i(x)tzn-2MFifty-four identities are derived which solve the problem for all cases except when both b amd m are odd; some special cases are given for that last possible case. Since fn(1)= Fn and zn(1)= Ln,thenth Fibonacci and Lucas numbers respectively, all of the ...
... fi/W = x, z2(x) = x +2, zn(x)-= xzn-i(x)tzn-2MFifty-four identities are derived which solve the problem for all cases except when both b amd m are odd; some special cases are given for that last possible case. Since fn(1)= Fn and zn(1)= Ln,thenth Fibonacci and Lucas numbers respectively, all of the ...
ON FINITE SUMS OF RECIPROCALS OF DISTINCT
... formally distinct sums in Pr_χ are separated by a distance of more than σ and hence, each element π of Pr-± gives rise to a half-open interval [TΓ, π + σ) which is disjoint from any other interval [TΓ', TΓ' + σ) for TΓ Φ τr'eP r _!. Therefore Ac(S) = \Jnepr-.x\.π9π + σ) is the disjoint union of exac ...
... formally distinct sums in Pr_χ are separated by a distance of more than σ and hence, each element π of Pr-± gives rise to a half-open interval [TΓ, π + σ) which is disjoint from any other interval [TΓ', TΓ' + σ) for TΓ Φ τr'eP r _!. Therefore Ac(S) = \Jnepr-.x\.π9π + σ) is the disjoint union of exac ...
The fractional part of n+ ø and Beatty sequences
... make it easy to get the actual Beatty sequences. The paper [5] is complete on the theory, but only the limited periodic case yields Beatty sequences. The paper [8] provides a very effective description of Beatty sequences, but regrettably, it is now known that the results do not match the facts (see ...
... make it easy to get the actual Beatty sequences. The paper [5] is complete on the theory, but only the limited periodic case yields Beatty sequences. The paper [8] provides a very effective description of Beatty sequences, but regrettably, it is now known that the results do not match the facts (see ...
The stronger mixing variables method
... f (a, b, c, d) − f ( ac, b, ac, d) = (a2 − c2 )2 − (b2 + d2 )(a − c)2 ≥ 0. By S.M.V. theorem with the transformation ∆ of (a, b, c), we only need to prove the inequality when a = b = c = t ≥ d. In this case, the problem becomes 3t4 + d4 + 2t3 d ≥ 3t4 + 3t2 d2 ⇔ d4 + t3 d + t3 d ≥ 3t2 d2 . By AM − GM ...
... f (a, b, c, d) − f ( ac, b, ac, d) = (a2 − c2 )2 − (b2 + d2 )(a − c)2 ≥ 0. By S.M.V. theorem with the transformation ∆ of (a, b, c), we only need to prove the inequality when a = b = c = t ≥ d. In this case, the problem becomes 3t4 + d4 + 2t3 d ≥ 3t4 + 3t2 d2 ⇔ d4 + t3 d + t3 d ≥ 3t2 d2 . By AM − GM ...
Arithmetic in Metamath, Case Study: Bertrand`s Postulate
... integer representations but rather base-4 representation. It is of course necessary to commit to a base to work in for practical purposes, and the most logical decision for a system such as set.mm which prioritizes abstract reasoning over numerical calculation is base 10. The reason this choice was ...
... integer representations but rather base-4 representation. It is of course necessary to commit to a base to work in for practical purposes, and the most logical decision for a system such as set.mm which prioritizes abstract reasoning over numerical calculation is base 10. The reason this choice was ...
Super-Isolated Elliptic Curves and Abelian Surfaces in Cryptography
... super-isolated elliptic curves and abelian surfaces. Second, we prove that only two super-isolated surfaces of cryptographic size and near-prime order exist, see Examples 35 and 36. Finally, we give some heuristics on the number of super-isolated varieties. Our results suggest that, unlike the case ...
... super-isolated elliptic curves and abelian surfaces. Second, we prove that only two super-isolated surfaces of cryptographic size and near-prime order exist, see Examples 35 and 36. Finally, we give some heuristics on the number of super-isolated varieties. Our results suggest that, unlike the case ...
A FOOTNOTE TO THE LEAST NON ZERO DIGIT OF n! IN BASE 12
... OF n! IN BASE 12 Jean-Marc DESHOUILLERS ABSTRACT. We continue the work initiated with Imre Ruzsa, showing that for any a ∈ {3, 6, 9}, there exist infinitely many integers n such that the least non zero digit of n! in base twelve is equal to a. ...
... OF n! IN BASE 12 Jean-Marc DESHOUILLERS ABSTRACT. We continue the work initiated with Imre Ruzsa, showing that for any a ∈ {3, 6, 9}, there exist infinitely many integers n such that the least non zero digit of n! in base twelve is equal to a. ...
Congruent Numbers - American Institute of Mathematics
... Along with the Riemann Hypothesis the Birch and Swinnerton-Dyer Conjecture is one of the seven Millenium Prize Problems posed by the Clay Institute with a prize of one million dollars for the solution of each. For each positive square-free number n another number is computed by Tunnell’s formula. If ...
... Along with the Riemann Hypothesis the Birch and Swinnerton-Dyer Conjecture is one of the seven Millenium Prize Problems posed by the Clay Institute with a prize of one million dollars for the solution of each. For each positive square-free number n another number is computed by Tunnell’s formula. If ...
Full text
... Clearly, we must show that V>i = V2 for all p. In fact, we show that both assumptions V>i < V>2 and V2 < V i lead to contradictions, so the desired equality must hold. Actually, the proof is not elegant. Since we can use neither symmetry nor rotation arguments, it is necessary to consider individual ...
... Clearly, we must show that V>i = V2 for all p. In fact, we show that both assumptions V>i < V>2 and V2 < V i lead to contradictions, so the desired equality must hold. Actually, the proof is not elegant. Since we can use neither symmetry nor rotation arguments, it is necessary to consider individual ...
On Sets Which Are Measured bar Multiples of Irrational Numbers
... fr In : n satisfies 0) = fim 1 in : n satisfies 0, n C N} N provided this limit exists . (A denotes the power of A) . We say that a set A (A C [0, 1)) belongs to the class ű if for every irrational 1 the frequency fr In : n~ e A (mod 1)} exists and does not depend on the choice of ~ . It is well-kno ...
... fr In : n satisfies 0) = fim 1 in : n satisfies 0, n C N} N provided this limit exists . (A denotes the power of A) . We say that a set A (A C [0, 1)) belongs to the class ű if for every irrational 1 the frequency fr In : n~ e A (mod 1)} exists and does not depend on the choice of ~ . It is well-kno ...