ODD PERFECT NUMBERS, DIOPHANTINE EQUATIONS, AND
... One of the oldest unsolved problems in mathematics is whether there exists an odd perfect number N . There are many roadblocks to the existence of such a number. For instance, from [14] we now know that N > 101500 and N has at least 101 prime factors (counting multiplicity). If k is the number of k ...
... One of the oldest unsolved problems in mathematics is whether there exists an odd perfect number N . There are many roadblocks to the existence of such a number. For instance, from [14] we now know that N > 101500 and N has at least 101 prime factors (counting multiplicity). If k is the number of k ...
Odd prime values of the Ramanujan tau function
... Remark 4 Considering the list p = 11, 17, 29, 41, 47, 59, 79, 89, 97, . . . for which we know LR (probable) primes, it is remarkable that the six first values correspond exactly to the odd values in the sequence of the Ramanujan primes: 2, 11, 17, 29, 41, 47, 59, 67, 71, 97, . . . . This sequence wa ...
... Remark 4 Considering the list p = 11, 17, 29, 41, 47, 59, 79, 89, 97, . . . for which we know LR (probable) primes, it is remarkable that the six first values correspond exactly to the odd values in the sequence of the Ramanujan primes: 2, 11, 17, 29, 41, 47, 59, 67, 71, 97, . . . . This sequence wa ...
here - Clemson University
... The most natural way to calculate this is to say: there are 20 choices for who goes first, 19 choices for who goes next, 18 choices for the third person, and so on. But we can also consider it from the student’s perspective. There are 20 slots. There are 20 slots for Adam, then 19 slots for Beth, th ...
... The most natural way to calculate this is to say: there are 20 choices for who goes first, 19 choices for who goes next, 18 choices for the third person, and so on. But we can also consider it from the student’s perspective. There are 20 slots. There are 20 slots for Adam, then 19 slots for Beth, th ...
FOR STARTERS
... To isolate the variable, simplify each side if applicable, and then undo the operations. ...
... To isolate the variable, simplify each side if applicable, and then undo the operations. ...
Lecture Notes - Department of Mathematics
... calculus class are tame. Let Ω be the set of all tame sets. (Ω being the last letter of the Greek alphabet is often used to denote “large” sets.) Is Ω tame? If it is, then it belongs to Ω and hence, Ω is wild. On the other hand, if Ω is wild then Ω is its element. Since by definition, all elements o ...
... calculus class are tame. Let Ω be the set of all tame sets. (Ω being the last letter of the Greek alphabet is often used to denote “large” sets.) Is Ω tame? If it is, then it belongs to Ω and hence, Ω is wild. On the other hand, if Ω is wild then Ω is its element. Since by definition, all elements o ...
MULTIPLICATIVE SEMIGROUPS RELATED TO THE 3x + 1
... that after adding a finite number of generators to Wq , we can obtain the entire group Q∗ . We can conceive of a “qx + 1 conjecture” as stating that sufficient iteration of this function, starting from any positive integer n, eventually reaches 1. As in the case q = 3, the fact that Wq is large woul ...
... that after adding a finite number of generators to Wq , we can obtain the entire group Q∗ . We can conceive of a “qx + 1 conjecture” as stating that sufficient iteration of this function, starting from any positive integer n, eventually reaches 1. As in the case q = 3, the fact that Wq is large woul ...
Proof of a theorem of Fermat that every prime number of the form 4n
... not divisible by the prime number 4n + 1, from which it follows in turn that certainly not all first differences are divisible by 4n + 1. 6. By virtue of which, the power of this proof is better observed, it is to be noted that the difference of order 2n is produced from 2n + 1 terms of the given s ...
... not divisible by the prime number 4n + 1, from which it follows in turn that certainly not all first differences are divisible by 4n + 1. 6. By virtue of which, the power of this proof is better observed, it is to be noted that the difference of order 2n is produced from 2n + 1 terms of the given s ...
Some simple continued fraction expansions for an infinite product
... fraction is a continued fraction in which a0 is an integer, all the partial numerators are equal to 1 and each partial denominator is a positive integer. We recall (see for example [2, Theorem 14]) that every positive irrational real number has ...
... fraction is a continued fraction in which a0 is an integer, all the partial numerators are equal to 1 and each partial denominator is a positive integer. We recall (see for example [2, Theorem 14]) that every positive irrational real number has ...
Orders of Growth - UConn Math
... Some notation to convey dominanting rates of growth will be convenient. For two sequences xn and yn , write xn ≺ yn to mean xn /yn → 0 as n → ∞. In other words, xn grows substantially slower than yn (if it just grew at half the rate, √ for instance, then xn /yn would be around 1/2 rather than ...
... Some notation to convey dominanting rates of growth will be convenient. For two sequences xn and yn , write xn ≺ yn to mean xn /yn → 0 as n → ∞. In other words, xn grows substantially slower than yn (if it just grew at half the rate, √ for instance, then xn /yn would be around 1/2 rather than ...
Pythagorean Triples Challenge - Virtual Commons
... If you know a lot about Pythagorean triples, feel free to skip right to the 5 challenge problems at the end of this article. Otherwise, here is some background information. A Pythagorean triple (a, b, c) is a triple of positive integers that can be used to form the sides of a right triangle with leg ...
... If you know a lot about Pythagorean triples, feel free to skip right to the 5 challenge problems at the end of this article. Otherwise, here is some background information. A Pythagorean triple (a, b, c) is a triple of positive integers that can be used to form the sides of a right triangle with leg ...
On the b-ary Expansion of an Algebraic Number.
... where k0 0, a k0 6 0 if k0 > 0, the ak 's are integers from f0; 1; . . . ; b 1g and ak is non-zero for infinitely many indices k. The sequence (ak )k k0 is uniquely determined by u: it is its b-ary expansion. We then define the function nbdc, `number of digit changes', by nbdc(n; u; b) Cardf1 ...
... where k0 0, a k0 6 0 if k0 > 0, the ak 's are integers from f0; 1; . . . ; b 1g and ak is non-zero for infinitely many indices k. The sequence (ak )k k0 is uniquely determined by u: it is its b-ary expansion. We then define the function nbdc, `number of digit changes', by nbdc(n; u; b) Cardf1 ...
Chapter 8.1 – 8.5 - MIT OpenCourseWare
... had computed. After simplifying, we were left with a linear combination of a and b equal to rem.x; y/, as desired. The final solution is boxed. This should make it pretty clear how and why the Pulverizer works. If you have doubts, it may help to work through Problem 8.13, where the Pulverizer is for ...
... had computed. After simplifying, we were left with a linear combination of a and b equal to rem.x; y/, as desired. The final solution is boxed. This should make it pretty clear how and why the Pulverizer works. If you have doubts, it may help to work through Problem 8.13, where the Pulverizer is for ...
Sample Chapter
... the number of nodes is one hence the statement is true for the base case. ii. Induction Hypothesis: It is assumed that S is true for some value k (i.e.) if a tree contains k edges then it contains (k + 1) nodes. iii. Induction Step: It is to be proved that if a tree contains (k + 1) edges then it co ...
... the number of nodes is one hence the statement is true for the base case. ii. Induction Hypothesis: It is assumed that S is true for some value k (i.e.) if a tree contains k edges then it contains (k + 1) nodes. iii. Induction Step: It is to be proved that if a tree contains (k + 1) edges then it co ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".