Diophantine Representation of the Fibonacci Numbers
... using Lemma 4, to obtain the lower inequality. Since x and y are integers, (6) implies that equation (4) must hold. According to Theorem 1, y must be a Fibonacci number. Equations (4) and (5) imply that w = y. Therefore w is a Fibonacci number. This completes the proof of the theorem. (Putnam's meth ...
... using Lemma 4, to obtain the lower inequality. Since x and y are integers, (6) implies that equation (4) must hold. According to Theorem 1, y must be a Fibonacci number. Equations (4) and (5) imply that w = y. Therefore w is a Fibonacci number. This completes the proof of the theorem. (Putnam's meth ...
2-1: Patterns and Inductive Reasoning 45, 40, 35, 30,... 25, 20
... What are the next two terms in each sequence? a. 45, 40,35,30,... ...
... What are the next two terms in each sequence? a. 45, 40,35,30,... ...
Lecture22 – Finish Knaves and Fib
... Assume not, that is, assume B is a knave. Then what B says is false, so it is false that at most two are knaves. So it must be that all three are knaves. We didn’t Then A is a knave. need this step So what A says is false, and so there are zero knaves. because we But all three are knaves and zero ar ...
... Assume not, that is, assume B is a knave. Then what B says is false, so it is false that at most two are knaves. So it must be that all three are knaves. We didn’t Then A is a knave. need this step So what A says is false, and so there are zero knaves. because we But all three are knaves and zero ar ...
Pseudoprimes and Carmichael Numbers, by Emily Riemer
... (n − 1) − (p − 1) and also divides n − 1, then it must divide p − 1. But we have shown that q − 1 - p − 1. Therefore, we have a contradiction, and n cannot be a Carmichael number. It was not proven until 1994 that there are infinitely many Carmichael numbers, despite the fact that mathematicians hav ...
... (n − 1) − (p − 1) and also divides n − 1, then it must divide p − 1. But we have shown that q − 1 - p − 1. Therefore, we have a contradiction, and n cannot be a Carmichael number. It was not proven until 1994 that there are infinitely many Carmichael numbers, despite the fact that mathematicians hav ...
Full text
... regarded as an anomalous numerical curiosity, possibly related to the fact that 89 is a Fibonacci number (see Remark in [5]), but not generalizing to other fractions in an obvious manner. In 1980, C. F. Winans [6] showed that the sums £ 10"(/c+ 1)Fa^ approximate 1/71, 2/59, and 3/31 for a = 2, 3, an ...
... regarded as an anomalous numerical curiosity, possibly related to the fact that 89 is a Fibonacci number (see Remark in [5]), but not generalizing to other fractions in an obvious manner. In 1980, C. F. Winans [6] showed that the sums £ 10"(/c+ 1)Fa^ approximate 1/71, 2/59, and 3/31 for a = 2, 3, an ...
1 Lecture 1
... Undefined terms: set of numbers N = {1, 2, 3, ...} with the property that if n, m ∈ N then n + m ∈ N and nm ∈ N. 1. Axiom 1 1 belongs to N : there is a number 1 such that 1 ∗ n ∈ N for any n ∈ N. 2. Axiom2 There is no number n such that n + 1 = 1. 3. Axiom 3 If S is a subset of N with the property t ...
... Undefined terms: set of numbers N = {1, 2, 3, ...} with the property that if n, m ∈ N then n + m ∈ N and nm ∈ N. 1. Axiom 1 1 belongs to N : there is a number 1 such that 1 ∗ n ∈ N for any n ∈ N. 2. Axiom2 There is no number n such that n + 1 = 1. 3. Axiom 3 If S is a subset of N with the property t ...
On the error term in a Parseval type formula in the theory of Ramanujan expansions,
... b) For fixed n, cr (n) is a multiplicative function i.e. for r1 , r2 with gcd(r1 , r2 ) = 1 we have cr1 r2 (n) = cr1 (n)cr2 (n). This is essentially due to the fact that, for r1 , r2 with gcd(r1 , r2 ) = 1, the fields Q(ζr1 ) and Q(ζr2 ) are linearly disjoint. c) cr (·) is a periodic function with per ...
... b) For fixed n, cr (n) is a multiplicative function i.e. for r1 , r2 with gcd(r1 , r2 ) = 1 we have cr1 r2 (n) = cr1 (n)cr2 (n). This is essentially due to the fact that, for r1 , r2 with gcd(r1 , r2 ) = 1, the fields Q(ζr1 ) and Q(ζr2 ) are linearly disjoint. c) cr (·) is a periodic function with per ...
Logarithmic concave measures with application to stochastic programming
... From the point of view of numerical solution it is enough to suppose that h1 (x), . . . , hM (x) are quasiconcave. A function h(x) defined in a convex set L is quasi-concave if for any x1 , x2 ∈ L and 0 < λ < 1 we have h(λx1 + (1 − λ)x2 ) ≥ min{h(x1 ), h(x2 )}. ...
... From the point of view of numerical solution it is enough to suppose that h1 (x), . . . , hM (x) are quasiconcave. A function h(x) defined in a convex set L is quasi-concave if for any x1 , x2 ∈ L and 0 < λ < 1 we have h(λx1 + (1 − λ)x2 ) ≥ min{h(x1 ), h(x2 )}. ...
1. Introduction A fundamental problem in statistical and solid
... derived for the model (1.2), and it is shown that as the number of N of particles gets large, the empirical measure µN associated with any sequence of microscopic minimizers converges after re-scaling to the characteristic function of the unique Wulff shape. Further insight into the uniqueness quest ...
... derived for the model (1.2), and it is shown that as the number of N of particles gets large, the empirical measure µN associated with any sequence of microscopic minimizers converges after re-scaling to the characteristic function of the unique Wulff shape. Further insight into the uniqueness quest ...
Mathematica 2014
... number is evenly divisible by 6. If you want, you can keep adding numbers until only one digit remains and do the same thing. So in this case, 3 + 9 = 12 and 1 + 2 = 3, and 3 is evenly divisible by 3! ...
... number is evenly divisible by 6. If you want, you can keep adding numbers until only one digit remains and do the same thing. So in this case, 3 + 9 = 12 and 1 + 2 = 3, and 3 is evenly divisible by 3! ...
Questions about Powers of Numbers
... • Catalan, 1844 (cf. [R], [B2]). The numbers 8 and 9 are the only consecutive perfect powers. • Waring Problems. This is a host of problems having to do with the number of ways an integer can be written as a sum of k “perfect” n-th powers. One does not have to go far to come to an unsolved problem a ...
... • Catalan, 1844 (cf. [R], [B2]). The numbers 8 and 9 are the only consecutive perfect powers. • Waring Problems. This is a host of problems having to do with the number of ways an integer can be written as a sum of k “perfect” n-th powers. One does not have to go far to come to an unsolved problem a ...
Df-pn: Depth-first Proof Number Search
... • Masahiro Seo: The C* Algorithm for AND/OR Tree Search and its Application to a Tsume-Shogi Program, M.Sc. Thesis, Department of Information Science, University of Tokyo, 1995 • Ayumu Nagai and Hiroshi Imai: Proof for the Equivalence Between Some Best-First Algorithms and Depth-First Algorithms for ...
... • Masahiro Seo: The C* Algorithm for AND/OR Tree Search and its Application to a Tsume-Shogi Program, M.Sc. Thesis, Department of Information Science, University of Tokyo, 1995 • Ayumu Nagai and Hiroshi Imai: Proof for the Equivalence Between Some Best-First Algorithms and Depth-First Algorithms for ...
(1) (a) Prove that if an integer n has the form 6q + 5 for some q ∈ Z
... Of course these steps were not expressed in symbols but only applied to specific numbers. Nevertheless, a general method is implicit in the many specific cases solved. ...
... Of course these steps were not expressed in symbols but only applied to specific numbers. Nevertheless, a general method is implicit in the many specific cases solved. ...
On Determining the Irrationality of the Mean of a Random Variable.
... consider a stopping rule,3 largely because we envision the physicist taking action on the basis of his current knowledge while his research for the laws of nature continues. It is futile to stop this process and declare a given law fixed forever. In the context of the problem below it will be shown ...
... consider a stopping rule,3 largely because we envision the physicist taking action on the basis of his current knowledge while his research for the laws of nature continues. It is futile to stop this process and declare a given law fixed forever. In the context of the problem below it will be shown ...
MAA245 NUMBERS 1 Natural Numbers, N
... “=” (equals) is an equivalence relation; it partitions N into the sets {1}, {2}, {3}, . . .. We now define “>” (greater than) and “<” (less than). Given m, n ∈ N: m > n if ∃ k ∈ N such that m = n + k; m < n if ∃ p ∈ N such that m + p = n. It follows immediately that m > n if and only if n < m. The f ...
... “=” (equals) is an equivalence relation; it partitions N into the sets {1}, {2}, {3}, . . .. We now define “>” (greater than) and “<” (less than). Given m, n ∈ N: m > n if ∃ k ∈ N such that m = n + k; m < n if ∃ p ∈ N such that m + p = n. It follows immediately that m > n if and only if n < m. The f ...
![[Part 1]](http://s1.studyres.com/store/data/008795787_1-e53528a6e716c9914682cf7824ec41a5-300x300.png)
[Part 1]
... c. Apparently (due to multiple-precision requirements on digital computers) these representation formulas do not give any special advantage to their u s e r in computing automorphic numbers with large numbers of digits. ...
... c. Apparently (due to multiple-precision requirements on digital computers) these representation formulas do not give any special advantage to their u s e r in computing automorphic numbers with large numbers of digits. ...
On perfect and multiply perfect numbers
... (a(n), n) > f (x) is less than c ;xl(f(x)lcs for some c, > 0 and c, > 0. The same result hold if a ; n) is replaced by Euler' s :p function . We are not going to give the proof of Theorem 3. It can further be shown that Theorem 3 is best possible in the following sense : Let f (x) = o((log x), ) for ...
... (a(n), n) > f (x) is less than c ;xl(f(x)lcs for some c, > 0 and c, > 0. The same result hold if a ; n) is replaced by Euler' s :p function . We are not going to give the proof of Theorem 3. It can further be shown that Theorem 3 is best possible in the following sense : Let f (x) = o((log x), ) for ...
Primes. - Elad Aigner
... We are now in a position to prove Theorem 1.2. Proof of Theorem 1.2. The argument consists of two parts. First we will show that for every positive integer there exists at least one way to express the number as a product of primes. Second, we will show that this expression is in fact unique. Existen ...
... We are now in a position to prove Theorem 1.2. Proof of Theorem 1.2. The argument consists of two parts. First we will show that for every positive integer there exists at least one way to express the number as a product of primes. Second, we will show that this expression is in fact unique. Existen ...
Full text
... 2N +1 -» 2N +1 by T(x) = ^ - , where 2J 13x +1 and 2j+l \ 3x +1. The famous 3x +1 Conjecture asserts that, for any x e 2 N + l , there exists I : G N satisfying Tk(x) = 1. Define the least whole number k for which Tk(x) = 1 as the total stopping time a(x) of x, and call the sequence of iterates (x, ...
... 2N +1 -» 2N +1 by T(x) = ^ - , where 2J 13x +1 and 2j+l \ 3x +1. The famous 3x +1 Conjecture asserts that, for any x e 2 N + l , there exists I : G N satisfying Tk(x) = 1. Define the least whole number k for which Tk(x) = 1 as the total stopping time a(x) of x, and call the sequence of iterates (x, ...
EULER’S THEOREM 1. Introduction
... sequence, the 12 numerators fall into two sets of size 6: {1, 3, 4, 9, 10, 12} and {2, 5, 6, 7, 8, 11}. Is there some significance to these two sets of numbers? More generally, can you explain how many digit sequences (up to shifting) will occur among all the reduced fractions with a given denominat ...
... sequence, the 12 numerators fall into two sets of size 6: {1, 3, 4, 9, 10, 12} and {2, 5, 6, 7, 8, 11}. Is there some significance to these two sets of numbers? More generally, can you explain how many digit sequences (up to shifting) will occur among all the reduced fractions with a given denominat ...