Balancing Numbers and Application
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POWER FIBONACCI SEQUENCES 1. Introduction Let G Be a Bi-Infinite Integer Sequence Satisfying the Recurrence Relation G If
POWER FIBONACCI SEQUENCES JOSHUA IDE AND MARC S. RENAULT Abstract. We examine integer sequences G satisfying the Fibonacci recurrence relation 2 3 Gn = Gn−1 + Gn−2 that also have the property that G ≡ 1; a; a ; a ;::: (mod m) for some modulus m. We determine those moduli m for which these power Fibonacci sequences exist and the number of such sequences for a given m. We also provide formulas for the periods of these sequences, based on the period of the Fibonacci sequence F modulo m. Finally, we establish certain sequence/subsequence relationships between power Fibonacci sequences. 1. Introduction Let G be a bi-infinite integer sequence satisfying the recurrence relation Gn = Gn−1 +Gn−2. If G ≡ 1; a; a2; a3;::: (mod m) for some modulus m, then we will call G a power Fibonacci sequence modulo m. Example 1.1. Modulo m = 11, there are two power Fibonacci sequences: 1; 8; 9; 6; 4; 10; 3; 2; 5; 7; 1; 8 ::: and 1; 4; 5; 9; 3; 1; 4;::: Curiously, the second is a subsequence of the first. For modulo 5 there is only one such sequence (1; 3; 4; 2; 1; 3; :::), for modulo 10 there are no such sequences, and for modulo 209 there are four of these sequences. In the next section, Theorem 2.1 will demonstrate that 209 = 11·19 is the smallest modulus with more than two power Fibonacci sequences. In this note we will determine those moduli for which power Fibonacci sequences exist, and how many power Fibonacci sequences there are for a given modulus. -
Triangular Numbers /, 3,6, 10, 15, ", Tn,'" »*"
TRIANGULAR NUMBERS V.E. HOGGATT, JR., and IVIARJORIE BICKWELL San Jose State University, San Jose, California 9111112 1. INTRODUCTION To Fibonacci is attributed the arithmetic triangle of odd numbers, in which the nth row has n entries, the cen- ter element is n* for even /?, and the row sum is n3. (See Stanley Bezuszka [11].) FIBONACCI'S TRIANGLE SUMS / 1 =:1 3 3 5 8 = 2s 7 9 11 27 = 33 13 15 17 19 64 = 4$ 21 23 25 27 29 125 = 5s We wish to derive some results here concerning the triangular numbers /, 3,6, 10, 15, ", Tn,'" »*". If one o b - serves how they are defined geometrically, 1 3 6 10 • - one easily sees that (1.1) Tn - 1+2+3 + .- +n = n(n±M and (1.2) • Tn+1 = Tn+(n+1) . By noticing that two adjacent arrays form a square, such as 3 + 6 = 9 '.'.?. we are led to 2 (1.3) n = Tn + Tn„7 , which can be verified using (1.1). This also provides an identity for triangular numbers in terms of subscripts which are also triangular numbers, T =T + T (1-4) n Tn Tn-1 • Since every odd number is the difference of two consecutive squares, it is informative to rewrite Fibonacci's tri- angle of odd numbers: 221 222 TRIANGULAR NUMBERS [OCT. FIBONACCI'S TRIANGLE SUMS f^-O2) Tf-T* (2* -I2) (32-22) Ti-Tf (42-32) (52-42) (62-52) Ti-Tl•2 (72-62) (82-72) (9*-82) (Kp-92) Tl-Tl Upon comparing with the first array, it would appear that the difference of the squares of two consecutive tri- angular numbers is a perfect cube. -
Problems in Relating Various Tasks and Their Sample Solutions to Bloomâ
The Mathematics Enthusiast Volume 14 Number 1 Numbers 1, 2, & 3 Article 4 1-2017 Problems in relating various tasks and their sample solutions to Bloom’s taxonomy Torsten Lindstrom Follow this and additional works at: https://scholarworks.umt.edu/tme Part of the Mathematics Commons Let us know how access to this document benefits ou.y Recommended Citation Lindstrom, Torsten (2017) "Problems in relating various tasks and their sample solutions to Bloom’s taxonomy," The Mathematics Enthusiast: Vol. 14 : No. 1 , Article 4. Available at: https://scholarworks.umt.edu/tme/vol14/iss1/4 This Article is brought to you for free and open access by ScholarWorks at University of Montana. It has been accepted for inclusion in The Mathematics Enthusiast by an authorized editor of ScholarWorks at University of Montana. For more information, please contact [email protected]. TME, vol. 14, nos. 1,2&3, p. 15 Problems in relating various tasks and their sample solutions to Bloom’s taxonomy Torsten Lindstr¨om Linnaeus University, SWEDEN ABSTRACT: In this paper we analyze sample solutions of a number of problems and relate them to their level as prescribed by Bloom’s taxonomy. We relate these solutions to a number of other frameworks, too. Our key message is that it remains insufficient to analyze written forms of these tasks. We emphasize careful observations of how different students approach a solution before finally assessing the level of tasks used. We take the arithmetic series as our starting point and point out that the objective of the discussion of the examples here in no way is to indicate an optimal way towards a solution. -
Perfect Gaussian Integer Sequences of Arbitrary Length Soo-Chang Pei∗ and Kuo-Wei Chang† National Taiwan University∗ and Chunghwa Telecom†
Perfect Gaussian Integer Sequences of Arbitrary Length Soo-Chang Pei∗ and Kuo-Wei Chang† National Taiwan University∗ and Chunghwa Telecom† m Objectives N = p or N = p using Legendre N = pq where p and q are coprime Conclusion sequence and Gauss sum To construct perfect Gaussian integer sequences Simple zero padding method: We propose several methods to generate zero au- of arbitrary length N: Legendre symbol is defined as 1) Take a ZAC from p and q. tocorrelation sequences in Gaussian integer. If the 8 2 2) Interpolate q − 1 and p − 1 zeros to these signals sequence length is prime number, we can use Leg- • Perfect sequences are sequences with zero > 1, if ∃x, x ≡ n(mod N) n ! > autocorrelation. = < 0, n ≡ 0(mod N) to get two signals of length N. endre symbol and provide more degree of freedom N > 3) Convolution these two signals, then we get a than Yang’s method. If the sequence is composite, • Gaussian integer is a number in the form :> −1, otherwise. ZAC. we develop a general method to construct ZAC se- a + bi where a and b are integer. And the Gauss sum is defined as N−1 ! Using the idea of prime-factor algorithm: quences. Zero padding is one of the special cases of • A Perfect Gaussian sequence is a perfect X n −2πikn/N G(k) = e Recall that DFT of size N = N1N2 can be done by this method, and it is very easy to implement. sequence that each value in the sequence is a n=0 N taking DFT of size N1 and N2 seperately[14]. -
Physics 403, Spring 2011 Problem Set 2 Due Thursday, February 17
Physics 403, Spring 2011 Problem Set 2 due Thursday, February 17 1. A Differential Equation for the Legendre Polynomials (15 pts): In this prob- lem, we will derive a second order linear differential equation satisfied by the Legendre polynomials (−1)n dn P (x) = [(1 − x2)n] : n 2nn! dxn (a) Show that Z 1 2 0 0 Pm(x)[(1 − x )Pn(x)] dx = 0 for m 6= n : (1) −1 Conclude that 2 0 0 [(1 − x )Pn(x)] = λnPn(x) 0 for λn a constant. [ Prime here and in the following questions denotes d=dx.] (b) Perform the integral (1) for m = n to find a relation for λn in terms of the leading n coefficient of the Legendre polynomial Pn(x) = knx + :::. What is kn? 2. A Generating Function for the Legendre Polynomials (20 pts): In this problem, we will derive a generating function for the Legendre polynomials. (a) Verify that the Legendre polynomials satisfy the recurrence relation 0 0 0 Pn − 2xPn−1 + Pn−2 − Pn−1 = 0 : (b) Consider the generating function 1 X n g(x; t) = t Pn(x) : n=0 Use the recurrence relation above to show that the generating function satisfies the first order differential equation d (1 − 2xt + t2) g(x; t) = tg(x; t) : dx Determine the constant of integration by using the fact that Pn(1) = 1. (c) Use the generating function to express the potential for a point particle q jr − r0j in terms of Legendre polynomials. 1 3. The Dirac delta function (15 pts): Dirac delta functions are useful mathematical tools in quantum mechanics. -
Polynomial Sequences Generated by Linear Recurrences
Innocent Ndikubwayo Polynomial Sequences Generated by Linear Recurrences: Location and Reality of Zeros Polynomial Sequences Generated by Linear Recurrences: Location and Reality of Zeros Linear Recurrences: Location by Sequences Generated Polynomial Innocent Ndikubwayo ISBN 978-91-7911-462-6 Department of Mathematics Doctoral Thesis in Mathematics at Stockholm University, Sweden 2021 Polynomial Sequences Generated by Linear Recurrences: Location and Reality of Zeros Innocent Ndikubwayo Academic dissertation for the Degree of Doctor of Philosophy in Mathematics at Stockholm University to be publicly defended on Friday 14 May 2021 at 15.00 in sal 14 (Gradängsalen), hus 5, Kräftriket, Roslagsvägen 101 and online via Zoom, public link is available at the department website. Abstract In this thesis, we study the problem of location of the zeros of individual polynomials in sequences of polynomials generated by linear recurrence relations. In paper I, we establish the necessary and sufficient conditions that guarantee hyperbolicity of all the polynomials generated by a three-term recurrence of length 2, whose coefficients are arbitrary real polynomials. These zeros are dense on the real intervals of an explicitly defined real semialgebraic curve. Paper II extends Paper I to three-term recurrences of length greater than 2. We prove that there always exist non- hyperbolic polynomial(s) in the generated sequence. We further show that with at most finitely many known exceptions, all the zeros of all the polynomials generated by the recurrence lie and are dense on an explicitly defined real semialgebraic curve which consists of real intervals and non-real segments. The boundary points of this curve form a subset of zero locus of the discriminant of the characteristic polynomial of the recurrence. -
Linear Recurrence Relations: the Theory Behind Them
Linear Recurrence Relations: The Theory Behind Them David Soukup October 29, 2019 Linear Recurrence Relations Contents 1 Foreword ii 2 The matrix diagonalization method 1 3 Generating functions 3 4 Analogies to ODEs 6 5 Exercises 8 6 References 10 i Linear Recurrence Relations 1 Foreword This guide is intended mostly for students in Math 61 who are looking for a more theoretical background to the solving of linear recurrence relations. A typical problem encountered is the following: suppose we have a sequence defined by an = 2an−1 + 3an−2 where a0 = 0; a1 = 8: Certainly this recurrence defines the sequence fang unambiguously (at least for positive integers n), and we can compute the first several terms without much problem: a0 = 0 a1 = 8 a2 = 2 · a1 + 3 · a0 = 2 · 8 + 3 · 0 = 16 a3 = 2 · a2 + 3 · a1 = 2 · 16 + 3 · 8 = 56 a4 = 2 · a3 + 3 · a2 = 2 · 56 + 3 · 16 = 160 . While this does allow us to compute the value of an for small n, this is unsatisfying for several reasons. For one, computing the value of a100 by this method would be time-consuming. If we solely use the recurrence, we would have to find the previous 99 values of the sequence first. Moreover, until we compute it we have no good idea of how large a100 will be. We can observe from the table above that an grows quite rapidly as n increases, but we do not know how to extrapolate this rate of growth to all n. Most importantly, we would like to know the order of growth of this function. -
Arxiv:1804.04198V1 [Math.NT] 6 Apr 2018 Rm-Iesequence, Prime-Like Hoy H Function the Theory
CURIOUS CONJECTURES ON THE DISTRIBUTION OF PRIMES AMONG THE SUMS OF THE FIRST 2n PRIMES ROMEO MESTROVIˇ C´ ∞ ABSTRACT. Let pn be nth prime, and let (Sn)n=1 := (Sn) be the sequence of the sums 2n of the first 2n consecutive primes, that is, Sn = k=1 pk with n = 1, 2,.... Heuristic arguments supported by the corresponding computational results suggest that the primes P are distributed among sequence (Sn) in the same way that they are distributed among positive integers. In other words, taking into account the Prime Number Theorem, this assertion is equivalent to # p : p is a prime and p = Sk for some k with 1 k n { ≤ ≤ } log n # p : p is a prime and p = k for some k with 1 k n as n , ∼ { ≤ ≤ }∼ n → ∞ where S denotes the cardinality of a set S. Under the assumption that this assertion is | | true (Conjecture 3.3), we say that (Sn) satisfies the Restricted Prime Number Theorem. Motivated by this, in Sections 1 and 2 we give some definitions, results and examples concerning the generalization of the prime counting function π(x) to increasing positive integer sequences. The remainder of the paper (Sections 3-7) is devoted to the study of mentioned se- quence (Sn). Namely, we propose several conjectures and we prove their consequences concerning the distribution of primes in the sequence (Sn). These conjectures are mainly motivated by the Prime Number Theorem, some heuristic arguments and related computational results. Several consequences of these conjectures are also established. 1. INTRODUCTION, MOTIVATION AND PRELIMINARIES An extremely difficult problem in number theory is the distribution of the primes among the natural numbers. -
Some Combinatorics of Factorial Base Representations
1 2 Journal of Integer Sequences, Vol. 23 (2020), 3 Article 20.3.3 47 6 23 11 Some Combinatorics of Factorial Base Representations Tyler Ball Joanne Beckford Clover Park High School University of Pennsylvania Lakewood, WA 98499 Philadelphia, PA 19104 USA USA [email protected] [email protected] Paul Dalenberg Tom Edgar Department of Mathematics Department of Mathematics Oregon State University Pacific Lutheran University Corvallis, OR 97331 Tacoma, WA 98447 USA USA [email protected] [email protected] Tina Rajabi University of Washington Seattle, WA 98195 USA [email protected] Abstract Every non-negative integer can be written using the factorial base representation. We explore certain combinatorial structures arising from the arithmetic of these rep- resentations. In particular, we investigate the sum-of-digits function, carry sequences, and a partial order referred to as digital dominance. Finally, we describe an analog of a classical theorem due to Kummer that relates the combinatorial objects of interest by constructing a variety of new integer sequences. 1 1 Introduction Kummer’s theorem famously draws a connection between the traditional addition algorithm of base-p representations of integers and the prime factorization of binomial coefficients. Theorem 1 (Kummer). Let n, m, and p all be natural numbers with p prime. Then the n+m exponent of the largest power of p dividing n is the sum of the carries when adding the base-p representations of n and m. Ball et al. [2] define a new class of generalized binomial coefficients that allow them to extend Kummer’s theorem to base-b representations when b is not prime, and they discuss connections between base-b representations and a certain partial order, known as the base-b (digital) dominance order. -
Analysis of Non-Linear Recurrence Relations for the Recurrence Coefficients of Generalized Charlier Polynomials
Journal of Nonlinear Mathematical Physics Volume 10, Supplement 2 (2003), 231–237 SIDE V Analysis of Non-Linear Recurrence Relations for the Recurrence Coefficients of Generalized Charlier Polynomials Walter VAN ASSCHE † and Mama FOUPOUAGNIGNI ‡ † Katholieke Universiteit Leuven, Department of Mathematics, Celestijnenlaan 200 B, B-3001 Leuven, Belgium E-mail: [email protected] ‡ University of Yaounde, Advanced School of Education, Department of Mathematics, P.O. Box 47, Yaounde, Cameroon E-mail: [email protected] This paper is part of the Proceedings of SIDE V; Giens, June 21-26, 2002 Abstract The recurrence coefficients of generalized Charlier polynomials satisfy a system of non- linear recurrence relations. We simplify the recurrence relations, show that they are related to certain discrete Painlev´e equations, and analyze the asymptotic behaviour. 1 Introduction Orthogonal polynomials on the real line always satisfy a three-term recurrence relation. For monic polynomials Pn this is of the form Pn+1(x)=(x − βn)Pn(x) − γnPn−1(x), (1.1) with initial values P0 = 1 and P−1 = 0. For the orthonormal polynomials pn the recurrence relation is xpn(x)=an+1pn+1(x)+bnpn(x)+anpn−1(x), (1.2) 2 where an = γn and bn = βn. The recurrence coefficients are given by the integrals 2 an = xpn(x)pn−1(x) dµ(x),bn = xpn(x) dµ(x) and can also be expressed in terms of determinants containing the moments of the orthog- onality measure µ [11]. For classical orthogonal polynomials one knows these recurrence coefficients explicitly, but when one uses non-classical weights, then often one does not Copyright c 2003 by W Van Assche and M Foupouagnigni 232 W Van Assche and M Foupouagnigni know the recurrence coefficients explicitly. -
Figurate Numbers
Figurate Numbers by George Jelliss June 2008 with additions November 2008 Visualisation of Numbers The visual representation of the number of elements in a set by an array of small counters or other standard tally marks is still seen in the symbols on dominoes or playing cards, and in Roman numerals. The word "calculus" originally meant a small pebble used to calculate. Bear with me while we begin with a few elementary observations. Any number, n greater than 1, can be represented by a linear arrangement of n counters. The cases of 1 or 0 counters can be regarded as trivial or degenerate linear arrangements. The counters that make up a number m can alternatively be grouped in pairs instead of ones, and we find there are two cases, m = 2.n or 2.n + 1 (where the dot denotes multiplication). Numbers of these two forms are of course known as even and odd respectively. An even number is the sum of two equal numbers, n+n = 2.n. An odd number is the sum of two successive numbers 2.n + 1 = n + (n+1). The even and odd numbers alternate. Figure 1. Representation of numbers by rows of counters, and of even and odd numbers by various, mainly symmetric, formations. The right-angled (L-shaped) formation of the odd numbers is known as a gnomon. These do not of course exhaust the possibilities. 1 2 3 4 5 6 7 8 9 n 2 4 6 8 10 12 14 2.n 1 3 5 7 9 11 13 15 2.n + 1 Triples, Quadruples and Other Forms Generalising the divison into even and odd numbers, the counters making up a number can of course also be grouped in threes or fours or indeed any nonzero number k. -
TRIANGULAR REPUNIT—THERE IS but 1 a Repunit Is Any Integer That
Czechoslovak Mathematical Journal, 60 (135) (2010), 1075–1077 TRIANGULAR REPUNIT—THERE IS BUT 1 John H. Jaroma, Ave Maria (Received June 27, 2009) Abstract. In this paper, we demonstrate that 1 is the only integer that is both triangular and a repunit. Keywords: repunit, triangular number MSC 2010 : 11A99 A repunit is any integer that can be written in decimal notation as a string of 1’s. Examples include 1, 11, 111, 1111, 11111,... Furthermore, if we denote rn to be the nth repunit then it follows that 10n − 1 (1) rn = . 9 Triangular numbers are those integers that can be represented by the number of dots evenly arranged in an equilateral triangle. More specifically, they are the numbers 1, 3, 6, 10, 15,... It follows that the nth triangular number, tn is the sum of the first n consecutive integers beginning with 1. In particular, for n > 1 n(n + 1) (2) tn = . 2 Now, the number 1 is both triangular and a repunit. It is the objective of this note to illustrate that 1 is the only such number. To this end, we state and prove the following lemma. It incorporates a result that [1] asserts had been known by the early Pythagoreans. Appearing in Platonic Questions, Plutarch states that eight times any triangular number plus one is a square. The result is actually necessary and sufficient. We demonstrate it as such here. 1075 Lemma. The integer n is triangular if and only if 8n +1 is a square. Proof. If n = tn, then by (2), n(n + 1) 8 +1=4n2 +4n +1=(2n + 1)2.