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Special Kinds of Numbers
Chapel Hill Math Circle: Special Kinds of Numbers 11/4/2017 1 Introduction This worksheet is concerned with three classes of special numbers: the polyg- onal numbers, the Catalan numbers, and the Lucas numbers. Polygonal numbers are a nice geometric generalization of perfect square integers, Cata- lan numbers are important numbers that (one way or another) show up in a wide variety of combinatorial and non-combinatorial situations, and Lucas numbers are a slight twist on our familiar Fibonacci numbers. The purpose of this worksheet is not to give an exhaustive treatment of any of these numbers, since that is just not possible. The goal is to, instead, introduce these interesting classes of numbers and give some motivation to explore on your own. 2 Polygonal Numbers The first class of numbers we'll consider are called polygonal numbers. We'll start with some simpler cases, and then derive a general formula. Take 3 points and arrange them as an equilateral triangle with side lengths of 2. Here, I use the term \side-length" to mean \how many points are in a side." The total number of points in this triangle (3) is the second triangular number, denoted P (3; 2). To get the third triangular number P (3; 3), we take the last configuration and adjoin points so that the outermost shape is an equilateral triangle with a side-length of 3. To do this, we can add 3 points along any one of the edges. Since there are 3 points in the original 1 triangle, and we added 3 points to get the next triangle, the total number of points in this step is P (3; 3) = 3 + 3 = 6. -
Catalan Numbers Modulo 2K
1 2 Journal of Integer Sequences, Vol. 13 (2010), 3 Article 10.5.4 47 6 23 11 Catalan Numbers Modulo 2k Shu-Chung Liu1 Department of Applied Mathematics National Hsinchu University of Education Hsinchu, Taiwan [email protected] and Jean C.-C. Yeh Department of Mathematics Texas A & M University College Station, TX 77843-3368 USA Abstract In this paper, we develop a systematic tool to calculate the congruences of some combinatorial numbers involving n!. Using this tool, we re-prove Kummer’s and Lucas’ theorems in a unique concept, and classify the congruences of the Catalan numbers cn (mod 64). To achieve the second goal, cn (mod 8) and cn (mod 16) are also classified. Through the approach of these three congruence problems, we develop several general properties. For instance, a general formula with powers of 2 and 5 can evaluate cn (mod k 2 ) for any k. An equivalence cn ≡2k cn¯ is derived, wheren ¯ is the number obtained by partially truncating some runs of 1 and runs of 0 in the binary string [n]2. By this equivalence relation, we show that not every number in [0, 2k − 1] turns out to be a k residue of cn (mod 2 ) for k ≥ 2. 1 Introduction Throughout this paper, p is a prime number and k is a positive integer. We are interested in enumerating the congruences of various combinatorial numbers modulo a prime power 1Partially supported by NSC96-2115-M-134-003-MY2 1 k q := p , and one of the goals of this paper is to classify Catalan numbers cn modulo 64. -
M-Gonal Number± N = Nasty Number, N =
International Journal of Research in Advent Technology, Vol.6, No.11, November 2018 E-ISSN: 2321-9637 Available online at www.ijrat.org M-Gonal number n = Nasty number, n = 1,2 Dr.K.Geetha Assistant Professor, Department of Mathematics, Dr.Umayal Ramanathan College for Women, Karaikudi. [email protected] Abstract: This paper concerns with study of obtaining the ranks of M-gonal numbers along with their recurrence relation such that M-gonal number n = Nasty number where n = 1,2. Key Words: Triangular number, Pentagonal number, Hexagonal number, Octagonal number and Dodecagonal number 1. INTRODUCTION may refer [3-7]. In this communication an attempt is Numbers exhibit fascinating properties, they form made to obtain the ranks of M-gonal numbers like patterns and so on [1]. In [3] the ranks of Triangular, Triangular, Pentagonal, Hexagonal, Octagonal and Pentagonal, Heptagonal, Nanogonal and Tridecagonal Dodecagonal numbers such that each of these M-gonal numbers such that each of these M-gonal number - number n = Nasty number, n = 1,2. Also the 2 = a perfect square and the ranks of Pentagonal and recurrence relations satisfied by the ranks of each of Heptagonal such that each of these M-gonal number + these M-gonal numbers are presented. 2= a perfect square are obtained. In this context one 2. METHOD OF ANALYSIS 2.1.Pattern 1: Let the rank of the nth Triangular number be A, then the identity 2 Triangular number – 2 = 6 x (1) is written as y22 48x 9 (2) where y 2A 1 (3) whose initial solution is x30 , y0 21 (4) Let xyss, be the general solution of the Pellian 22 yx48 1 (5) 1 ss11 where xs 7 48 7 48 2 48 1 ss11 ys 7 48 7 48 , s 0,1,... -
Ces`Aro's Integral Formula for the Bell Numbers (Corrected)
Ces`aro’s Integral Formula for the Bell Numbers (Corrected) DAVID CALLAN Department of Statistics University of Wisconsin-Madison Medical Science Center 1300 University Ave Madison, WI 53706-1532 [email protected] October 3, 2005 In 1885, Ces`aro [1] gave the remarkable formula π 2 cos θ N = ee cos(sin θ)) sin( ecos θ sin(sin θ) ) sin pθ dθ p πe Z0 where (Np)p≥1 = (1, 2, 5, 15, 52, 203,...) are the modern-day Bell numbers. This formula was reproduced verbatim in the Editorial Comment on a 1941 Monthly problem [2] (the notation Np for Bell number was still in use then). I have not seen it in recent works and, while it’s not very profound, I think it deserves to be better known. Unfortunately, it contains a typographical error: a factor of p! is omitted. The correct formula, with n in place of p and using Bn for Bell number, is π 2 n! cos θ B = ee cos(sin θ)) sin( ecos θ sin(sin θ) ) sin nθ dθ n ≥ 1. n πe Z0 eiθ The integrand is the imaginary part of ee sin nθ, and so an equivalent formula is π 2 n! eiθ B = Im ee sin nθ dθ . (1) n πe Z0 The formula (1) is quite simple to prove modulo a few standard facts about set par- n titions. Recall that the Stirling partition number k is the number of partitions of n n [n] = {1, 2,...,n} into k nonempty blocks and the Bell number Bn = k=1 k counts n k n n k k all partitions of [ ]. -
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. -
An Heptagonal Numbers
© April 2021| IJIRT | Volume 7 Issue 11 | ISSN: 2349-6002 An Heptagonal Numbers Dr. S. Usha Assistant Professor in Mathematics, Bon Secours Arts and Science College for Women, Mannargudi, Affiliated by Bharathidasan University, Tiruchirappalli, Tamil Nadu Abstract - Two Results of interest (i)There exists an prime factor that divides n is one. First few square free infinite pairs of heptagonal numbers (푯풎 , 푯풌) such that number are 1,2,3,5,6,7,10,11,13,14,15,17…… their ratio is equal to a non –zero square-free integer and (ii)the general form of the rank of square heptagonal Definition(1.5): ퟑ ퟐ풓+ퟏ number (푯 ) is given by m= [(ퟏퟗ + ퟑ√ퟒퟎ) + 풎 ퟐퟎ Square free integer: ퟐ풓+ퟏ (ퟏퟗ − ퟑ√ퟒퟎ) +2], where r = 0,1,2…….relating to A Square – free Integer is an integer which is divisible heptagonal number are presented. A Few Relations by no perfect Square other than 1. That is, its prime among heptagonal and triangular number are given. factorization has exactly one factors for each prime that appears in it. For example Index Terms - Infinite pairs of heptagonal number, the 10 =2.5 is square free, rank of square heptagonal numbers, square-free integer. But 18=2.3.3 is not because 18 is divisible by 9=32 The smallest positive square free numbers are I. PRELIMINARIES 1,2,3,5,6,7,10,11,13,14,15,17,19,21,22,23……. Definition(1.1): A number is a count or measurement heptagon: Definition(1.6): A heptagon is a seven –sided polygon. -
Deep Algebra Projects: Algebra 1/Algebra 2 Polygon Polynomials
Deep Algebra Projects: Algebra 1/Algebra 2 Polygon Polynomials Topics • Adding, subtracting, and multiplying polynomials • Factoring common factors out from a polynomial • Proving that polynomials are equivalent • Patterns and formulas for polygonal numbers • Arithmetic sequences and series In the Polygon Polynomials projects, students practice operations on polynomials as they investigate patterns and relationships in polygonal numbers—numbers used to create diagrams of regular polygons. By exploring geometric models built from squares and dots, students create formulas to describe the patterns that they find and prove that the expressions in the formulas are equivalent. I strongly suggest allowing students to share and compare their formulas and strategies in order to help them appreciate the great variety of possibilities and to explore the relationships between them. Sometimes, the greatest benefits of a project lie hidden beneath the surface. Apart from the obvious value in reinforcing and extending students’ content knowledge, the Polygon Polynomials project deepens students’ appreciation for the ways in which numbers, shapes, and algebra can unite to support discovery and understanding. The project also offers students many opportunities to enhance their skills in observation and questioning. There is more to each image than first meets the eye—especially as the images become progressively more complex. Allow plenty of time for open-ended conversation about each image before sharing the directions. Note: Dr. Ron Knott’s website on Polygonal numbers inspired some of the problems and images in this activity, especially in Problems #4 – 6. After students have completed these problems, they may enjoy exploring his site and the additional ideas and challenges that it presents. -
Integer Partitions
12 PETER KOROTEEV Remark. Note that out n! fixed points of T acting on Xn only for single point q we 12 PETER KOROTEEV get a polynomial out of Vq.Othercoefficient functions Vp remain infinite series which we 12shall further ignore. PETER One KOROTEEV can verify by examining (2.25) that in the n limit these Remark. Note that out n! fixed points of T acting on Xn only for single point q we coefficient functions will be suppressed. A choice of fixed point q following!1 the large-n limits get a polynomial out of Vq.Othercoefficient functions Vp remain infinite series which we Remark. Note thatcorresponds out n! fixed to zooming points of intoT aacting certain on asymptoticXn only region for single in which pointa q we a shall further ignore. One can verify by examining (2.25) that in the n | Slimit(1)| these| S(2)| ··· get a polynomial out ofaVSq(n.Othercoe) ,whereS fficientSn is functionsa permutationVp remain corresponding infinite to series the choice!1 which of we fixed point q. shall furthercoe ignore.fficient functions One| can| will verify be suppressed. by2 examining A choice (2.25 of) that fixed in point theqnfollowinglimit the large- thesen limits corresponds to zooming into a certain asymptotic region in which a!1 a coefficient functions will be suppressed. A choice of fixed point q following| theS(1) large-| | nSlimits(2)| ··· aS(n) ,whereS Sn is a permutation corresponding to the choice of fixed point q. q<latexit sha1_base64="gqhkYh7MBm9GaiVNDBxVo5M1bBY=">AAAB8HicbVBNS8NAEN3Ur1q/qh69LBbBU0lEUG9FLx4rWFtsQ9lsJ+3SzSbuTsQS+i+8eFDx6s/x5r9x2+agrQ8GHu/NMDMvSKQw6LrfTmFpeWV1rbhe2tjc2t4p7+7dmTjVHBo8lrFuBcyAFAoaKFBCK9HAokBCMxheTfzmI2gjYnWLowT8iPWVCAVnaKX7DsITBmH2MO6WK27VnYIuEi8nFZKj3i1/dXoxTyNQyCUzpu25CfoZ0yi4hHGpkxpIGB+yPrQtVSwC42fTi8f0yCo9GsbalkI6VX9PZCwyZhQFtjNiODDz3kT8z2unGJ77mVBJiqD4bFGYSooxnbxPe0IDRzmyhHEt7K2UD5hmHG1IJRuCN//yImmcVC+q3s1ppXaZp1EkB+SQHBOPnJEauSZ10iCcKPJMXsmbY5wX5935mLUWnHxmn/yB8/kDhq2RBA==</latexit> -
Algebra & Number Theory Vol. 7 (2013)
Algebra & Number Theory Volume 7 2013 No. 3 msp Algebra & Number Theory msp.org/ant EDITORS MANAGING EDITOR EDITORIAL BOARD CHAIR Bjorn Poonen David Eisenbud Massachusetts Institute of Technology University of California Cambridge, USA Berkeley, USA BOARD OF EDITORS Georgia Benkart University of Wisconsin, Madison, USA Susan Montgomery University of Southern California, USA Dave Benson University of Aberdeen, Scotland Shigefumi Mori RIMS, Kyoto University, Japan Richard E. Borcherds University of California, Berkeley, USA Raman Parimala Emory University, USA John H. Coates University of Cambridge, UK Jonathan Pila University of Oxford, UK J-L. Colliot-Thélène CNRS, Université Paris-Sud, France Victor Reiner University of Minnesota, USA Brian D. Conrad University of Michigan, USA Karl Rubin University of California, Irvine, USA Hélène Esnault Freie Universität Berlin, Germany Peter Sarnak Princeton University, USA Hubert Flenner Ruhr-Universität, Germany Joseph H. Silverman Brown University, USA Edward Frenkel University of California, Berkeley, USA Michael Singer North Carolina State University, USA Andrew Granville Université de Montréal, Canada Vasudevan Srinivas Tata Inst. of Fund. Research, India Joseph Gubeladze San Francisco State University, USA J. Toby Stafford University of Michigan, USA Ehud Hrushovski Hebrew University, Israel Bernd Sturmfels University of California, Berkeley, USA Craig Huneke University of Virginia, USA Richard Taylor Harvard University, USA Mikhail Kapranov Yale University, USA Ravi Vakil Stanford University, -
Mathematics 2016.Pdf
UNIVERSITY INTERSCHOLASTIC LEAGUE Making a World of Diference Mathematics Invitational A • 2016 DO NOT TURN THIS PAGE UNTIL YOU ARE INSTRUCTED TO DO SO! 1. Evaluate: 1 qq 8‚ƒ 1 6 (1 ‚ 30) 1 6 (A) qq 35 (B) 11.8 (C) qqq 11.2 (D) 8 (E) 5.0333... 2. Max Spender had $40.00 to buy school supplies. He bought six notebooks at $2.75 each, three reams of paper at $1.20 each, two 4-packs of highlighter pens at $3.10 per pack, twelve pencils at 8¢ each, and six 3-color pens at 79¢ each. How much money did he have left? (A) $4.17 (B) $5.00 (C) $6.57 (D) $7.12 (E) $8.00 3. 45 miles per hour is the same speed as _________________ inches per second. (A) 792 (B) 3,240 (C) 47,520 (D) 880 (E) 66 4. If P = {p,l,a,t,o}, O = {p,t,o,l,e,m,y}, and E = {e,u,c,l,i,d} then (P O) E = ? (A) { g } (B) {p,o,e} (C) {e, l } (D) {p,o,e,m} (E) {p,l,o,t} 5. An equation for the line shown is: (A) 3x qq 2y = 1 (B) 2x 3y = 2 (C) x y = q 1 (D) 3x 2y = qq 2 (E) x y = 1.5 6. Which of the following relations describes a function? (A) { (2,3), (3,3) (4,3) (5,3) } (B) { (qq 2,0), (0, 2) (0,2) (2,0) } (C) { (0,0), (2, q2) (2,2) (3,3) } (D) { (3,3), (3,4) (3,5) (3,6) } (E) none of these are a function 7. -
MAT344 Lecture 6
MAT344 Lecture 6 2019/May/22 1 Announcements 2 This week This week, we are talking about 1. Recursion 2. Induction 3 Recap Last time we talked about 1. Recursion 4 Fibonacci numbers The famous Fibonacci sequence starts like this: 1; 1; 2; 3; 5; 8; 13;::: The rule defining the sequence is F1 = 1;F2 = 1, and for n ≥ 3, Fn = Fn−1 + Fn−2: This is a recursive formula. As you might expect, if certain kinds of numbers have a name, they answer many counting problems. Exercise 4.1 (Example 3.2 in [KT17]). Show that a 2 × n checkerboard can be tiled with 2 × 1 dominoes in Fn+1 many ways. Solution: Denote the number of tilings of a 2 × n rectangle by Tn. We check that T1 = 1 and T2 = 2. We want to prove that they satisfy the recurrence relation Tn = Tn−1 + Tn−2: Consider the domino occupying the rightmost spot in the top row of the tiling. It is either a vertical domino, in which case the rest of the tiling can be interpreted as a tiling of a 2 × (n − 1) rectangle, or it is a horizontal domino, in which case there must be another horizontal domino under it, and the rest of the tiling can be interpreted as a tiling of a 2 × (n − 2) rectangle. Therefore Tn = Tn−1 + Tn−2: Since the number of tilings satisfies the same recurrence relation as the Fibonacci numbers, and T1 = F2 = 1 and T2 = F3 = 2, we may conclude that Tn = Fn+1. -
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.