Construction of the Figurate Numbers
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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. -
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> -
Input for Carnival of Math: Number 115, October 2014
Input for Carnival of Math: Number 115, October 2014 I visited Singapore in 1996 and the people were very kind to me. So I though this might be a little payback for their kindness. Good Luck. David Brooks The “Mathematical Association of America” (http://maanumberaday.blogspot.com/2009/11/115.html ) notes that: 115 = 5 x 23. 115 = 23 x (2 + 3). 115 has a unique representation as a sum of three squares: 3 2 + 5 2 + 9 2 = 115. 115 is the smallest three-digit integer, abc , such that ( abc )/( a*b*c) is prime : 115/5 = 23. STS-115 was a space shuttle mission to the International Space Station flown by the space shuttle Atlantis on Sept. 9, 2006. The “Online Encyclopedia of Integer Sequences” (http://www.oeis.org) notes that 115 is a tridecagonal (or 13-gonal) number. Also, 115 is the number of rooted trees with 8 vertices (or nodes). If you do a search for 115 on the OEIS website you will find out that there are 7,041 integer sequences that contain the number 115. The website “Positive Integers” (http://www.positiveintegers.org/115) notes that 115 is a palindromic and repdigit number when written in base 22 (5522). The website “Number Gossip” (http://www.numbergossip.com) notes that: 115 is the smallest three-digit integer, abc, such that (abc)/(a*b*c) is prime. It also notes that 115 is a composite, deficient, lucky, odd odious and square-free number. The website “Numbers Aplenty” (http://www.numbersaplenty.com/115) notes that: It has 4 divisors, whose sum is σ = 144. -
1 + 4 + 7 + 10 = 22;... the Nth Pentagonal Number Is Therefore
CHAPTER 6 1. The pentagonal numbers are 1; 1 + 4 = 5; 1 + 4 + 7 = 12; 1 + 4 + 7 + 10 = 22;... The nth pentagonal number is therefore n 1 2 − n(n 1) 3n n (3i +1) = n + 3 − = − . i 2 ! 2 X=0 Similarly, since the hexagonal numbers are 1; 1+5 = 6; 1+5+9 = 15; 1+5+9+13 = 28;... it follows that the nth hexagonal number is n 1 − n(n 1) (4i +1) = n + 4 − = 2n2 n. i 2 ! − X=0 2. The pyramidal numbers with triangular base are 1; 1+3 = 4; 1+3+6 = 10; 1+3+6+10 = 20; ... Therefore the nth pyramidal number with triangular base is n n k(k + 1) 1 1 n(n + 1)(2n + 1) n(n + 1) = (k2 + k)= + k 2 2 k 2 " 6 2 # X=1 X=1 . n(n + 1) 2n + 1 1 n(n + 1)(n + 2) = + = 2 6 2 6 The pyramidal numbers with square base are 1; 1+4 = 5; 1+4+9 = 14; 1+4+9+16= 30;... Thus the nth pyramidal number with square base is given by the sum of the squares from 1 to n, namely, n(n + 1)(2n + 1) . 6 3. In a harmonic proportion, c : a =(c b):(b a). It follows that ac ab = bc ac or that b(a + c) = 2ac. Thus the sum of− the extremes− multiplied by the mean− equals− twice the product of the extremes. 4. Since 6:3=(5 3) : (6 5), the numbers 3, 5, 6 are in subcontrary proportion. -
Newsletter 91
Newsletter 9 1: December 2010 Introduction This is the final nzmaths newsletter for 2010. It is also the 91 st we have produced for the website. You can have a look at some of the old newsletters on this page: http://nzmaths.co.nz/newsletter As you are no doubt aware, 91 is a very interesting and important number. A quick search on Wikipedia (http://en.wikipedia.org/wiki/91_%28number%29) will very quickly tell you that 91 is: • The atomic number of protactinium, an actinide. • The code for international direct dial phone calls to India • In cents of a U.S. dollar, the amount of money one has if one has one each of the coins of denominations less than a dollar (penny, nickel, dime, quarter and half dollar) • The ISBN Group Identifier for books published in Sweden. In more mathematically related trivia, 91 is: • the twenty-seventh distinct semiprime. • a triangular number and a hexagonal number, one of the few such numbers to also be a centered hexagonal number, and it is also a centered nonagonal number and a centered cube number. It is a square pyramidal number, being the sum of the squares of the first six integers. • the smallest positive integer expressible as a sum of two cubes in two different ways if negative roots are allowed (alternatively the sum of two cubes and the difference of two cubes): 91 = 6 3+(-5) 3 = 43+33. • the smallest positive integer expressible as a sum of six distinct squares: 91 = 1 2+2 2+3 2+4 2+5 2+6 2. -
Elementary School Numbers and Numeration
DOCUMENT RESUME ED 166 042 SE 026 555 TITLE Mathematics for Georgia Schools,' Volume II: Upper Elementary 'Grades. INSTITUTION Georgia State Dept. of Education, Atlanta. Office of Instructional Services. e, PUB DATE 78 NOTE 183p.; For related document, see .SE 026 554 EDRB, PRICE MF-$0.33 HC-$10.03 Plus Postage., DESCRIPTORS *Curriculm; Elementary Educatia; *Elementary School Mathematics; Geometry; *Instruction; Meadurement; Number Concepts; Probability; Problem Solving; Set Thory; Statistics; *Teaching. Guides ABSTRACT1 ' This guide is organized around six concepts: sets, numbers and numeration; operations, their properties and number theory; relations and functions; geometry; measurement; and probability and statistics. Objectives and sample activities are presented for.each concept. Separate sections deal with the processes of problem solving and computation. A section on updating curriculum includes discussion of continuing program improvement, evaluation of pupil progress, and utilization of media. (MP) ti #######*#####*########.#*###*######*****######*########*##########**#### * Reproductions supplied by EDRS are the best that can be made * * from the original document. * *********************************************************************** U S DEPARTMENT OF HEALTH, EDUCATION & WELFARE IS NATIONAL INSTITUTE OF.. EDUCATION THIS DOCUMENT HA4BEEN REPRO- DuCED EXACTLY AS- RECEIVEDFROM THE PERSON OR ORGANIZATIONORIGIN- ATING IT POINTS OF VIEWOR 01NIONS STATED DO NOT NECESSARILYEpRE SENT OFFICIAL NATIONAL INSTITUTEOF TO THE EDUCATION4 -
PENTAGONAL NUMBERS in the PELL SEQUENCE and DIOPHANTINE EQUATIONS 2X2 = Y2(3Y -1) 2 ± 2 Ve Siva Rama Prasad and B
PENTAGONAL NUMBERS IN THE PELL SEQUENCE AND DIOPHANTINE EQUATIONS 2x2 = y2(3y -1) 2 ± 2 Ve Siva Rama Prasad and B. Srlnivasa Rao Department of Mathematics, Osmania University, Hyderabad - 500 007 A.P., India (Submitted March 2000-Final Revision August 2000) 1. INTRODUCTION It is well known that a positive integer N is called a pentagonal (generalized pentagonal) number if N = m(3m ~ 1) 12 for some integer m > 0 (for any integer m). Ming Leo [1] has proved that 1 and 5 are the only pentagonal numbers in the Fibonacci sequence {Fn}. Later, he showed (in [2]) that 2, 1, and 7 are the only generalized pentagonal numbers in the Lucas sequence {Ln}. In [3] we have proved that 1 and 7 are the only generalized pentagonal numbers in the associated Pell sequence {Qn} delned by Q0 = Qx = 1 and Qn+2 = 2Qn+l + Q„ for n > 0. (1) In this paper, we consider the Pell sequence {PJ defined by P 0 =0,P 1 = 1, and Pn+2=2Pn+l+Pn for«>0 (2) and prove that P±l, P^, P4, and P6 are the only pentagonal numbers. Also we show that P0, P±1, P2, P^, P4, and P6 are the only generalized pentagonal numbers. Further, we use this to solve the Diophantine equations of the title. 2. PRELIMINARY RESULTS We have the following well-known properties of {Pn} and {Qn}: for all integers m and n, a a + pn = "-P" a n d Q = " P" w herea = 1 + V2 and p = 1-V2, (3) l P_„ = {-\r Pn and Q_„ = (-l)"Qn, (4) a2 = 2P„2+(-iy, (5) 2 2 e3„=a(e„ +6P„ ), (6> ^ + n = 2PmG„-(-l)"Pm_„. -
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. -
On Ternary Cubic Equation
ISSN: 2277-3754 ISO 9001:2008 Certified International Journal of Engineering and Innovative Technology (IJEIT) Volume 3, Issue 9, March 2014 ON TERNARY CUBIC EQUATION P. Thirunavukarasu, S. Sriram Assistant Professor -P.G & Research Department of Mathematics, Periyar E.V.R College Tiruchirappalli – 620 023, Tamilnadu, India Assistant Professor–P.G & Research Department of Mathematics, National College, Tiruchirappalli – 620 001, Tamilnadu, India equations are analyzed for the non-trivial integral Abstract we obtain the non-trivial integral solutions for the solutions. These results have motivated us to search for ternary cubic equation non-trivial integral solutions of their varieties of ternary cubic Diophantine equation. This paper concerns with the problem of determining non-trivial integral solutions of . the equation with three unknowns given by A few interesting relations among the solutions are presented. Index Terms: Ternary Cubic, integral solutions, Pell’s explicit integral solutions of the above equation are form, nasty numbers presented. A few interesting relations among the solutions Notations are obtained. Oblong number of rank n obln n n 1 II. METHOD OF ANALYSIS Tetrahedral number of rank The ternary cubic equation under consideration is n n12 n n Tet n 6 Triangular number of rank (1) Taking (2) Polygonal number of rank n with sides nm12 (3) m tnmn, 1 2 We get Square pyramidal number of rank n n1 2 n 1 (4) Again taking the transformation n Sqpn 6 Pentagonal pyramidal number of rank Star number = and apply in (4) we get (5) 2 Stella Octangula number = St. oct n 2 n 1 n It is well known that the general form of the integral solutions of the Pellian equation. -
PELL FACTORIANGULAR NUMBERS Florian Luca, Japhet Odjoumani
PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 105(119) (2019), 93–100 DOI: https://doi.org/10.2298/PIM1919093L PELL FACTORIANGULAR NUMBERS Florian Luca, Japhet Odjoumani, and Alain Togbé Abstract. We show that the only Pell numbers which are factoriangular are 2, 5 and 12. 1. Introduction Recall that the Pell numbers Pm m> are given by { } 1 αm βm P = − , for all m > 1, m α β − where α =1+ √2 and β =1 √2. The first few Pell numbers are − 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025,... Castillo [3], dubbed a number of the form Ftn = n!+ n(n + 1)/2 for n > 1 a factoriangular number. The first few factoriangular numbers are 2, 5, 12, 34, 135, 741, 5068, 40356, 362925,... Luca and Gómez-Ruiz [8], proved that the only Fibonacci factoriangular numbers are 2, 5 and 34. This settled a conjecture of Castillo from [3]. In this paper, we prove the following related result. Theorem 1.1. The only Pell numbers which are factoriangular are 2, 5 and 12. Our method is similar to the one from [8]. Assuming Pm = Ftn for positive integers m and n, we use a linear forms in p-adic logarithms to find some bounds on m and n. The resulting bounds are large so we run a calculation to reduce the bounds. This computation is highly nontrivial and relates on reducing the diophantine equation Pm = Ftn modulo the primes from a carefully selected finite set of suitable primes. 2010 Mathematics Subject Classification: Primary 11D59, 11D09, 11D7; Secondary 11A55.