(m) (-I)P-I(ap b. rn+s n. kpn kpn-1 AND LINEAR SECOND ORDER RECURRENCES SOME CONGRUENCE PROPERTIES OF BINOMIAL COEFFICIENTS NEVILLE ROBBINS (1.
|
|
- Corey Patrick
- 5 years ago
- Views:
Transcription
1 Iterat. J. Math. & Math. Sci. VOL. II NO. 4 (1988) SOME CONGRUENCE ROERTIES OF BINOMIAL COEFFICIENTS AND LINEAR SECOND ORDER RECURRENCES NEVILLE ROBBINS Departmet of athematlcs Sa Fracisco State Uiversity.Sa Fracisco, CA 9413 (Received May 1, 1987) ABSTRACT. Usig elemetary methods, the followig results are obtaled:(1) If p is -m prime, 0 m If r,s are the roots of x Ax-B, where (A,B) ad D A_4B > 0, if -, 0 < b < ap -m, ad p ab, the (m) (-I)-I(ap b (rood p). u v r+s, ad > O, the (II) v =- v (rood p). r-s p p-1 (III) If p is odd ad p D, the u (_D) u (rood p); p p -I (IV) u (_1)Bu ) -I (rood KEYWORDS AND HRASES. Biomial coefficiet, liear secod order recurrece AMS SUBJECT CLASSIFICATION CODE: loal0 IOA35. I. INTRODUCTION. Followig Lucas [I], let A, B be itegers such that (A,B) ad D-- A-4B > O. Let the roots of x Ax-B be: r 1/( ), s--i/ (A- ). Let 0. Let the sequeces u v be defied by: u r -s r-s (1.1) v r+s (1.) the u 0 0 u u Au Bu for -l- - v, v A, v Av -Bv for o -I - (1.4) r+s=a
2 744 N. ROBBINS rs B (1.6) U V UV _B v (1.8) (I.9) Um+ UmV + UVm (I.I0) Let p be prime. Let O (t) j if J t p 3+ p t. If 0 < < p, the (1.11) If 0, the (- () (1.1) -I a p a p (rood p) (1.13) If x +/- a (rood ), the x a (rood +l (1.14) e If 0 < < pe ad p, the Op(()) e (I.5) REMARK. (I.I) through (I.I0) appear i Lucas [I]. (I.II) through (1.14) are elemetary. (1.15) is theorem i Robbls [].. MAI N RE SULTS. Lemma ( I), If 0 m, 0 < b < ap -m ad p a b, -m b(pm-l) (mod p). the (m) (a b (-I) bp m -I ROOF. (pm) j--0 ap-j bpm-j I where 1 is the product of all such factors where pmlj, ad is the product of all such factors where Now s-m 1 bpm-i j jip b-1 a m b-i -m_i -m I _ap ap i--o bpm-ip m i=o b-i b a_ (_l)bpm_b (rood J pro_ am) (a b( I) (bp (-I) (rood p). Therefore ) while
3 SOME CONGRUENCE ROERTIES OF BINOMIAL COEFFICIENTS 745 -m THEOREM.1. If 0 m, 0 < b < ap ad p ab, -m bp (rood p). the (am) (-I)-I (a b ROOF. If p is odd, the b(pm-l) -- p-! 0 (rood ); if p, the by hypothesis, b is odd, so b(m-l) -1 (rood ). I either case, the coclusio ow follows from Lemma.1. LEMMA.. V+! A+l (+l B ROOF. By (1.) ad the biomial theorem, we have +I +l- +l +l +l s V+l r + s (r+s) [ r = so (1.5) implies V+ A+l-{ +l +l- +I +l-l.. )r s + (+l_)r s Now (1.1) implies +l) V+l A +l l (rs) (r+l-+s+i- ), so (1.) ad (1.6) imply V+l A+l (+l)bv LEMMA.3. v -- v (rood p) +l- ROOF. Lemma. ad (I.II) imply v p-- A p ); too, (1.4) implies v v (rood p). LEMMA.4. If i ) J, the vi+j viv j ROOF. By (1.) ad (1.6), BJv (rood p); (I.13) implles A p -- A (rood isj+r j i J viv j vi+j (ri+si)(rj+sj)-(ri+j+si+j) r s (rs) (rl-j+s LEMMA.5. If 0 m, y z (rood pro), w x (mod p), ad x 0 (rood p-m), the wy xz (rood p). --- ROOF. Hypothesis implies Y z+ip m, w x+jp so wy E xz+ipmx (rood p ). Hypothesis also implies p-mlx, so wy xz (rood p). LEMMA.6. If v (rood pm m m-i ), the v ---v (mod pm p m p m-1 ROOF. (Iductio o ). Lemma 6 holds trivially for =O, ad by hypothesis for. O, m ) ad v ---
4 746 N. ROBBINS m v =v =v v -B v =v v B v m m t m m m (+!)p m pm+p m p m p p -p p p (-l)p m by Lemma.4. Now iductio hypothesis ad (1.13) imply that v =v v -B p v (+ )pro p m- p (-I )p v v (mod pm) (+ )pro (+ )pro- m-1 (rood pro). Now Lemma.4 implies m-1 m-1 v (rood p ) -I ROOF. (Iductio o ) Lemma.7 holds for =1, by Lemma.3. Suppose Lemma LEMMA 7. If p is odd ad I, the v holds for all m <, where. The Lemma. implies -I I/ (p-l-l) -I v A p p )Biv -I i - p i=o p I-i I/ (p- I) v Ap p j=o ()BJVp_ ] If p j, the (I.15) implies 0 (rood p J also Therefore I/ (p-1-1) v Ap p j=o > (pp)biv p - ip Let i p hpm, where p h ad m <. Now Bi Bh Bh B i (rood pm), by (1.13); also v p _ ip m m-i V V V p_hpm (p-m_h)pm (p-m_h)pm- =- v -I p -i (rood pro) by iductio hypothesis ad Lemma.6. Therefore B ip v B i v (rood pm -1 p-ip p -i Also p -I -I -m _= (p) i hpm) -= (hp m-l) p ) i h implies ( 0 (rood p -m) Therefore Lemma 5 h implies (rood p by Theorem I, ad (I.15) -I (p)biv ( )Biv (mod -ip p ) Now (I 13) i implles p p -l_ i
5 SOME. CONGRUENCE ROERTIES OF BINOMIAL COEFFICIENTS / (p-l-l) v A p (p - )B i v (rood p ) that is i -I p i=o p -i v E v (mod p). -I LEMMA.8. If 1A, the v (rood +l) for 0. ROOF. (Iductio o ) v0 -I -I =- 4 (mod +l) Hypothesis implies B v A (rood by (1.4) ad hypothesis. Now iductio hypothesis implies v (rood ), so (1.14) implies v implies v v -1 -I -B 4-(I) (rood +l). -I B Now (1.9) so (rood ) LEMMA.9. If AB, the v -I (rood +l) for 0. ROOF. (Iductio o ) v0 v A =- -I (rood ) by (1.4) ad hypothesis. Now iductio hypothesis implies v =-I (rood ), so (1.14) implies -I v (rood +l).- Agai, B odd implies -I _B-1 v v =- I-(I) -I (rood +lj. -I LEMMA.10. If B, the v (rood +l)- for O. -I B -= (rood ), so (1.9) implies ROOF. (Iductio o ) Hypothesis implies A is odd, so (1.4) implies v 0 v A (rood ). By hypothesis, B 0 (rood ), so -1 B -= 0 (rood -1 Sice -I -I for we have B 0 (rood ) By (rood ), so (1.14) implies v (mod +l). iductio hypothesis, V_1 _l -I Now (1.9) implies v v -B E 1-(0) =- (rood +l)." -I LEMMA.11. V v (mod ) -I ROOF. Lemmas.8,.9,.10 imply v t (rood -I ), V where t or +/-I. Therefore V t V_l (rood ). E t (rood +l),
6 748 N. ROBBINS THEOREM.. I I, the v =- v (rood )- ROOF. Follows from Lemmas.7 ad.11. THEOREM.3. If ad 0, the v v (rood p) p p -I ROOF. Fol-los from Theorem. ad Lemma.6. LEMMA.1. U+l D + (-B)U+l_ + + ROOF. (i.i) ad (1.7) imply U+l r s +l (_l)(+l)r+l-. s so (1.7) implies DI/ Um+l D + I/-{ (-l)(+l)r+l- s. Settig j +l- i the secod sum, we obtai I/9 D U+ +I/_{ (-I) (+1)r s (-i) = D + I/-. D/- U+l D. (-l)(rs)( I) (r (+l +l + (-l)j( +1)r J=+ J +l- -( +l- +l- s by (1.1) (-B) U+l_ by (1.6) ad (I.7), so (-B) u + l- LEMMA.13. If p is odd, the u =- () (rood p) +l-j sj ROOF. Follows from Lemma.1, (I.II), ad Euler s criterio. +l +l- +l- r s LEMMA.14. If p is odd, pd, ad I, the DI/ (p)-- ( D) (mod p). ROOF. (Iductio o ) Lemma.14 holds for =l by Euler s criterio. Let () t 11. Now iductio hypothesis implies DI/(p)E t (rood p), I/(p). + so D t+ip. D I/( )= (D i/ (p ) ) (t+ip) p t p + pt p-i (ip) + (.) t-j(ip)j. Now t p t 3 j--
7 SOME CONGRUENCE ROERTIES OF BINOMIAL COEFFICIENTS 749 ad p+llpj for j ), so Dl/(p+l )-- (rood p+l). LEMMA.15. If ) 0, m > I, p is odd, (), ad u tu (rood pm ), the u tu (mod pm). m m-i ROOF. (Iductio o ) Lemma.15 is trivially true for =O, ad is true by hypothesis for. Now (I.I0) implies U(+ m u u v + u v pm m m m m l)p pm+ p p p p By iductio hypothesis, ), ad Upm --tupm_l pm pm- u E tu (rood pm m pm- Theorem. implies v v (rood pm m pm- m tu u( +1)p (rood pro); for O. Therefore m-i v + tu v 9tu (rood pro). Sice m-i m-i m p p p pm-i p p is odd, we have U(+l)p m -= tu (mod p m LEMMA.16. If p is odd, pd, ad ) I, the u -= (--D)u (mod p -I ROOF. (Iductio o ) Lemma.16 is true for by Lemma.13. -I DI/ (p-l_l) I/ (p -I) -I i Lemma.1 implies u -I (pff (-B) u p i p _l_ i - (p) 0 p). j (J) (-B)jup- j. If pj, the (1.15) implies (rood p). Therefore u DI/ (p-i) I/ (p-i-i) )-. (pp) (-B) iu - (rood p i= p ip Let ip hp m where ph ad m <. Let t (). m )hpm- Now (-B) ip s (-B) hp (-B (-B) i (rood pro) by (1.13). By iductio p). hypothesis ad Lemma.15, we have u E u E u =- tu m-i E tu -I m-i p-ip p-hpm (p-m-h) pm (p-m-h) p p -hp
8 750 N. ROBBINS tu _l_ (rood pro) Therefore (-B)iu t(-b)iu -l_ p i p-ip p i (mod pro). As i the proof of Lemma.7, we have -! (-B)iu t( p (-B)i () ip p-ip p -i u -I (rood p). Therefore u E DI/ (p-l) -t(d 1/ (p-l-l) -u (rood p) that is p ; -I u tu + DI/ (pr-l_l) (DI/ (p_p-l)_t) (rood p) -I Sice (p) -I - Lemma.14 implies u )u (rood p) -I LEMMA.17. If ad D is odd, the u =- (-l)bu_l (rood ). ROOF. By hypothesis ad by the defiitios of A, B, ad D, A must be odd. If B is odd, the Lemma.9 implies implies v -= (rood ). -1 Now (1.8) implies u (-l)bu_l (D) if p is odd B (-I) if p v -I -= -I (rood ); if B is eve the Lemma.10 Therefore, i either case, v (-I)B (rood ). -I (mod ). THEOREM.4. If ad p D, the u tu (rood p), where -l ROOF. Follows from Lemmas.16 ad.17. THEOREM.5. If ) 0, > I, ad pd, the u p where is defied as i Theorem.4. -= tu (mod p ) -I p ROOF. Follows from Lemma.15 ad Theorem.4. Cocludig Remars. Let be defied as i Theorem.4 as a result of Theorems.3 ad 5, the sequeces v t u determie p-adic itegers for each 0. p p REFERENCE S I. LUCAS, E.. ROBBI NS, N. Theorie des foctios umeriques simplemet periodiques, Amer. J. Math. (187 7) ; O the umber of biomial coefficiets which are divisible by their row umber, Caad. Math. Bull. 5 (3) 198,
Some identities involving Fibonacci, Lucas polynomials and their applications
Bull. Math. Soc. Sci. Math. Roumaie Tome 55103 No. 1, 2012, 95 103 Some idetities ivolvig Fiboacci, Lucas polyomials ad their applicatios by Wag Tigtig ad Zhag Wepeg Abstract The mai purpose of this paper
More informationand each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.
MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,
More informationFermat s Little Theorem. mod 13 = 0, = }{{} mod 13 = 0. = a a a }{{} mod 13 = a 12 mod 13 = 1, mod 13 = a 13 mod 13 = a.
Departmet of Mathematical Scieces Istructor: Daiva Puciskaite Discrete Mathematics Fermat s Little Theorem 43.. For all a Z 3, calculate a 2 ad a 3. Case a = 0. 0 0 2-times Case a 0. 0 0 3-times a a 2-times
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More informationDIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS
DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS VERNER E. HOGGATT, JR. Sa Jose State Uiversity, Sa Jose, Califoria 95192 ad CALVIN T. LONG Washigto State Uiversity, Pullma, Washigto 99163
More informationOn Divisibility concerning Binomial Coefficients
A talk give at the Natioal Chiao Tug Uiversity (Hsichu, Taiwa; August 5, 2010 O Divisibility cocerig Biomial Coefficiets Zhi-Wei Su Najig Uiversity Najig 210093, P. R. Chia zwsu@ju.edu.c http://math.ju.edu.c/
More informationAN ALMOST LINEAR RECURRENCE. Donald E. Knuth Calif. Institute of Technology, Pasadena, Calif.
AN ALMOST LINEAR RECURRENCE Doald E. Kuth Calif. Istitute of Techology, Pasadea, Calif. form A geeral liear recurrece with costat coefficiets has the U 0 = a l* U l = a 2 " ' " U r - l = a r ; u = b, u,
More informationMath 2112 Solutions Assignment 5
Math 2112 Solutios Assigmet 5 5.1.1 Idicate which of the followig relatioships are true ad which are false: a. Z Q b. R Q c. Q Z d. Z Z Z e. Q R Q f. Q Z Q g. Z R Z h. Z Q Z a. True. Every positive iteger
More informationAbstract. 1. Introduction This note is a supplement to part I ([4]). Let. F x (1.1) x n (1.2) Then the moments L x are the Catalan numbers
Abstract Some elemetary observatios o Narayaa polyomials ad related topics II: -Narayaa polyomials Joha Cigler Faultät für Mathemati Uiversität Wie ohacigler@uivieacat We show that Catala umbers cetral
More informationDifferent kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationPERIODS OF FIBONACCI SEQUENCES MODULO m. 1. Preliminaries Definition 1. A generalized Fibonacci sequence is an infinite complex sequence (g n ) n Z
PERIODS OF FIBONACCI SEQUENCES MODULO m ARUDRA BURRA Abstract. We show that the Fiboacci sequece modulo m eriodic for all m, ad study the eriod i terms of the modulus.. Prelimiaries Defiitio. A geeralized
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationLINEAR RECURSION RELATIONS - LESSON FOUR SECOND-ORDER LINEAR RECURSION RELATIONS
LINEAR RECURSION RELATIONS - LESSON FOUR SECOND-ORDER LINEAR RECURSION RELATIONS BROTHER ALFRED BROUSSEAU St. Mary's College, Califoria Give a secod-order liear recursio relatio (.1) T. 1 = a T + b T 1,
More informationADVANCED PROBLEMS AND SOLUTIONS
ADVANCED PROBLEMS AND SOLUTIONS EDITED BY FLORIAN LUCA Please sed all commuicatios cocerig ADVANCED PROBLEMS AND SOLUTIONS to FLORIAN LUCA, SCHOOL OF MATHEMATICS, UNIVERSITY OF THE WITWA- TERSRAND, WITS
More informationIn number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.
Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers
More informationEISENSTEIN S CRITERION, FERMAT S LAST THEOREM, AND A CONJECTURE ON POWERFUL NUMBERS arxiv: v6 [math.ho] 13 Feb 2018
EISENSTEIN S CRITERION, FERMAT S LAST THEOREM, AND A CONJECTURE ON POWERFUL NUMBERS arxiv:174.2885v6 [math.ho] 13 Feb 218 PIETRO PAPARELLA Abstract. Give itegers l > m >, moic polyomials X, Y, ad Z are
More information+ au n+1 + bu n = 0.)
Lecture 6 Recurreces - kth order: u +k + a u +k +... a k u k 0 where a... a k are give costats, u 0... u k are startig coditios. (Simple case: u + au + + bu 0.) How to solve explicitly - first, write characteristic
More informationECEN 644 HOMEWORK #5 SOLUTION SET
ECE 644 HOMEWORK #5 SOUTIO SET 7. x is a real valued sequece. The first five poits of its 8-poit DFT are: {0.5, 0.5 - j 0.308, 0, 0.5 - j 0.058, 0} To compute the 3 remaiig poits, we ca use the followig
More informationSolutions to Problem Set 7
8.78 Solutios to Problem Set 7. If the umber is i S, we re doe sice it s relatively rime to everythig. So suose S. Break u the remaiig elemets ito airs {, }, {4, 5},..., {, + }. By the Pigeohole Pricile,
More informationA LIMITED ARITHMETIC ON SIMPLE CONTINUED FRACTIONS - II 1. INTRODUCTION
A LIMITED ARITHMETIC ON SIMPLE CONTINUED FRACTIONS - II C. T. LONG J. H. JORDAN* Washigto State Uiversity, Pullma, Washigto 1. INTRODUCTION I the first paper [2 ] i this series, we developed certai properties
More information1. INTRODUCTION. P r e s e n t e d h e r e is a generalization of Fibonacci numbers which is intimately connected with the arithmetic triangle.
A GENERALIZATION OF FIBONACCI NUMBERS V.C. HARRIS ad CAROLYN C. STYLES Sa Diego State College ad Sa Diego Mesa College, Sa Diego, Califoria 1. INTRODUCTION P r e s e t e d h e r e is a geeralizatio of
More informationON MONOTONICITY OF SOME COMBINATORIAL SEQUENCES
Publ. Math. Debrece 8504, o. 3-4, 85 95. ON MONOTONICITY OF SOME COMBINATORIAL SEQUENCES QING-HU HOU*, ZHI-WEI SUN** AND HAOMIN WEN Abstract. We cofirm Su s cojecture that F / F 4 is strictly decreasig
More information(6), (7) and (8) we have easily, if the C's are cancellable elements of S,
VIOL. 23, 1937 MA THEMA TICS: H. S. VANDIVER 555 where the a's belog to S'. The R is said to be a repetitive set i S, with respect to S', ad with multiplier M. If S cotais a idetity E, the if we set a,
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationExam 2 CMSC 203 Fall 2009 Name SOLUTION KEY Show All Work! 1. (16 points) Circle T if the corresponding statement is True or F if it is False.
1 (1 poits) Circle T if the correspodig statemet is True or F if it is False T F For ay positive iteger,, GCD(, 1) = 1 T F Every positive iteger is either prime or composite T F If a b mod p, the (a/p)
More informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
More informationSolutions to Problem Set 8
8.78 Solutios to Problem Set 8. We ow that ( ) ( + x) x. Now we plug i x, ω, ω ad add the three equatios. If 3 the we ll get a cotributio of + ω + ω + ω + ω 0, whereas if 3 we ll get a cotributio of +
More informationTHE p-adic VALUATION OF LUCAS SEQUENCES
THE p-adic VALUATION OF LUCAS SEQUENCES CARLO SANNA Abstract. Let (u n) n 0 be a nondegenerate Lucas sequence with characteristic polynomial X 2 ax b, for some relatively prime integers a and b. For each
More informationSuper congruences concerning Bernoulli polynomials. Zhi-Hong Sun
It J Numer Theory 05, o8, 9-404 Super cogrueces cocerig Beroulli polyomials Zhi-Hog Su School of Mathematical Scieces Huaiyi Normal Uiversity Huaia, Jiagsu 00, PR Chia zhihogsu@yahoocom http://wwwhytceduc/xsjl/szh
More informationarxiv: v1 [math.nt] 10 Dec 2014
A DIGITAL BINOMIAL THEOREM HIEU D. NGUYEN arxiv:42.38v [math.nt] 0 Dec 204 Abstract. We preset a triagle of coectios betwee the Sierpisi triagle, the sum-of-digits fuctio, ad the Biomial Theorem via a
More informationSome p-adic congruences for p q -Catalan numbers
Some p-adic cogrueces for p q -Catala umbers Floria Luca Istituto de Matemáticas Uiversidad Nacioal Autóoma de México C.P. 58089, Morelia, Michoacá, México fluca@matmor.uam.mx Paul Thomas Youg Departmet
More informationarxiv: v1 [math.nt] 28 Apr 2014
Proof of a supercogruece cojectured by Z.-H. Su Victor J. W. Guo Departmet of Mathematics, Shaghai Key Laboratory of PMMP, East Chia Normal Uiversity, 500 Dogchua Rd., Shaghai 0041, People s Republic of
More informationModern Algebra 1 Section 1 Assignment 1. Solution: We have to show that if you knock down any one domino, then it knocks down the one behind it.
Moder Algebra 1 Sectio 1 Assigmet 1 JOHN PERRY Eercise 1 (pg 11 Warm-up c) Suppose we have a ifiite row of domioes, set up o ed What sort of iductio argumet would covice us that ocig dow the first domio
More informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More informationThe 4-Nicol Numbers Having Five Different Prime Divisors
1 2 3 47 6 23 11 Joural of Iteger Sequeces, Vol. 14 (2011), Article 11.7.2 The 4-Nicol Numbers Havig Five Differet Prime Divisors Qiao-Xiao Ji ad Mi Tag 1 Departmet of Mathematics Ahui Normal Uiversity
More informationBertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n.
Bertrad s Postulate Our goal is to prove the followig Theorem Bertrad s Postulate: For every positive iteger, there is a prime p satisfyig < p We remark that Bertrad s Postulate is true by ispectio for,,
More informationThe value of Banach limits on a certain sequence of all rational numbers in the interval (0,1) Bao Qi Feng
The value of Baach limits o a certai sequece of all ratioal umbers i the iterval 0, Bao Qi Feg Departmet of Mathematical Scieces, Ket State Uiversity, Tuscarawas, 330 Uiversity Dr. NE, New Philadelphia,
More informationA Simple Derivation for the Frobenius Pseudoprime Test
A Simple Derivatio for the Frobeius Pseudoprime Test Daiel Loebeberger Bo-Aache Iteratioal Ceter for Iformatio Techology March 17, 2008 Abstract Probabilistic compositeess tests are of great practical
More informationBasic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S.
Basic Sets Example 1. Let S = {1, {2, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 (b) {1} S (c) {2, 3} S (d) {1, 4} S (e) 2 S. (f) S = {1, 4, {2, 3}} (g) S Example 2. Compute the
More informationBINOMIAL PREDICTORS. + 2 j 1. Then n + 1 = The row of the binomial coefficients { ( n
BINOMIAL PREDICTORS VLADIMIR SHEVELEV arxiv:0907.3302v2 [math.nt] 22 Jul 2009 Abstract. For oegative itegers, k, cosider the set A,k = { [0, 1,..., ] : 2 k ( ). Let the biary epasio of + 1 be: + 1 = 2
More informationLecture 7: Properties of Random Samples
Lecture 7: Properties of Radom Samples 1 Cotiued From Last Class Theorem 1.1. Let X 1, X,...X be a radom sample from a populatio with mea µ ad variace σ
More informationANOTHER GENERALIZED FIBONACCI SEQUENCE 1. INTRODUCTION
ANOTHER GENERALIZED FIBONACCI SEQUENCE MARCELLUS E. WADDILL A N D LOUIS SACKS Wake Forest College, Wisto Salem, N. C., ad Uiversity of ittsburgh, ittsburgh, a. 1. INTRODUCTION Recet issues of umerous periodicals
More informationA 2nTH ORDER LINEAR DIFFERENCE EQUATION
A 2TH ORDER LINEAR DIFFERENCE EQUATION Doug Aderso Departmet of Mathematics ad Computer Sciece, Cocordia College Moorhead, MN 56562, USA ABSTRACT: We give a formulatio of geeralized zeros ad (, )-discojugacy
More informationSOLUTION SET VI FOR FALL [(n + 2)(n + 1)a n+2 a n 1 ]x n = 0,
4. Series Solutios of Differetial Equatios:Special Fuctios 4.. Illustrative examples.. 5. Obtai the geeral solutio of each of the followig differetial equatios i terms of Maclauri series: d y (a dx = xy,
More informationOn Some Properties of Digital Roots
Advaces i Pure Mathematics, 04, 4, 95-30 Published Olie Jue 04 i SciRes. http://www.scirp.org/joural/apm http://dx.doi.org/0.436/apm.04.46039 O Some Properties of Digital Roots Ilha M. Izmirli Departmet
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationM A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O
M A T H 2 4 0 F A L L 2 0 1 4 HOMEWORK ASSIGNMENT #4 CORRECTION Algebra I 1 4 / 1 0 / 2 0 1 4 U N I V E R S I T Y O F T O R O N T O P r o f e s s o r : D r o r B a r - N a t a Correctio Homework Assigmet
More informationFactors of sums and alternating sums involving binomial coefficients and powers of integers
Factors of sums ad alteratig sums ivolvig biomial coefficiets ad powers of itegers Victor J. W. Guo 1 ad Jiag Zeg 2 1 Departmet of Mathematics East Chia Normal Uiversity Shaghai 200062 People s Republic
More informationRegent College Maths Department. Further Pure 1. Proof by Induction
Reget College Maths Departmet Further Pure Proof by Iductio Further Pure Proof by Mathematical Iductio Page Further Pure Proof by iductio The Edexcel syllabus says that cadidates should be able to: (a)
More informationAssignment 5: Solutions
McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece
More informationA Study on Some Integer Sequences
It. J. Cotemp. Math. Scieces, Vol. 3, 008, o. 3, 03-09 A Study o Some Iteger Sequeces Serpil Halıcı Sakarya Uiversity, Departmet of Mathematics Esetepe Campus, Sakarya, Turkey shalici@sakarya.edu.tr Abstract.
More informationProc. Amer. Math. Soc. 139(2011), no. 5, BINOMIAL COEFFICIENTS AND THE RING OF p-adic INTEGERS
Proc. Amer. Math. Soc. 139(2011, o. 5, 1569 1577. BINOMIAL COEFFICIENTS AND THE RING OF p-adic INTEGERS Zhi-Wei Su* ad Wei Zhag Departmet of Mathematics, Naig Uiversity Naig 210093, People s Republic of
More informationNATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION ADVANCED CALCULUS II. November 2003 Time allowed :
NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER EXAMINATION 003-004 MA08 ADVANCED CALCULUS II November 003 Time allowed : hours INSTRUCTIONS TO CANDIDATES This examiatio paper cosists of TWO
More informationEnumerative & Asymptotic Combinatorics
C50 Eumerative & Asymptotic Combiatorics Notes 4 Sprig 2003 Much of the eumerative combiatorics of sets ad fuctios ca be geeralised i a maer which, at first sight, seems a bit umotivated I this chapter,
More informationON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS
ON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS NORBERT KAIBLINGER Abstract. Results of Lid o Lehmer s problem iclude the value of the Lehmer costat of the fiite cyclic group Z/Z, for 5 ad all odd. By complemetary
More informationThe Structure of Z p when p is Prime
LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z
More informationCERTAIN GENERAL BINOMIAL-FIBONACCI SUMS
CERTAIN GENERAL BINOMIAL-FIBONACCI SUMS J. W. LAYMAN Virgiia Polytechic Istitute State Uiversity, Blacksburg, Virgiia Numerous writers appear to have bee fasciated by the may iterestig summatio idetitites
More informationThe log-concavity and log-convexity properties associated to hyperpell and hyperpell-lucas sequences
Aales Mathematicae et Iformaticae 43 2014 pp. 3 12 http://ami.etf.hu The log-cocavity ad log-covexity properties associated to hyperpell ad hyperpell-lucas sequeces Moussa Ahmia ab, Hacèe Belbachir b,
More informationAN APPLICATION OF HYPERHARMONIC NUMBERS IN MATRICES
Hacettepe Joural of Mathematic ad Statitic Volume 4 4 03, 387 393 AN APPLICATION OF HYPERHARMONIC NUMBERS IN MATRICES Mutafa Bahşi ad Süleyma Solak Received 9 : 06 : 0 : Accepted 8 : 0 : 03 Abtract I thi
More informationQ-BINOMIALS AND THE GREATEST COMMON DIVISOR. Keith R. Slavin 8474 SW Chevy Place, Beaverton, Oregon 97008, USA.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 2008, #A05 Q-BINOMIALS AND THE GREATEST COMMON DIVISOR Keith R. Slavi 8474 SW Chevy Place, Beaverto, Orego 97008, USA slavi@dsl-oly.et Received:
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationA METHOD TO SOLVE THE DIOPHANTINE EQUATION ax 2 by 2 c 0
A METHOD TO SOLVE THE DIOPHANTINE EQUATION ax by c Floreti Smaradache, Ph D Associate Professor Chair of Departmet of Math & Scieces Uiversity of New Mexico College Road Gallup, NM 87, USA E-mail:smarad@um.edu
More informationGENERALIZATIONS OF ZECKENDORFS THEOREM. TilVIOTHY J. KELLER Student, Harvey Mudd College, Claremont, California
GENERALIZATIONS OF ZECKENDORFS THEOREM TilVIOTHY J. KELLER Studet, Harvey Mudd College, Claremot, Califoria 91711 The Fiboacci umbers F are defied by the recurrece relatio Fi = F 2 = 1, F = F - + F 0 (
More informationChain conditions. 1. Artinian and noetherian modules. ALGBOOK CHAINS 1.1
CHAINS 1.1 Chai coditios 1. Artiia ad oetheria modules. (1.1) Defiitio. Let A be a rig ad M a A-module. The module M is oetheria if every ascedig chai!!m 1 M 2 of submodules M of M is stable, that is,
More informationHomework 9. (n + 1)! = 1 1
. Chapter : Questio 8 If N, the Homewor 9 Proof. We will prove this by usig iductio o. 2! + 2 3! + 3 4! + + +! +!. Base step: Whe the left had side is. Whe the right had side is 2! 2 +! 2 which proves
More informationSolutions to Tutorial 5 (Week 6)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial 5 (Wee 6 MATH2962: Real ad Complex Aalysis (Advaced Semester, 207 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
More informationds n SOME APPLICATIONS OF LEGENDRE NUMBERS ps(x) (i x2)s/2dsp (x), KEY WORDS ANY PHRASES. Aoed Legenre functions and polynomials, Legenre polynomials,
Iterat. J. Math. & Math. Sci. VOL. II NO. (1988) 45-41 45 SOME APPLICATIONS OF LEGENDRE NUMBERS PAUL W. HAGGARD Departmet of Mathematics, East Carolia Uiversity Greeville, North Carolia 7858 U.S.A. (Received
More informationMath 140A Elementary Analysis Homework Questions 1
Math 14A Elemetary Aalysis Homewor Questios 1 1 Itroductio 1.1 The Set N of Natural Numbers 1 Prove that 1 2 2 2 2 1 ( 1(2 1 for all atural umbers. 2 Prove that 3 11 (8 5 4 2 for all N. 4 (a Guess a formula
More informationWorksheet on Generating Functions
Worksheet o Geeratig Fuctios October 26, 205 This worksheet is adapted from otes/exercises by Nat Thiem. Derivatives of Geeratig Fuctios. If the sequece a 0, a, a 2,... has ordiary geeratig fuctio A(x,
More informationPrinciple of Strong Induction
Strog Iductio Pricile of Strog Iductio Let P() be a statemet about the th iteger. If the followig hyotheses hold: i. P(1) is True. ii. The statemet P(1) P(2) P() P( +1) is True for all 1. The we ca coclude
More informationLECTURE NOTES, 11/10/04
18.700 LECTURE NOTES, 11/10/04 Cotets 1. Direct sum decompositios 1 2. Geeralized eigespaces 3 3. The Chiese remaider theorem 5 4. Liear idepedece of geeralized eigespaces 8 1. Direct sum decompositios
More informationWeek 5-6: The Binomial Coefficients
Wee 5-6: The Biomial Coefficiets March 6, 2018 1 Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( 1 1 + 1 The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers
More informationON SOME DIOPHANTINE EQUATIONS RELATED TO SQUARE TRIANGULAR AND BALANCING NUMBERS
Joural of Algebra, Number Theory: Advaces ad Applicatios Volume, Number, 00, Pages 7-89 ON SOME DIOPHANTINE EQUATIONS RELATED TO SQUARE TRIANGULAR AND BALANCING NUMBERS OLCAY KARAATLI ad REFİK KESKİN Departmet
More informationAn analog of the arithmetic triangle obtained by replacing the products by the least common multiples
arxiv:10021383v2 [mathnt] 9 Feb 2010 A aalog of the arithmetic triagle obtaied by replacig the products by the least commo multiples Bair FARHI bairfarhi@gmailcom MSC: 11A05 Keywords: Al-Karaji s triagle;
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More informationGenerating Functions II
Geeratig Fuctios II Misha Lavrov ARML Practice 5/4/2014 Warm-up problems 1. Solve the recursio a +1 = 2a, a 0 = 1 by usig commo sese. 2. Solve the recursio b +1 = 2b + 1, b 0 = 1 by usig commo sese ad
More informationFLOOR AND ROOF FUNCTION ANALOGS OF THE BELL NUMBERS. H. W. Gould Department of Mathematics, West Virginia University, Morgantown, WV 26506, USA
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A58 FLOOR AND ROOF FUNCTION ANALOGS OF THE BELL NUMBERS H. W. Gould Departmet of Mathematics, West Virgiia Uiversity, Morgatow, WV
More informationk-generalized FIBONACCI NUMBERS CLOSE TO THE FORM 2 a + 3 b + 5 c 1. Introduction
Acta Math. Uiv. Comeiaae Vol. LXXXVI, 2 (2017), pp. 279 286 279 k-generalized FIBONACCI NUMBERS CLOSE TO THE FORM 2 a + 3 b + 5 c N. IRMAK ad M. ALP Abstract. The k-geeralized Fiboacci sequece { F (k)
More informationMATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1
MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The
More informationOn the Jacobsthal-Lucas Numbers by Matrix Method 1
It J Cotemp Math Scieces, Vol 3, 2008, o 33, 1629-1633 O the Jacobsthal-Lucas Numbers by Matrix Method 1 Fikri Köke ad Durmuş Bozkurt Selçuk Uiversity, Faculty of Art ad Sciece Departmet of Mathematics,
More informationSome Extensions of the Prabhu-Srivastava Theorem Involving the (p, q)-gamma Function
Filomat 31:14 2017), 4507 4513 https://doi.org/10.2298/fil1714507l Published by Faculty of Scieces ad Mathematics, Uiversity of Niš, Serbia Available at: http://www.pmf.i.ac.rs/filomat Some Extesios of
More informationOn the Inverse of a Certain Matrix Involving Binomial Coefficients
It. J. Cotemp. Math. Scieces, Vol. 3, 008, o. 3, 5-56 O the Iverse of a Certai Matrix Ivolvig Biomial Coefficiets Yoshiari Iaba Kitakuwada Seior High School Keihokushimoyuge, Ukyo-ku, Kyoto, 60-0534, Japa
More informationSOME FIBONACCI AND LUCAS IDENTITIES. L. CARLITZ Dyke University, Durham, North Carolina and H. H. FERNS Victoria, B. C, Canada
SOME FIBONACCI AND LUCAS IDENTITIES L. CARLITZ Dyke Uiversity, Durham, North Carolia ad H. H. FERNS Victoria, B. C, Caada 1. I the usual otatio, put (1.1) F _
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationINEQUALITIES BJORN POONEN
INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad
More informationMath 680 Fall Chebyshev s Estimates. Here we will prove Chebyshev s estimates for the prime counting function π(x). These estimates are
Math 680 Fall 07 Chebyshev s Estimates Here we will prove Chebyshev s estimates for the prime coutig fuctio. These estimates are superseded by the Prime Number Theorem, of course, but are iterestig from
More informationRamanujan s Famous Partition Congruences
Ope Sciece Joural of Mathematics ad Applicatio 6; 4(): - http://wwwopescieceoliecom/joural/osjma ISSN:8-494 (Prit); ISSN:8-494 (Olie) Ramauja s Famous Partitio Cogrueces Md Fazlee Hossai, Nil Rata Bhattacharjee,
More informationInduction proofs - practice! SOLUTIONS
Iductio proofs - practice! SOLUTIONS 1. Prove that f ) = 6 + + 15 is odd for all Z +. Base case: For = 1, f 1) = 41) + 1) + 13 = 19. Sice 19 is odd, f 1) is odd - base case prove. Iductive hypothesis:
More informationJournal of Ramanujan Mathematical Society, Vol. 24, No. 2 (2009)
Joural of Ramaua Mathematical Society, Vol. 4, No. (009) 199-09. IWASAWA λ-invariants AND Γ-TRANSFORMS Aupam Saikia 1 ad Rupam Barma Abstract. I this paper we study a relatio betwee the λ-ivariats of a
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationPROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1.
Math 7 Sprig 06 PROBLEM SET 5 SOLUTIONS Notatios. Give a real umber x, we will defie sequeces (a k ), (x k ), (p k ), (q k ) as i lecture.. (a) (5 pts) Fid the simple cotiued fractio represetatios of 6
More informationA L A BA M A L A W R E V IE W
A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N
More informationMath 609/597: Cryptography 1
Math 609/597: Cryptography 1 The Solovay-Strasse Primality Test 12 October, 1993 Burt Roseberg Revised: 6 October, 2000 1 Itroductio We describe the Solovay-Strasse primality test. There is quite a bit
More informationA New Sifting function J ( ) n+ 1. prime distribution. Chun-Xuan Jiang P. O. Box 3924, Beijing , P. R. China
A New Siftig fuctio J ( ) + ω i prime distributio Chu-Xua Jiag. O. Box 94, Beijig 00854,. R. Chia jiagchuxua@vip.sohu.com Abstract We defie that prime equatios f (, L, ), L, f (, L ) (5) are polyomials
More informationSOME TRIBONACCI IDENTITIES
Mathematics Today Vol.7(Dec-011) 1-9 ISSN 0976-38 Abstract: SOME TRIBONACCI IDENTITIES Shah Devbhadra V. Sir P.T.Sarvajaik College of Sciece, Athwalies, Surat 395001. e-mail : drdvshah@yahoo.com The sequece
More informationInverse Matrix. A meaning that matrix B is an inverse of matrix A.
Iverse Matrix Two square matrices A ad B of dimesios are called iverses to oe aother if the followig holds, AB BA I (11) The otio is dual but we ofte write 1 B A meaig that matrix B is a iverse of matrix
More informationSome Trigonometric Identities Involving Fibonacci and Lucas Numbers
1 2 3 47 6 23 11 Joural of Iteger Sequeces, Vol. 12 (2009), Article 09.8.4 Some Trigoometric Idetities Ivolvig Fiboacci ad Lucas Numbers Kh. Bibak ad M. H. Shirdareh Haghighi Departmet of Mathematics Shiraz
More informationCOMPLEX FACTORIZATIONS OF THE GENERALIZED FIBONACCI SEQUENCES {q n } Sang Pyo Jun
Korea J. Math. 23 2015) No. 3 pp. 371 377 http://dx.doi.org/10.11568/kjm.2015.23.3.371 COMPLEX FACTORIZATIONS OF THE GENERALIZED FIBONACCI SEQUENCES {q } Sag Pyo Ju Abstract. I this ote we cosider a geeralized
More informationHomework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger.
Homewor 1 Solutios Math 171, Sprig 2010 Hery Adams The exercises are from Foudatios of Mathematical Aalysis by Richard Johsobaugh ad W.E. Pfaffeberger. 2.2. Let h : X Y, g : Y Z, ad f : Z W. Prove that
More information