ADVANCED PROBLEMS AND SOLUTIONS

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1 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 2050, JOHANNESBURG, SOUTH AFRICA or by at the address as files of the type te, dvi, ps, doc, html, pdf, etc. This departmet especially welcomes problems believed to be ew or etedig old results. Proposers should submit solutios or other iformatio that will assist the editor. To facilitate their cosideratio, all solutios set by regular mail should be submitted o separate siged sheets withi two moths after publicatio of the problems. PROBLEMS PROPOSED IN THIS ISSUE H-755 Proposed by D. M. Bătieţu-Giurgiu, Bucharest ad Neculai Staciu, Buzău, Romaia. Let be a iteger. Prove that ( If R for,...,, the ( 2 (2 If m, the L si ( L cos (L L + 2. m m (+L 2 m+ (m+ m+ (L H-756 Proposed by Russell J. Hedel, Towso Uiversity. We see to geeralize a ow problem which states that #{,..., + : } F +3, where i are Boolea variables for i,...,. To geeralize the above formula, we (i fi itegers d, i with d > i ; (ii let D j be products of d Boolea variables with cosecutive idices such that D j ad D j+ have i variables i commo; (iii let m be the total umber of variables occurrig i D,...,D ad (iv let S #{,..., m : D D 2 D 0}. Determie the coefficiets of the miimal recursio satisfied by the {S }. AUGUST

2 THE FIBONACCI QUARTERLY H-757 Proposed by H. Ohtsua, Saitama, Japa. For a odd prime p prove that p { 2( (p+/4 (mod F L p if p (mod 4, ( (p /4 F p 3 (mod F p if p (mod 4. H-758 Proposed by D. M. Bătieţu-Giurgiu, Bucharest ad Neculai Staciu, Buzău, Romaia. Compute: lim ( ( ( ( F m! F m+ π(+ + + (2!! ta 4 H-759 Proposed by H. Ohtsua, Saitama, Japa. Let r 2 be a iteger. Defie the sequece {G } by G G + +G r ( with arbitrary G 0,G,...,G r+. For a iteger, prove that r G 2 (r +2 (G +i G +i G i G i. 2(r i SOLUTIONS A Idetity Ivolvig Middle Biomial Coefficiets H-727 Proposed by Bassem Ghalayii, Louaize, Lebao. (Vol. 50, No. 3, August 202 Let be a atural umber. Prove that ( 2 (2+ Solutio by Eduardo Brietze. 0 i,j, i+j+ 0 ( 2i i ( 2j j ( 2. Fm+2. The geeratig fuctio for the cetral biomial coefficiets is ( 2. ( 4 Replacig by 2 ad multiplyig by we obtai Differetiatig we get 4 2 ( ( ( 2 ( VOLUME 52, NUMBER 3

3 Replacig bac 2 by ad usig ( yields ( ( 2 3 or, i,j, i+j+ ad the desired idetity follows. ( 2i i ( 2j j ADVANCED PROBLEMS AND SOLUTIONS ( 2 (2+ 0 ( 2 0, ( 2 (2+, Also solved by Paul S. Brucma, Keeth B. Daveport, Ágel Plaza, ad the proposer. Iequalities With Cosecutive Fiboacci Numbers, Square Roots ad Powers H-728 Proposed by D. M. Bătieţu-Giurgiu, Bucharest ad Neculai Staciu, Buzău, Romaia. (Vol. 50, No. 4, November 202 Let a,b,c,m be positive real umbers ad be a positive real umber. Prove that: F F (a + F , F 2 +af + F +2 F + 2 +bf +2F F cf F + (b provided that a+b+c 24; a 3m 3 (F b+f + c m+ + b 3m 3 (F c+f + a m+ + c 3m 3 (F a+f + b m+ 3 provided that abc. Solutio by Ágel Plaza. F+2 m+ Part (a is a direct cosequece of a more geeral iequality: Let,y,z,a,b,c be positive real umbers, with a+b+c 24. The 2 +ayz + y y 2 +bz + z z 2 +cy. By Hölder s iequality ( ( ( ( 2 +µyz +µyz 2 +µyz (+y +z 3, 2 where the sums are cyclic ad the coefficiet µ meas the correspodig coefficiet a,b, or c. Therefore, we eed oly to show that which is equivalet to (+y +z 3 3 +y 3 +z 3 +(a+b+cyz, (+y(y +z(z + (a+b+c yz. 3 Ad this is true due to the AM-GM iequality: ( ( ( +y y +z z + 8 a+b+c. y yz z 3 AUGUST ,

4 THE FIBONACCI QUARTERLY For part (b it is eough to prove the followig more geeral iequality: Let,y,a,b,c be positive real umbers, with abc. The a 3 (b+yc + b 3 (c+ya + c 3 (a+yb 3 +y, which is a cosequece of (b+yc(c+ya(a+yb (+y 3, sice by the AM-GM iequality a 2 b+b 2 c+c 2 a 3. Also solved by Paul S. Brucma, Dmitry Fleischma, ad the proposers. O the Sequece of Rotatioal Numbers H-729 Proposed by Paul S. Brucma, Naaimo, BC. (Vol. 50, No. 4, November 202 Defieasequece{a } 0 ofratioalumbersbytherecurrece δ i,j is the Kroecer symbol which equals if i j ad 0; otherwise. a (a Prove that γ, the Euler costat; 0 0 a + δ,0, where (b Prove that a + + u a for, where u m 2(H m (m+2 H m m for all m. Solutio by Aastasios Kotrois. We start provig a lemma. Lemma. For 0, a. Proof. From the recurrece we get a 0 ad a /2. Assume that a m for m. From the recurrece, we get (+a (+2a ad a + + a a + 0, ( a + +2 a a + 0. (2 2 3 Now usig the iductio hypothesis ad subtractig ( from (2, we get: a + ( + a ( ( H H VOLUME 52, NUMBER 3

5 ADVANCED PROBLEMS AND SOLUTIONS Net, we compute the geeratig fuctio of a. Multiplyig by ad summig for the recurrece defiig a for 0, we get a + δ,0, 0 0 or, equivaletly, or, equivaletly, therefore, 0 a l( +, a, A( : a l(, from where we get that 0 a has radius of covergece. Now for (,, we have so, A( + a, A( a, which implies that A(t dt a 0 t. But from Lemma, we have that a / / so from a ow theorem of Hardy ad Littlewood (see [, p. 65, Theorem 8.4], we have that a coverges ad (see [2, p. ]. This proves part (a. a A(t dt 0 t 0 lt + t dt γ (b With the covetio that the sum over a empty set of idices is zero, we get that H 0 0, so u 0 ad what we wat to prove is equivalet to a u, (. + 0 Deotig by U( the geeratig fuctio of u, multiplyig by ad summig for, we get u 0a +, AUGUST

6 THE FIBONACCI QUARTERLY therefore, so A(U( a 0 u 0 (A(, U( l2 ( 2 ad it suffices to show that ( l [ 2 ( ] 2 +2 l( +2 l(, 2(H, +2 where [ ]f( deotes the coefficiet of i the Taylor epasio of f(. But [ ] l2 ( 2 ( +( + 0 ( so 2H + +2, ( l [ 2 ( ] 2 +2 l( 0 2H ( 2 H H (+( (H. +2 Refereces [] A. M. Odlyzo, Asymptotic eumeratio methods, [2] P. Sebah ad X. Gourdo, Collectio of formulae for Euler s costat γ, Also solved by G. C. Greubel ad the proposer. Idetities Ivolvig Fiboacci, Lucas ad Pell Numbers H-730 Proposed by N. Gauthier, Kigsto, ON. (Vol. 5, No., February 203 Let be the largest iteger less tha or equal to ad put ε ( +( /2. The, with P the th Pell umber prove the followig idetities: (a ( 2 25 [ 2 5 /2 ε (L L + +( ε ] 5(F F + ; 6 (b 6 ( [P +]; 286 VOLUME 52, NUMBER 3

7 ADVANCED PROBLEMS AND SOLUTIONS (c (d ( / ( 4 5 ( ( 2 2 [ 5 /2 ε (L 2 +L 2(+( /2 ] + ( ε 5(F 2 +F 5 /2 54 [ε ((45 20F 2 5L 2 + ( ε ] 5((9 4L 2 5F 2. ; Solutio by Ágel Plaza. (a I order to prove the equality we will show that both sides of the equality preset the same geeratig fuctio. For the left-had side we use the Sae Oil Method [] applied to ( 2. a 25 2 Let A( be the geeratig fuctio of {a }. That is, A( ( ( 2+ ( 25( 25( 2 4, where we have used the idetity ( r r r 0 ( + 0 ( ( 2 ( 0, 2 (see eq. (4.3., page 20 i []. The, the geeratig fuctio for the eve terms of the cosidered sequece is A e ( A(+A( 25( , ad the geeratig fuctio for the odd terms of the sequece is A o ( A( A( 2 25( Fortheright-hadsideof(awecosiderthetwocases: eve, adodd, sicerespectively ε ad ε 0. By usig the Biet s formulas for Lucas ad Fiboacci umbers ad the sum of geometric series it is a routie to get the same geeratig fuctios obtaied for the left-had side of (a, A e ( ad A o (. AUGUST

8 THE FIBONACCI QUARTERLY (b Let A( be the geeratig fuctio of the sequece at the left-had side. That is, A( ( ( ( ( 2+ ( 4 6( 2 6( 6( 2 4. For the right-had side of (b we may use the Biet s formula for the Pell umbers, or more directly its geeratig fuctio P( 2 2. The P 2 /2 2(/2 (/ Also, sice we the have Fially, sice the result follows , 2 d ( 2 d 2 2 ( (2 2 6( 6( 2 4, Idetities (c ad (d may be proved by similar argumets to the used for idetity (a. Refereces [] H. S. Wilf, Geeratigfuctioology, Academic Press, Ic., Secod ed Also solved by Paul S. Brucma, G. C. Greubel, ad the proposer. 288 VOLUME 52, NUMBER 3