Determinants Containing Powers of Generalized Fibonacci Numbers

Transcription

1 Journal of Integer Sequences, Vol 19 (2016), Artcle 1671 Determnants Contanng Powers of Generalzed Fbonacc Numbers Aram Tangboonduangjt and Thotsaporn Thanatpanonda Mahdol Unversty Internatonal College Nakhonpathom Thaland aramtan@mahdolacth thotsaporntha@mahdolacth Abstract We study the determnants of matrces whose entres are powers of the Fbonacc numbers We then extend the results to nclude entres that are powers of a secondorder lnear recurrence relaton These results motvate a fundamental dentty of determnants whose entres are powers of lnear polynomals Fnally, we dscuss the determnants of matrces whose entres are products of the general second-order lnear recurrence relatons 1 Introducton In the frst ssue of the Fbonacc Quarterly, Alfred posed the followng problem [1]: Prove Fn 2 Fn+1 2 F 2 n+2 Fn+1 2 Fn+2 2 Fn+3 2 = 2( 1) n+1, Fn+2 2 Fn+3 2 Fn+4 2 where F n s the nth Fbonacc number 1

2 In the second volume of the Fbonacc Quarterly, Parker [7] posed a smlar problem wth the exponent of each entry changed to 4 and the dmenson of the matrx changed to 5 5 These two results naturally suggest the followng queston: what would be the determnant of an analogous form where the dmenson of the matrx and the exponent of each entry are arbtrary? Carltz [4] answered ths queston by showng that the determnant of the form D(r,n) =, where,j = 0,1,,r, s gven by F r n++j D(r,n) = ( 1) (n+1)(r+1 2 ) (F r 1 F r 1 2 F r ) 2 (1) In ths paper we generalze the entres even further by consderng the determnant of the form D(r,s,k,n) =, where,j = 0,1,,r We show that F r s+k(n++j) D(r,s,k,n) = ( 1) (s+kn+1)(r+1 2 ) (F r k F r 1 2k F rk ) 2 (2) Carltz based hs proof of formula (1) on the Bnet form of the Fbonacc numbers; whereas we employ the matrx methods, such as the factorzaton method of Krattenthaler, to prove formula (2) In addton, we requre a generalzed form of the Catalan dentty, proved by Melham and Shannon [6] In Secton 2, we present an alternatve proof of the generalzed Catalan dentty usng a matrx representaton of the sequence and the propertes of the matrx multplcaton In Secton 3, we then present the proof of formula (2) as a specal case of the determnant wth entres nvolvng the powers of the numbers n a second-order lnear recurrence wth constant coeffcents In the last secton, we present the determnant whose entres are the products of the numbers defned as a second-order lnear recurrence wth constant coeffcents usng the Desnanot-Jacob dentty The methodology used for ths work reles on a computer programmng developed by the second author [8] 2 Generalzed Catalan dentty The well-known Catalan dentty states that for all nonnegatve ntegers s and, F 2 s+ F s F s+2 = ( 1) s F 2 A generalzaton of ths dentty useful for ths work s gven by Melham and Shannon [6] We shall, however, present an alternatve proof of ths generalzaton For ntegers a,b,c 1, and c 2 wth c 2 0, let W n = W n (a,b;c 1,c 2 ) denote the second-order lnear recurrence wth constant coeffcents, defned by W 0 = a, W 1 = b and W n = c 1 W n 1 +c 2 W n 2 for n 2 Wth ths notaton, the Fbonacc sequence (F n ) and the Lucas sequence (U n ) correspond to F n = W n (0,1;1,1) and U n = W n (0,1;c 1,c 2 ), respectvely Moreover, we can use ths 2

3 recurrence to extend the defnton of a sequence to the terms wth negatve ndces Usually, we can explctly fnd the relatonshp between the negatve-ndexed terms and the postvendexed terms For example, for the Fbonacc sequence and the Lucas sequence, we have F n = ( 1) n+1 F n and U n = ( 1) n+1 c n 2 U n for n 1 Proposton 1 (Generalzed Catalan Identty) Let W n = W n (a 0,a 1 ;c 1,c 2 ) and Y n = W n (b 0,b 1 ;c 1,c 2 ) be second-order lnear recurrences Then for all ntegers s,j, and W s+ Y s+j W s Y s++j = ( c 2 ) s (W 1 Y j W 0 Y j+1 ) U, (3) Proof The proof s by nducton on For the case when = 0, the dentty s trval For the case when = 1, we have ( ) ( )( ) ( ) s ( ) W s+1 Y s+j+1 c = 1 c 2 W s Y s+j c = 1 c 2 W 1 Y j+1 for s 0, W s Y s+j 1 0 W s 1 Y s+j W 0 Y j and ( ) ( ) 1 ( ) ( ) s ( ) W s+1 Y s+j+1 c = 1 c 2 W s+2 Y s+j+2 c = 1 c 2 W 1 Y j+1 W s Y s+j 1 0 W s+1 Y s+j W 0 Y j for s < 0, where the second equalty n both equatons follows from repeated applcaton of the matrx representaton n the frst equalty Takng the determnant of both sdes of any one equaton yelds W s+1 Y s+j W s Y s+j+1 = ( c 2 ) s (W 1 Y j W 0 Y j+1 ) (4) Now, consder two cases Case > 1: Assume that the dentty holds for some ntegers 1 and 2 Then W s+ Y s+j W s Y s++j = W s+ Y s++j W s Y s+j = c 1 W s+( 1) +c 2 W s+( 2) c 1 Y s+( 1)+j +c 2 Y s+( 2)+j W s Y s+j W = c s+( 1) Y s+( 1)+j 1 W s Y s+j +c 2 W s+( 2) Y s+( 2)+j W s Y s+j = ( c 2 ) s (W 1 Y j W 0 Y j+1 )(c 1 U 1 +c 2 U 2 ) = ( c 2 ) s (W 1 Y j W 0 Y j+1 )U, 3

4 Case < 0: Assume that the dentty holds for some ntegers +1 and +2 Then W W s+ Y s+j W s Y s++j = s+ Y s++j W s Y s+j c 1 = W s+(+1) + 1 c 1 W s+(+2) Y s+(+1)+j + 1 Y s+(+2)+j c 2 c 2 c 2 c 2 W s Y s+j = c 1 W s+(+1) Y s+(+1)+j c 2 W s Y s+j + 1 W s+(+2) Y s+(+2)+j c 2 W s Y s+j = ( c 2 ) s (W 1 Y j W 0 Y j+1 )( c 1 c 2 U c 2 U +2 ) = ( c 2 ) s (W 1 Y j W 0 Y j+1 )U, where we apply the nducton hypothess n the penultmate equalty n both cases Hence, the proof s complete and We note some specal cases of Proposton 1 useful n later sectons: U s+ U s+j U s U s++j = ( c 2 ) s U U j, (5) U s+ W s+j U s W s++j = ( c 2 ) s U W j, (6) W s+ W s+j W s W s++j = ( c 2 ) s (W 1 W j W 0 W j+1 )U = ( c 2 ) s U U j, (7) W where = 1 W 2 W 0 W 1 = a2 1 c 1 a 0 a 1 c 2 a 2 0 We justfy the second equalty of (7) by applyng Proposton 1 as follows: In (3), let Y n = W n, substtute j = 1 and s = 0, respectvely, and rename the ndex by j The dentty (7) can be restated as follows: Corollary 2 Let k,n,r,s, and t be ntegers Then W s+k(n+t) = A(t)W s+kn +B(t)U s+k(n+r), (8) where A(t) = W k(t r) W kr and B(t) = ( c 2) kr U kt W kr Proof Let ntegers k,n,r,s, and t be gven Applyng (7) wth s = k r, = k t, and j = s +k (n +r ), we have W k r W s +k (n +t ) = W k (t r )W s +k n ( c 2) k r U k t U s +k (n +r ) Dvdng by W k r on both sdes and renamng the varables yeld the dentty (8), as requred 4

5 3 Determnants nvolvng powers of terms of secondorder recurrence Our goal n ths secton s to gve the closed form of the determnant of the (r+1) (r+1) matrx whose entres are Ws+k(n++j) r, where,j = 0,1,,r, and s and k are any ntegers Ths matrx s Ws+kn r Ws+k(n+1) r Ws+k(n+r) r W A s,k n (r) = s+k(n+1) r Ws+k(n+2) r W r s+k(n+r+1) (9) Ws+k(n+r) r Ws+k(n+r+1) r Ws+k(n+2r) r We begn wth the followng proposton on the determnant whose entres are some power of lnear polynomals Lemma 3 Let c 0,,c r and x 0,,x r be real numbers Then det((c j x +1) r ) 0,j r = (x x j ) (c c j ) (10) Proof We prove Lemma 3 by usng the factorzaton method [5] The determnant wll be 0 f x 0 s replaced by any x for 0 < r, snce some two rows of the matrx would be equal Ths mples that (x 0 x ) s a factor of the determnant for each = 1,2,,r Smlarly, we have that (x 1 x ) s a factor of the determnant for each = 2,,r, and so on In a smlar manner, we see that f c 0 s replaced by any c j for 0 < j r, then two columns of the matrx wll be the same yeldng the zero determnant Ths mples that (c 0 c j ) s a factor of the determnant for each j = 1,2,,r Smlarly, we have that (c 1 c j ) s a factor of the determnant for each j = 2,,r, and so on Therefore (x x j ) (c c j ) (11) s a factor of ths determnant As a functon of x for some fxed or a functon of c j for some fxed j the determnant s a polynomal of degree r Ths mples that the factor (11) and the requred determnant have the same degree Therefore, we can wrte the determnant as det((c j x +1) r ) 0,j r = C (x x j ) (c c j ), for some constant C To fnd C, we compare on both sdes the coeffcents of the monomal (c r x r ) r (c r 1 x r 1 ) r 1 (c 0 x 0 ) 0 (12) 5

6 On the rght-hand sde, the monomal (12) appears as ( x j )( c j ) = 0<<j r 0<<j r Hence, the coeffcent of the monomal (12) on the rght-hand sde s just equal to C We see that for each 0 r, the term (c x ) appears n (c x +1) r Hence, on the left-hand sde, the monomal (12) appears as ± (c x +1) r 0 r x j c j By the defnton of the determnant, the sgn n front of the expresson s determned by the party of the dentty permutaton (0)(1) (r) Snce (0)(1) (r) = (01)(01), t follows that the dentty permutaton s even Hence, the sgn s determned to be + Snce, for each 0 r, the coeffcent of (c x ) n (c x +1) r s ( r ), t follows that Ths completes the proof of Lemma 3 C = Corollary 4 Let A(j),B(j),X,Y be real numbers for 0,j r Then det((a(j)x +B(j)Y ) r ) 0,j r = (X Y j X j Y ) (A()B(j) A(j)B()) (13) Proof We prove n the case of B(j) 0 and Y 0 for all 0,j r Applyng Lemma 3 wth c j = A(j)/B(j) and x = X /Y for all 0,j r and clearng the denomnators, we obtan (13) The proof of the other case when some of B(j) or Y are 0 follows from the fact that the determnant wth polynomal entres s a contnuous functon Thus, ths allows us to prove one of the man results of ths paper Theorem 5 The determnant of the matrx A s,k n (r) s gven by deta s,k n (r) = ( 1) (s+kn+1)(r+1 2 ) (s+kn)( c r+1 2 (r+1 2 ) 2 )+2k( r+1 3 ) U 2(r ) (+1)k 6

7 Proof By (9), (8), (13), and (6) respectvely, we have deta s,k n (r) = det(ws+k(n++j)) r 0,j r = det( )0,j r A(j)W s+k(n+) +B(j)U s+k(n+r+) = ( ) r (W s+k(n+) U s+k(n+r+j) W s+k(n+j) U s+k(n+r+) ) A()B(j) A(j)B() = = (( c 2 ) s+k(n+r+) U k(j ) W kr ) ( ( c 2 ) s+k(n+2) U 2 k(j ) ) ( r ( ) ( c 2 ) kr ( c W kr 2 2 ) k U k(j ) W kr ), ( r ) Rearrangng the last expresson allows us to obtan the desred dentty Remark 6 By lettng W n = F n n Theorem 5 and notng that for the Fbonacc sequence c 2 = 1 and = 1, we then derve (2) 4 Determnants nvolvng products of terms of secondorder recurrence The followng lemma was mentoned by Krattenthaler[5] as part of the factorzaton method We provde a dfferent proof of ths lemma usng the Desnanot-Jacob dentty [3] Lemma 7 Let X 0,,X r,d 1,,D r, and E 1,,E r be ndetermnates Then j det (X +D l ) (X +E m ) = (X j X ) (D j E ) l=j+1 m=1 0,j r An alternatve way of wrtng ths dentty would be det l=j+1 (A(d l )X +B(e l )Y ) = j (A(e m )X +B(e m )Y ) m=1 (X Y j X j Y ) The man result of ths secton s as follows: 1 j r 0,j r 1 j r (B(e )A(d j ) A(e )B(d j )) (14) 7

8 Theorem 8 Let d 1,,d r and e 1,,e r be sequences of ntegers Then det l=j+1 W s+k(n++dl ) j m=1 W s+k(n++em) 0,j r = ( ) (r+1 2 ) ( c2 ) (s+kn)(r+1 2 )+k( r+1 3 ) l=1 U r+1 l kl 1 j r ( c 2 ) kd j U k(e d j ) Proof We respectvely apply the denttes (8),(14), and (7), and the detals of the proof are smlar to those of Theorem 5 The followng slght varaton of the result by Alfred [2] called basc power determnant s a specal case of ths theorem Corollary 9 Let r,s, and k be ntegers wth r 0 Then Fs r Fs r 1 F s+k F r s+k Fs+k r F r 1 s+k F s+2k Fs+2k r = ( 1) (s+1)(r+1 2 )+k( r+1 3 ) ( F r+1 k Fs+rk r F r 1 s+rk F s+(r+1)k Fs+(r+1)k r 2 ) l=1 F r+1 l kl Proof Ths dentty follows mmedately from Theorem 8 by lettng d 1 = d 2 = = d r = 0, e 1 = e 2 = = e r = 1, n = 0 together wth the recurrence and the ntal values of the Fbonacc numbers Another nterestng case arses when we let (d j ) and (e j ) n Theorem 8 be n some specfc forms Corollary 10 Let s,k,n, and p be ntegers Let d j = p 1+j and e j = j 1 for 1 j r Then det l=j+1 W s+k(n++dl ) j m=1 5 Acknowledgments W s+k(n++em) 0,j r = (r+1 2 ) ( c2 ) (s+kn)(r+1 2 )+2k( r+1 3 ) l=1 r 1 U r+1 l lk U r l k(p+l) We thank the anonymous referee and Ramesh Boonratana for ther metculous readng and helpful comments 8 l=0

9 References [1] Brother U Alfred, Advanced problems and solutons, Fbonacc Quart 1 (1963), 48 [2] Brother U Alfred, Some determnants nvolvng powers of Fbonacc numbers, Fbonacc Quart 2 (1964), [3] T Amdeberhan and D Zelberger, Determnants through the lookng glass, Adv Appl Math 27 (2001), [4] L Carltz, Some determnants contanng powers of Fbonacc numbers, Fbonacc Quart 4 (1966), [5] C Krattenthaler, Advanced determnant calculus: a complement, Lnear Algebra Appl 411 (2005), [6] R S Melham and A G Shannon, A generalzaton of the Catalan dentty and some consequences, Fbonacc Quart 33 (1995), [7] F D Parker, Advanced problems and solutons, Fbonacc Quart 2 (1964), 303 [8] T Thanatpanonda, Personal webste, Mathematcs Subject Classfcaton: Prmary 11B39; Secondary 11B20 Keywords: Fbonacc number, determnant, second-order recurrence, Catalan dentty (Concerned wth sequence A000045) Receved February ; revsed versons receved June ; July Publshed n Journal of Integer Sequences, August Return to Journal of Integer Sequences home page 9