SOME 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 0, 1, 1,, 4, 7, 13, 4, 44, is kow as Triboacci sequece ad the related recurrece relatio is T +3 = T + + T +1 + T ; for 1 with T 1 = 0, T = T 3 = 1. The purpose of this paper is to develop some iterestig idetities related with {T }. We first derive idetities for i=1 T i, i=1 T i, i=1 T i 1, i=0 T 4+t (t = 1,, 3, 4) ad i=1 T i. We the preset some iequalities coectig Triboacci umbers. We ext prove reductio formula for T m+ ad T m. We also itroduce a Q-matrix ad show its coectio with the sequece {T }. Fially we use it to derive some more reductio formulae for Triboacci umbers. Key Words : Triboacci umbers, Q-matrix, Idetities. AMS subject classificatio (000) : 11B39 1. Itroductio: By defiitio, a Fiboacci sequece cosists of umbers equal to the sum of the precedig two. Symbolically it meas that ay term F which satisfy F 1 = 0, F = 1 ad F + = F +1 + F for 1. This defiitio ca be exteded to defie ay term as the sum of the precedig three terms. It is the purpose of this paper to examie this sequece called Triboacci sequece (the ame obviously resultig from tri meaig three). This sequece cosists of terms T 1, T, T 3, T 4,, where we defie T 1 = 0, T = 1, T 3 = 1 ad ay followig terms give by T +3 = T + + T +1 + T, for 1. (1.1) First few Triboacci umbers are T 1 = 0, T = 1, T 3 = 1, T 4 =, T 5 = 4, T 6 = 7, T 7 = 13, T 8 = 4, T 9 = 44,. We cosider T 0 = 0. Numerous results are available i the literature for the sequece {F()} of Fiboacci umbers. But othig much is kow about the sequece {T()} of Triboacci umbers. Despite its simple appearace, the Triboacci sequece cotais a wealth of subtle ad fasciatig properties. I this paper we explore some fudametal idetities related with {T()}. We begi by demostratig the followig result.

Mathematics Today Vol.7(Dec-011) 1-9 Lemma.1.1: gcd(t, T +1, T + ) = 1 ; for all = 0, 1,,. Proof: Let positive iteger d be the required greatest commo divisor. The it is clear that all of T, T +1 ad T + are divisible by d. I this case T 1 = T + T +1 T will also be divisible by d. Cotiuig thus we ca see that d T, d T 3 ad so o. Evetually, we must have d T. Sice T = 1, we have d = 1. This says that 1 is the oly positive iteger which divides ay three successive terms of the Triboacci sequece. This proves the result.. Some fudametal idetities: Oe of the purposes of this paper is to develop some of the idetities related with the sequece {T()}. We use the techique of iductio as a useful tool i provig these idetities. We deote the sum of first Triboacci umbers by S ad first prove some idetities related with the summatio of Triboacci umbers. Lemma.1: S = T i = 1 i=1 (T + + T 1). Proof: We have T 1 = T 4 T 3 T T = T 5 T 4 T 3 T 3 = T 6 T 5 T 4 T 1 = T + T +1 T T = T +3 T + T +1. Addig all these equatios term by term we get, T 1 + T + T 3 + + T = (T +3 T +1 T 3 ) (T + T 3 + + T ). Addig ad subtractig T 1 we get S = (T + + T T 3 + T 1 ) S. Sice T 1 = 0 ad T 3 = 1, we get S = 1 (T + + T 1), as required. We here ote that T + + T is always a odd iteger, which makes the value of S a iteger. This is because of the followig result: Lemma.: T + ± T 1(mod ). Proof: We use iductio to prove the result. For = 1, we have T 3 ± T 1 = 1 + 0 1 (mod ).

D V Shah - Some Triboacci Idetities 3 Suppose it is true for = r. i.e. let T r+ ± T r 1(mod ) be true. Now, T r+3 ± T r+1 = (T r+ + T r+1 + T r ) ± T r+1 T r+ + T r (mod ) 1(mod ). So the result is true for = r + 1 also. This proves the result for all itegers. A alterate method of provig lemma..1 is to apply the priciple of mathematical iductio. Usig the same process or by iductio we ca derive formulae for the sum of the first Triboacci umbers with various subscripts. Lemma.3: (i) T i 1 = 1 i=1 (T + T 1 1) (ii) T i = 1 i=1 (T +1 + T ) (iii) T 4i+1 = 1 i=0 S 4+1 (iv) T 4i+ = 1 i=0 (S 4+ + 1) (v) T 4i+3 = 1 i=0 S 4+3 (vi) T 4i+4 = 1 i=0 S 4+4. Proof: All these results ca be proved easily by iductio. For the sake of brevity we provide the proof of first two results oly. Remaiig results ca be proved accordigly. (i) We have T 1 = 0 = 1 (1 + 0 1) = 1 (T + T 1 1). So the result is true for = 1. Let it r be true for some positive iteger = r. That is, we assume that T i 1 = 1 i=1 (T r + T r 1 1) is true. r+1 Now i=1 T i 1 = i=1 T i 1 + T r+1 r for all positive itegers. = 1 (T r + T r 1 1) + T r+1 = 1 (T r+1 + T r + T r 1 + T r+1 1) = 1 (T r+ + T r+1 1) = 1 (T (r+1) + T (r+1) 1 1). This proves the result for = r + 1 also. Hece by iductio, the result is true (ii) i=1 T i = i=1 T i T i 1 i=1. By usig lemma..1 ad first part of this lemma, this becomes T i = i=1 1 (T + + T 1) 1 (T + T 1 1)

4 Mathematics Today Vol.7(Dec-011) 1-9 = 1 (T + T 1 ) = 1 (T +1 + T + T 1 T 1 ) = 1 (T +1 + T ), as required. We ow prove a iequality which coects two cosecutive Triboacci umbers. Lemma.4: T +1 T for >. Proof: Clearly we have 1, i.e. T 4 T 3. So the result is true for = 3. Suppose it is true for all itegers such that 3 r. Thus all of T r 1 T r, T r T r 1 ad T r+1 T r are true. Now addig them we get T r 1 + T r + T r+1 (T r + T r 1 + T r ), i.e. T r+ T r+1. This shows that the result is true for = r + 1 also. Hece T +1 T is true for all >. We ext prove the ice iequality which shows how the value of S is placed betwee cosecutive Triboacci umbers. Lemma.5: T +1 < S < T + for > 4. Proof: We have 7 < 8 < 13, i.e. T 6 < S 5 < T 7. So the result is true for = 5. Let it be true for all itegers such that 5 r. So by iductio hypothesis we have T r < S r 1 < T r+1. But the we have T r 1 < T r < S r 1 < T r+1 < T r+. This ow shows that T r 1 < S r 1 < T r+. Addig T r + T r+1 throughout we get T r 1 + T r + T r+1 < S r 1 + T r + T r+1 < T r + T r+1 + T r+. Thus T r+ < S r+1 < T r+3, which shows that the result is true for = r + 1 also. This proves the result by iductio. Corollary.6: T 6 < S 5 < T 7 < < S < T < S 1 < T +1 < S <. Proof: Follows immediately by successive applicatio of lemma..5. Agroomof [1] derived the followig results for T. (i) T +m = T m+1 T + (T m + T m 1 )T 1 + T m T. (ii) T = T 1 + T (T +1 + T 1 + T ). (iii) T 1 = T + T 1 (T 1 + T ). Here we derive some other forms for T +m, T ad T 1.

D V Shah - Some Triboacci Idetities 5 Theorem.7: T m+ = T m+1 T +1 + T m T 1 + T m 1 T +T m T. Proof: Let m be a fixed positive iteger ad we will proceed by iductig o. Whe = 1, we have T m+1 = T m+1 T + T m T 0 + T m 1 T 1 +T m T 1. Sice T 0 = T 1 = 0, T = 1 we have T m+1 = T m+1, which is obviously true. Now let us assume that the idetity is true for = 1,, 3,, r ad we will show that it holds for = r + 1. By assumptio, T m+r = T m+1 T r+1 + T m T r 1 + T m 1 T r + T m T r, T m+(r 1) = T m+1 T r + T m T r + T m 1 T r 1 + T m T r 1 ad T m+(r ) = T m+1 T r 1 + T m T r 3 + T m 1 T r + T m T r. Addig them we get, T m+r + T m+(r 1) + T m+(r ) = T m+1 (T r+1 + T r + T r 1 ) + T m (T r 1 + T r + T r 3 ) + T m 1 (T r + T r 1 + T r ) +T m (T r + T r 1 + T r ). Thus T m+(r+1) = T m+1 T r+ + T m T r + T m 1 T r+1 + T m T r+1. This is precisely our idetity whe is replaced by r + 1. Hece by iductio, result is true for all positive itegers. As a example of this idetity we observe that T 15 = T 8+7 = T 9 T 8 + T 8 T 7 + T 7 T 7 +T 8 T 6 = 44 4 + 4 13 + 13 13 + 4 7 = 1705. We ow deduce some idetities usig theorem.7. The double argumet formula is as uder. Corollary.8: T = T +1 + T + T 1 T. Proof: By takig m = i theorem..7 the result follows easily. Corollary.9 (a) T 1 = T 1 + T T +1 T (b) T +1 = T + T +1 (T + + T + T 1 ). Proof: By cosiderig 1 ad + 1 i place of m i theorem..7, both the results follow immediately. Triboacci umbers: Zeitli [3] had proved the followig result for the summatio of square of first T i = 1 i=0 {T 8 +5 T +4 4T +3 10T + 9T +1 T + } Here we derive much more simple result for T i i=0.

6 Mathematics Today Vol.7(Dec-011) 1-9 Lemma.10: T i = 1 i=1 { 1 + 4T 4 T +1 (T +1 T 1 ) }. Proof: We prove the result by iductio o. Sice T 1 = 0 = 1 4 (1 + 4 0 1 (1 0) ) = 1 4 (1 + 4T 1T (T T 0 ) ), result is true for = 1. Suppose it holds for = r, i.e. as a iductio hypothesis we assume that r T i = 1 i=1 { 1 + 4T 4 r T r+1 (T r+1 T r 1 ) }. r+1 r Now i=1 T i = T i + by iductio. i=1 T r+1 = 1 { 1 + 4T 4 rt r+1 (T r+1 T r 1 ) } + T r+1 = 1 { 1 + 4T 4 r+1 + 4T r T r+1 + 4T r+1 T r 1 T r+1 T r+1 T r 1 T r 1 } = 1 4 { 1 + 4T r+1(t r+1 + T r + T r 1 ) (T r+1 + T r 1 ) } = 1 4 { 1 + 4T r+1t r+ (T r+ T r ) }. This is our required idetity whe is replaced by r + 1. This proves the result 3. Triboacci umbers with egative subscripts: It is ofte useful to exted the Triboacci sequece backward with egative subscripts. I fact if we try to exted the Triboacci sequece backwards still keepig to the rule that Triboacci umber is the sum of the three umbers o its left, we get the followig : : 8 7 6 5 4 3 1 0 1 3 4 5 T : 8 4 1 3 0 1 1 0 0 1 1 4 We the defie T = T +1 T T + = T +1 T +. (3.1) T T +1 It ca be easily see that this formula is cosistet with above table havig egative values of. So ow we ca thik of T beig defied o all iteger values of, both positive ad egative ad the Triboacci sequece extedig ifiitely i both the positive ad egative directios. We observe here that T (3 ) = T +. This helps us to prove that T (3+1) = T (3 ) + T (3 5) + T (3 8). Now we ca combie theorem.7 ad (3.1) to obtai the followig: T m+1 T m T m 1 Theorem 3.1: T m = T +1 T T 1. T + T +1 T Proof: By cosiderig ( ) i place of i lemma.7, we get T m = T m+1 T ( 1) + T m T (+1) + T m 1 T +T m T.

D V Shah - Some Triboacci Idetities 7 Usig (3.1) we get 4. The Q-matrix: T m = T m+1 (T T 1 T +1 ) + T m (T + T +1 T +3 ) +T m 1 (T +1 T T + )+T m (T +1 T T + ). T m = T m+1 (T T 1 T +1 ) + T m 1 (T +1 T T + ) T m (T +1 T +3 + T T + T + T +1 ). Usig recurrece relatio of the Triboacci sequece for T +3, we get T m = T m+1 (T T 1 T +1 ) + T m 1 (T +1 T T + ) T m (T +1 T + + T +1 T + T T + T + ) = T m+1 (T T 1 T +1 ) + T m 1 (T +1 T T + ) T m (T +1 T T + (T + T +1 T )) = T m+1 (T T 1 T +1 ) + T m 1 (T +1 T T + ) T m (T +1 T T + T 1 ) T m+1 T m T m 1 = T +1 T T 1. T + T +1 T Robiso [] used the matrices to discover facts about the Fiboacci sequece. We ow demostrate a close lik betwee matrices ad Triboacci umbers. We defie a 1 1 1 importat 3 3 matrix Q = [ 1 0 0] as Q-matrix, which plays a sigificat role i 0 1 0 discussios cocerig Triboacci sequece. Note that det(q) = 1. We prove the followig iterestig result. T + T +1 + T T +1 Theorem 4.1: Q = [ T +1 T T + T 1 T 1 + T T ] ; = 1,, 3,. T 1 Proof: We prove the result by usig iductio. T 3 T + T 1 T 1 1 1 For = 1, we have Q 1 = [ T T 1 + T 0 T 1 ] = [ 1 0 0]. T 1 T 0 + T 1 T 0 0 1 0 Thus result to be proved is true for = 1. Suppose it is true for = r. T r+ T r+1 + T r T r+1 1 1 1 Now Q r+1 = Q r Q = [ T r+1 T r + T r 1 T r ] [ 1 0 0] T r T r 1 + T r T r 1 0 1 0

8 Mathematics Today Vol.7(Dec-011) 1-9 T r+3 T r+ + T r+1 T r+ = [ T r+ T r+1 + T r T r+1 ]. T r+1 T r + T r 1 T r Thus result is true for = r + 1 also. This proves the result by iductio. 0 1 0 Remark: Sice Q 1 = [ 0 0 1 ], takig ( ) i place of ad usig iductio, theorem 1 1 1.4.1 ca also be proved to be true for egative values of. We ow metio a iterestig relatio of T with Q-matrix. T +3 1 Corollary 4.: [ T + ] = Q [ 1]. T +1 0 Proof: Follows immediately from above theorem. We ext show that both Q-matrix ad its traspose satisfy the equatio x +1 = x + x 1 + x for every positive iteger. Lemma 4.3: (i) Q +1 = Q + Q 1 + Q. (ii) (Q T ) +1 = (Q T ) + (Q T ) 1 + (Q T ). Proof: We have Q + Q 1 + Q = Q (Q + Q + I). of above proof. 1 1 1 1 1 0 0 = Q { [ 1 1 1] + [ 1 0 0] + [ 0 1 0] } 1 0 0 0 1 0 0 0 1 4 3 = Q [ 1] = Q Q 3 = Q +1. 1 1 1 The secod part follows easily by takig traspose of every matrix at each step Remark: Proceedig o the same lie we ca also show that (Q 1 ) 3 = (Q 1 ) + (Q 1 ) 1 + (Q 1 ). For the sequece of Fiboacci umbers there is a well kow idetity called Simpso formula F +1 F 1 F = ( 1), which ca be writte as F +1 F F F 1 = ( 1). We geeralize it to Triboacci umbers ad derive a idetity coectig five cosecutive T s.

D V Shah - Some Triboacci Idetities 9 T T +1 T + Lemma 4.4: T 1 T T +1 = 1 ; for all itegers. T T 1 T Proof: By lemma 4.1 we have T + T +1 + T T +1 det(q ) = T +1 T T + T 1 T 1 + T T T 1 T + T +1 T +1 T + T T +1 (det(q)) = T +1 T T + T +1 T 1 T T T 1 T 1 T T T 1 T T +1 T + (det(q)) = T 1 T T +1. T T 1 T Sice det Q = 1, we get the required result. We ca apply Theorem 4.1 to derive four ew Triboacci idetities as the ext corollary shows, although they are basically the same. Corollary 4.5: T m++ = T m+ T + + T m+1 T +1 + T m T +1 +T m+1 T T m++1 = T m+1 T + + T m T +1 + T m 1 T +1 + T m T T m+ = T m T + + T m 1 T +1 + T m T +1 +T m 1 T T m+ 1 = T m T +1 + T m 1 T + T m T + T m 1 T 1. Proof: Sice Q m Q = Q m+, we have T m+ T m+1 + T m T m+1 T + T +1 + T T +1 [ T m+1 T m + T m 1 T m ] [ T +1 T + T 1 T ] T m T m 1 + T m T m 1 T T 1 + T T 1 T m++ T m++1 + T m+ T m++1 = [ T m++1 T m+ + T m+ 1 T m+ ]. T m+ T m+ 1 + T m+ T m+ 1 By multiplyig the two matrices ad equatig the correspodig etries, all the idetities follow immediately. Refereces: [1] Agroomof M. (1914) : Sur ue récurrete, Mathesis (Series 4), 4 15 16. [] Robiso D. W. (1963) : The Fiboacci matrix modulo m, The Fiboacci Quarterly,1, 9 36. [3] Zeitli D. (1967) : O summatio formulas ad idetities for Fiboacci umbers, The Fiboacci Quarterly, 5(1), 1 43.