F. T. HOWARD AND CURTIS COOPER

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1 SOME IDENTITIES FOR r-fibonacci NUMBERS F. T. HOWARD AND CURTIS COOPER Abstract. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n} is defined as 0, if 0 n < r 1; G n = 1, if n = r 1; G n 1 +G n + +G n r, if n r. We will present several identities and congruences involving r-generalized Fibonacci numbers. 1. Introduction Definition 1.1. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n } is defined as 0, if 0 n < r 1; G n = 1, if n = r 1; G n 1 +G n + +G n r, if n r. This definition is not new. According to Dickson [, p. 409], M. d Ocagne, in a series of papers from 1883 to 1890, considered (with slightly different notation) the sequence {U i } with U n = c 1 U n 1 +c U n + +c r U n r and U 0,...,U r 1 arbitrary. He also considered the sequence {u n } satisfying the same recurrence, with u i = 0 (i = 0,...,r ), u r 1 = 1, and he found a relationship between {U n } and {u n }. Evidently r and the c i s are fixed in these definitions. According to Dickson, for each sequence d Ocagne found the sum of any fixed number of consecutive terms and the limit of that sum. In one of the papers involving continued fractions each c i = 1. More recently (1960), Miles [6] used Definition 1.1 in a paper involving k k matrices with k-generalized Fibonacci numbers for elements. Kessler and Schiff [4] stated that the Miles article seems to be the oldest well-known paper on the generalized numbers. Since 1960 many more papers involving the r-generalized Fibonacci numbers have appeared, including several in The Fibonacci Quarterly. See [3] and [4] for example. Kessler and Schiff [4] gave many interesting and relevant references, and they noted that an exhaustive bibliography of these numbers would cover pages. Though Definition 1.1 is not new, the authors believe that most of the results in this paper are new, or at least not well-known. All of the proofs are original. In Section we will prove an identity that enables us to find congruences modulo k for the r-generalized Fibonacci sequence. In Section 3 we prove several identities, including a formula for the sum of the squares of the G n s. In Section 4 we find formulas for G n and G n+1, and in Section 5 we prove some miscellaneous results. A table for the r-generalized Fibonacci numbers, for r 8 can be found in Section 6. Finally, in Section 7 we present possible topics for future research. AUGUST

2 THE FIBONACCI QUARTERLY Theorem.1. For n r +1, Proof.. An Identity and Congruences for {G n } G n = G n 1 G n r 1. G n = G n 1 +G n + +G n r +0 = G n 1 +G n + +G n r +G n 1 G n G n r G n r 1 = G n 1 G n r 1. We note that Gabai [3] stated Theorem.1 without proof. Theorem.. For r n r 1, For 0 n r, G n = n r. G r+n = n+r (n+) n 1. Proof. From Theorem.1, we have, for r n r 1, We now note that G n = G n 1 = G n = = n r G r = n r. G r = G r 1 G r 1 = ( r 1 ) 1 = r 1. Assumen > 0, andassumethetheoremistrueforg r+n 1. Ifr n, thenr 1 r+n 1 r, so G r+n 1 = n 1 by the first part of Theorem.. We have G r+n = G r+n 1 G r+n 1 = [ n+r (n+1) n 1 ] n 1 = n+r (n+) n 1. In a similar way we can prove the following results. If 0 n r+1, then [( n+ G 3r+n = r+n (r +n+) r+n 1 + ) 1 ] n. If 0 n r+, then [( ) ] [( ) ] r+n+ n+ G 4r+n = 3r+n (r +n+) r+n r+n n n 3. 3 The general theorem is Theorem.3. Let m = kr +n, with 0 n r +k, and k. Then k 1 G m = m r + ( 1) j a m,j m (j+1)r j j=1 where a m,j = a m 1,j +a m r 1,j 1 with a i,0 = 1 for all i, a i,1 = 0 for i < r, and a r,1 =. 3 VOLUME 49, NUMBER 3

3 SOME IDENTITIES FOR r-fibonacci NUMBERS For example, in the Fibonacci case (r = ) the recurrence for the coefficients is a m,j = a m 1,j +a m 3,j 1 and we can construct a triangle (like Pascal s) to quickly get Theorem.4. For r and m r, G m = m r + a 10,0 = 1, a 10,1 = 8, a 10, = 14, a 10,3 =, so F 10 = G 10 = 8 (8) 5 +(14) () 1 = 55. m+r r+1 1 j=1 ( 1) j [( m rj r + j ) ( m rj r j )] m (r+1)j r. Before looking at the proof, here are some congruences we get by looking at the last term in the summation. We assume k > 1 in the proofs of congruences (.1) (.4), and we can use Theorem. to verify the cases k = 0 and k = 1. ( ) Case 1: m = (r + 1)k. Then m+r r+1 1 = k 1, and m (r +1) m+r r+1 1 r = 1. In the last term ( ) ( m kr+ k 1 m kr ) ( k 3 = k+ ) ( 3 k ) 3 = k. The exponent of in the next to last term is r +. Thus: G (r+1)k ( 1) k 1 k (mod r+ ). (.1) ( Case : m = (r+1)k+t, 0 < t < r+1. Then m+r r+1 1 = k, and m (r+1) m+r r+1 ) 1 r = t r. In the last term ( m rk r+) ( k m rk r ) k = 0 if 0 < t < r 1. In this case the exponent of in the next to last term is t+1. Thus G (r+1)k+t 0 (mod t+1 ), if 0 < t < r 1. (.) If t = r 1, then ( m rk r+) ( k m rk r ) k = (k +1) (k 1) =. Since the exponent of in the next to last term is r, we have G (r+1)k+(r 1) ( 1) k (mod r ). (.3) If t = r, then ( m rk r+) ( k m rk r ) ( k = k+ ) ( k k k ) = k +1. Thus G (r+1)k+r ( 1) k (k +1) (mod r+1 ). (.4) Proof of Theorem.4. We first prove that in Theorem.3, the upper limit of the summation, k 1, can be replaced by m+r r+1 1. Case 1: m = (r + 1)j + t, with 1 t r. In this case m = r(j + 1) + (j r + t), so k = j +1. It is easy to verify that m+r r+1 = (r+1)j+r+t r+1 = j +1 = k. Case : m = (r + 1)j. In this case, m = rj + j, so k = j. It is easy to verify that m+r r+1 = (r+1)j+r r+1 = j = k. Now we will use Theorem.3 to prove Theorem.4. Recall a n,0 = 1 for all n, and a n,1 = 0 for n < r. We know that G r = r 1 = r a r,1 1, so a r,1 =. Using the recurrence a m,j = a m 1,j +a m r 1,j 1 AUGUST

4 THE FIBONACCI QUARTERLY it is easy to see that ( ) ( ) m r + m r a m,1 = m r + = 1 1 Note that the second binomial coefficient is 0. Now we use induction on j. Suppose we know that for h = 1,...,j ( ) ( ) m rh r + m rh r a m,h =. h h Then i=j+1 (m r). a m,j+1 = a m 1,j+1 +a m r 1,j ( ) ( ) m r(j +1) r+1 m r(j +1) r 1 = a m 1,j+1 + j j ( ) ( ) m r(j +1) r+1 m r(j +1) r 1 = a m,j+1 + j j ( ) ( ) m r(j +1) r m r(j +1) r + j j m r(j+1) r+1 [( ) ( )] i i = a (j+)r+j 1,j+1 +. j j Since a (j+)r+j 1,j+1 = 0, we have a m,j+1 = m r(j+1) r+1 i=j [( ) i j Here we have used the identity ( )] i = j n i=j This completes the induction argument. ( ) i = j ( ) ( m r(j +1) r + m r(j +1) r j +1 j 1 ( ) n+1. j +1 Notice that the induction argument works in the case j = 1, since in that case we have m 3r+1 ( ) m 3r+1 i ( ) ( ) ( ) i m 3r + m 3r a m, = = 1 = i= i=1 We have used the fact that a m,j = 0 for m < (j + 1)r + (j 1). For example a m,1 = 0 if m < r; a m, = 0 if m < 3r +1, etc. This follows from the recurrence for a m,j. Note that we have used the convention ( a b) = 0 if a < b or if either a or b is negative. When r = in Theorem.4, we have the following formula for the Fibonacci numbers. Theorem.5. For m, F m = m + m+ 3 1 j=1 ( 1) j [( m j j ) ( m j j )] m 3j. 34 VOLUME 49, NUMBER 3 ).

5 SOME IDENTITIES FOR r-fibonacci NUMBERS Corresponding to (.1), (.3), (.4), the congruences for the Fibonacci numbers (for k 0) are: F 3k ( 1) k 1 k (mod 16), (.5) F 3k+1 ( 1) k (mod 4), (.6) F 3k+ ( 1) k (k +1) (mod 8). (.7) If we look at the last two terms of the summation in Theorem.5, we see that Thus, if k 0 or 3 (mod 4), If k 1 or (mod 4), F 3k+1 ( 1) k +( 1) k (10k)(k +1)(k +1) (mod 3). F 3k+1 ( 1) k (mod 8). F 3k+1 ( 1) k 5 (mod 8). The A array in Theorem.3 and Theorem.4 is interesting. For r =, the Fibonacci case, the A array is m\j a m,j for r = (Fibonacci case). For r = 3, the Tribonacci case, the Tribonacci A array is similar to the Fibonacci A array; there are just more 0 s at the top of each column. Condensing the A array in the Fibonacci case to a B array by defining b m,j = a m+3j,j we have the following B array. AUGUST

6 THE FIBONACCI QUARTERLY m\j B array. The first column of the B array is 1 and after the first entry, the first row is. Entries in the middle of the B array can be determined by adding the elements in the B array immediately to its left and directly above the entry, i.e., A closed form formula for b m,j is Here, we assume the j = 1 column is b m,j = b m,j 1 +b m 1,j. m(m+1) (m+j )(m+j 1). j! m+1. 1! The entries in the j = column of B are A in Sloane s OEIS [7]. The j = 3 column of B is A in Sloane s OEIS [7] and the j = 4 column of B is A00558 in Sloane s OEIS [7]. 3. The Sum of the Squares and Other Identities Theorem 3.1. For n r 1, r 1 G n = r 1 G n r+1 k 1 G n r k. 36 VOLUME 49, NUMBER 3

7 SOME IDENTITIES FOR r-fibonacci NUMBERS Proof. We iterate r 1 times the recurrence in Theorem.1: G n = G n 1 G n r 1 = (G n G n r ) G n r 1 = G n G n r G n r 1 = (G n 3 G n r 3 ) G n r G n r 1 = 3 G n 3 G n r 3 G n r G n r 1 r 1 = r 1 G n r+1 k 1 G n r k. Theorem 3.. For r and n r 1, r 1 G n = r 1 G n r + For the Fibonacci sequence, this identity is ( r 1 )G i 1 n r k. i=k F n = F n +F n 3. Listing this identity for r =,3,4,5, and 6 we have the resulting formulas. r = : G n = G n +G n 3 r = 3 : G n = 4G n 3 +3G n 4 +G n 5 r = 4 : G n = 8G n 4 +7G n 5 +6G n 6 +4G n 7 r = 5 : G n = 16G n 5 +15G n 6 +14G n 7 +1G n 8 +8G n 9 r = 6 : G n = 3G n 6 +31G n 7 +30G n 8 +8G n 9 +4G n G n 11. Proof of Theorem 3.. From Theorem 3.1, we have r 1 G n = r 1 G n r+1 k 1 G n r k. Thus, by Definition 1.1 applied to G n r+1, we have (note that the 1 s cancel in the first equation): r 1 r 1 G n = r 1 G n r + ( r 1 1)G n r k ( k 1 1)G n r k = r 1 G n r + = r 1 G n r + ( r 1 r 1 i=1 ( r 1 r 1 r 1 )G i 1 n r k )G i 1 n r k. i=k ( k 1 )G i 1 n r k AUGUST i=1

8 THE FIBONACCI QUARTERLY Theorem 3.3. For r, n 0, n r 1 G k + n i G k G k+i = G n G n+1. k=0 i= k=0 The special case of this identity for the Fibonacci sequence was discovered by Lucas in He discovered that for n 0, Its proof can be found in [5, pp ]. n i=1 Proof. For j = n,n 1,...,r 1, we start with and multiply by G j to get F i = F n F n+1. G j = G j+1 G j 1 G j G j r+1 G n = G ng n+1 G n G n 1 G n G n i G n G n r+1, G n 1 = G n 1 G n G n 1 G n G n 1 G n 1 i G n 1 G n r, G r 1 = G r 1 G r G r 1 G r G r 1 G r 1 i G r 1 G 0. When we add, we get n k=0 G k on the left side, and all of the nonzero terms in columns 1 and on the right side cancel except for G n G n+1. The i+1 column is n i k=0 G kg k+i and since i r 1, n r 1 G k + n i G k G k+i = G n G n+1. k=0 i= k=0 We note that Gabai [3] proved a result equivalent to Theorem 3.3. Theorem 4.1. For r, n > 0, m > 0, where A i = 4. Formulas for G n and G n+1 r G n+m = G n G m +G n G m 1 +G n 1 G m + G n+i A i, i=1 { G m+r i 1 G m+r i G m+1 (i < r ); G m+1 (i = r ). Before giving the proof, here are some applications. If we let m = n, we have: r = : F n = F n +F n 1 F n r = 3 : G n = G n +G n+1 +G n 1 G n r = 4 : G n = G n G n+1 +G n 1 G n +G n+1 G n+ r = 5 : G n = G n G n+1 +G n+ +G n 1G n G n+1 G n+ +G n+1 G n VOLUME 49, NUMBER 3

9 SOME IDENTITIES FOR r-fibonacci NUMBERS If we let m = n+1, we have (after a little manipulation) r = : F n+1 = F n +F n+1 r = 3 : G n+1 = G n +G n+1 +G n 1G n+1 +G n G n+1 r = 4 : G n+1 = G n +G n+1 +G n+ +G n 1 G n+1 +G n G n+1 r = 5 : G n+1 = G n +G n+1 G n+ +G n 1 G n+1 +G n G n+1 +G n+ G n+3. Proof. We will use Zhou s Theory of Constructing Identities (TCI) [8]. Let F r (x) = x r x r 1 x 1. Then clearly That is, modulo F r (x): m 1 F r (x) G n+i x m 1 i 0 (mod F r (x)). 0 (x r x r 1 x 1)(G n x m 1 +G n+1 x m + +G n+m r x r 1 + +G n+m 1 ) G n x m+r 1 +(G n+1 G n )x m+r +(G n+ G n+1 G n )x m+r 3 + +(G n+r G n+r 3 G n )x m+1 +(G n+r 1 G n+r G n )x m +0 x m 1 +0 x m + +0 x r (G n+m 1 +G n+m + +G n+m r )x r 1 G n+m 1. By TCI, we can replace x k by G k and congruence is changed to equality. Thus G n G m+r 1 +(G n+1 G n )G m+r +(G n+ G n+1 G n )G m+r 3 + (4.1) Notice we have used the identities +(G n+r G n+r 3 G n )G m+1 +G n 1 G m G n+m = 0. G n+r 1 G n+r G n = G n 1 (in the x m term), (G n+m 1 +G n+m + +G n+m r ) = G n+m (in the x r 1 term), G r 1 = 1; G i = 0 for i < r 1. To simplify further, use G m+r 1 = G m+r +G m+r 3 + +G m+1 +G m +G m 1 in the first term of (4.1) and notice that all the other G n terms are (G m+r +G m+r 3 + +G m+1 )G n. Thus the G n terms are reduced to G n G m +G n G m 1. Now for i = 1,...,r 3, we get all the G n+i terms together: (G m+r i 1 G m+r i+ G m+1 )G n+i and we note that the only term with G n+r is G m+1 G n+r. This completes the proof. AUGUST

10 THE FIBONACCI QUARTERLY (a) Lucas numbers. 5. Miscellaneous Results Define K n = 0 for 0 n r 3, K r = a, K r 1 = b and K n = n 1 i=n r K i if n r. Let K(x) be the generating function K(x) = K ix i. It is easy to see that K(x) = axr +(b a)x r 1 1 x x x r. Note that if a = 0 and b = 1, then K n = G n, and we use the notation K(x) = G(x). From the generating functions, it is easy to see that K n = ag n+1 +(b a)g n. In particular, if a =, b = 1, we get the Lucas sequence {L n }: which gives x Since xl(x) = G(x), we have x r x r 1 L(x) = 1 x x x r = L i x i L n = G n+1 G n. n 1 ( ) 1 n i G n = L i. (b) Relationship between G (r) n and G (r+1) n. For fixed r, write G(x) = G r (x) and G n = G (r) n ; likewise L(x) = L r (x) and L n = L (r) n. Since [ ] 1, we have which gives For example, when r = Similarly we have so G r (x) G r+1 (x) = 1 x x r+1 1 x x x r xg r (x) = G r+1 (x) x G r (x)g r+1 (x), G (r+1) n+1 = G (r) G (3) n + n 1 G (r) i G (r+1) n 1 i. n 1 n+1 = F n + F i G (3) n 1 i. L r+1 (x) = xl r (x)+x L r+1 (x)g r (x) L (r+1) n 1 n+1 = L (r) n + L (r+1) i G (r) n 1 i. 40 VOLUME 49, NUMBER 3

11 The case r = 1 gives the well-known formulas SOME IDENTITIES FOR r-fibonacci NUMBERS n 1 n 1 F n+1 = 1+ F i, L n+1 = 1+ L i. (c) Another example using Zhou s TCI. If we use Zhou s Theory of Constructing Identities on G n = G n 1 G n 1 r, we have (x r+1 x r +1)(G n x m 1 +G n+1 x m + +G n m 1 ) 0 (mod x r+1 x r +1). Simplifying as we did before, we get r 1 G n+m = G n+m r +G n G m+r G n 1 i G m+i, which could be written or r 1 G n+m+1 G n+m = G n G m+r G n 1 i G m+i, r 1 G n = G n r +G n G n+r G n 1 i G n+i. 6. Table of r-generalized Fibonacci Numbers The first few terms of the r-generalized Fibonacci sequence for r 8 are given in the following table. r-generalized Fibonacci Sequences r\n The r-generalized Fibonacci sequences for r =,3,4,5,6,7,8 can be found in Sloane [7] as sequences A000045, A000073, A000078, A001591, A00159, A1189, and A0796, respectively. AUGUST

12 THE FIBONACCI QUARTERLY 7. Topics for Future Study Many of the theorems, like Theorem 3.3, can undoubtedly be proved using combinatorial arguments in the manner of Benjamin and Quinn [1]. In fact, G n+r 1 (for n 0) counts the number of tilings of an n-board with tiles of length at most r. It would be interesting to see different approaches to our theorems. The r-generalized Lucas sequence could obviously be examined more thoroughly, and more relationships to the r-generalized Fibonacci numbers can probably be found. The more general recurrence G n = c 1 G n 1 +c G n + +c r G n r, where the c i s are constants, can certainly be studied in more detail. We leave all of these topics for future research. Acknowledgment The authors wish to express their appreciation to the anonymous referee for the careful reading and the helpful comments and suggestions that improved the quality of this paper. References [1] A. T. Benjamin and J. J. Quinn, Proofs That Really Count-The Art of Combinatorial Proof, Mathematical Association of America, Washington, DC, 003, 8 9. [] L. E. Dickson, History of the Theory of Numbers, Volume 1, Chelsea Publishing Company, New York, [3] H. Gabai, Generalized Fibonacci k-sequences, The Fibonacci Quarterly, 8.1 (1970), [4] D. Kessler and J. Schiff, A combinatoric proof and generalization of Ferguson s formula for k-generalized Fibonacci numbers, The Fibonacci Quarterly, 4.3 (004), [5] T. Koshy, Fibonacci and Lucas Numbers with Application, John Wiley & Sons, Inc., New York, 001. [6] E. P. Miles, Jr. Generalized Fibonacci numbers and associated matrices, American Mathematical Monthly, 67 (1960), [7] N. J. A. Sloane (Ed.), The On-Line Encyclopedia of Integer Sequences (008), njas/sequences/. [8] C. Zhou, Constructing identities involving kth-order F-L numbers by using the characteristic polynomial, Applications of Fibonacci Numbers, Volume 8, Edited F. T. Howard, 1999, MSC010: 11B39 Department of Mathematics, Box 7388, Reynolda Station, Wake Forest University, Winston- Salem, NC address: howard@wfu.edu Department of Mathematics and Computer Science, University of Central Missouri, Warrensburg, MO address: cooper@ucmo.edu 4 VOLUME 49, NUMBER 3

1. Introduction Definition 1.1. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n } is defined as

1. Introduction Definition 1.1. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n } is defined as SOME IDENTITIES FOR r-fibonacci NUMBERS F. T. HOWARD AND CURTIS COOPER Abstract. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n} is defined as 8 >< 0, if 0 n < r 1; G n = 1, if n = r