Impulse Response Sequences and Construction of Number Sequence Identities

Transcription

1 Journal of Integer Sequences, Vol. 16 (013), Article Impulse Response Sequences and Construction of Number Sequence Identities Tian-Xiao He Department of Mathematics Illinois Wesleyan University Bloomington, IL USA the@iwu.edu Abstract In this paper, we investigate impulse response sequences over the integers by presenting their generating functions and expressions. We also establish some of the corresponding identities. In addition, we give the relationship between an impulse response sequence and all linear recurring sequences satisfying the same linear recurrence relation, which can be used to transfer the identities among different sequences. Finally, we discuss some applications of impulse response sequences to the structure of Stirling numbers of the second kind, the Wythoff array, and the Boustrophedon transform. 1 Introduction Many number and polynomial sequences can be defined, characterized, evaluated, and classified by linear recurrence relations with certain orders. A number sequence {a n } is called a sequence of order r if it satisfies the linear recurrence relation of order r a n r p a n, n r, (1) for some constants p, 1,,...,r, p r 0, and initial conditions a ( 0,1,...,r 1). Linear recurrence relations with constant coefficients are important in subects including 1

2 combinatorics, pseudo-random number generation, circuit design, and cryptography, and they have been studied extensively. To construct an explicit formula of the general term of a number sequence of order r, one may use a generating function, a characteristic equation, or a matrix method (See Comtet [4], Niven, Zuckerman, and Montgomery [1], Strang [15], Wilf [16], etc.) Let A r be the set of all linear recurring sequences defined by the homogeneous linearrecurrencerelation(1)withcoefficientsete r {p 1,p,...,p r }. Tostudythestructure ofa r withrespecttoe r, weconsidertheimpulse response sequence(or, IRSforabbreviation) in A r, which is a particular sequence in A r with initials a 0 a 1 a r 0 and a r 1 1. The study of impulse response sequences in a finite field has been considerably active in the theory of finite fields. A summary work on this subect can be found in Lidl and Niederreiter [10]. In this paper, we investigate the same concept in Z. If r, the IRS in Z is the well known Lucas sequence. In this sense, an IRS of order r > is an extension of a Lucas sequence in a high order setting. In the next section, we will give the generating function and the expression of the IRS of A r and find out the relationship between the IRS and other sequences in the set A r. Several examples of IRS in sets A and A 3 will be given in Section 3. The work shown in Sections and 3 is partially motivated by Shiue and the author [8], which presented a method for the construction of general term expressions and identities for the linear recurring sequences of order r by using the reduction order method. However, the reduction method is too complicated for the linear recurring sequences of order r >, although our method can be applied readily to linear recurring sequences with arbitrary order by using their relationship with impulse response sequences. In Section 4, we shall give some applications of IRS in the discussion of the structure of Stirling numbers of the second kind, the Wythoff array, and the Boustrophedon transform. Impulse response sequences Among all the homogeneous linear recurring sequences satisfying rth order homogeneous linear recurrence relation (1) with a nonzero p r and arbitrary initial conditions {a } r 1 0, we definetheimpulseresponsesequence(irs)withrespecttoe r {p } r asthesequencewith initial conditions a 0 a r 0 and a r 1 1. In particular, when r, the homogeneous linear recurring sequences with respect to E {p 1,p } satisfy a n p 1 a n 1 +p a n, n, () with arbitrary initial conditions a 0 and a 1, or, initial vector (a 0,a 1 ). If the initial vector (a 0,a 1 ) (0,1), the corresponding sequence generated by using () is the IRS with respect to E, is called the Lucas sequence. For instance, the Fibonacci sequence {F n } is the IRS with respect to {1,1}, the Pell number sequence {P n } is the IRS with respect to {,1}, and the Jacobsthal number sequence {J n } is the IRS with respect to {1,}. In the following, we will present the structure of the linear recurring sequences defined by (1) using their characteristic polynomial. This structure provides a way to find the

3 relationship between those sequences and the IRS. Let A r be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation (1) with coefficient set E r {p 1,p,...,p r }, and let n be the IRS of A r, namely, n satisfies (1) with initial conditions 0 r 0 and r 1 1. Suppose {a n } A r. Let B p 1 + p + + p r, and for n N 0 let M n max{ a 0, a 1,..., a n }. Thus, for n r, M n BM n 1. By induction, a n CB n for some constant A and n 0,1,,... A sequence satisfying a bound of this kind is called a sequence having at most exponential growth (see [1]). Therefore, the generating function 0 a t of such sequence has a positive radius of convergence, 0 < c < 1/B. More precisely, if t c, then by comparison with the convergent geometric series 0 CB c, the series 0 a t converges absolutely. In the following, we always assume that {a n } has at most exponential growth. Proposition 1. Let {a n } A r, i.e., let {a n } be the linear recurring sequence defined by (1). Then its generating function P r (t) can be written as ( r 1 n r P r (t) {a 0 + a n p a n )t n }/{1 p t }. (3) n1 Hence, the generating function for the IRS with respect to {p } is P r (t) t r 1 p t. (4) Proof. Multiplying t n on the both sides of (1) and summing from n r to infinity yields n t n ra n r r p a n t n p a n t n n r n r ( ) r 1 r 1 p a n t n a n t n +p r a n r t n n n n r ( ) r 1 p t r 1 n t n 0a n a n t n +p r t r a n t n n n 0 Denote P r (t) n 0 a nt n. Then the equation of the leftmost and the rightmost sides can be written as r 1 r 1 r 1 r 1 P r (t) a n t n P r (t) p t p a n t n +p r t r P r (t), n0 n 3

4 which implies P r (t) ( 1 ) r r 1 p t a n t n n0 r 1 n1 n p a n t n. Therefore, (4) follows immediately. By substituting a 0 a r 0 and a r 1 1 into (3), we obtain (4). We now give the explicit expression of n polynomial of recurrence relation shown in (1). in terms of the zeros of the characteristic Theorem. Sequence { n } n is the IRS with respect to {p } defined by (1) with 1 r 0 ane r 1 1 if and only if n ( l n ) n m m 1 α +1 l k1,k (α α k ) 0 (5) for n r, where α 1,α,...,α l are the distinct real or complex zeros of characteristic polynomial, p(t) t r p 1 t r 1 p r, of the recurrence relation (1) with the multiplicities m 1,m,...,m l, respectively, and m 1 +m + +m l r. In particular, if all α 1,α,..., α r are of multiplicity one, then (5) becomes to n r α n r k1,k (α, n r, (6) α k ) Proof. First,weprove(6). Thenweuseittoprovethenecessityof(5). Letα, 1,,...,l, be l zeros of the characteristic polynomial t n r i1 p it n i. Then from (4) P r (t) t r 1 p t 1/t (1/t) r r p (1/t) n 1/t r (1/t α ) t r 1 r (1 α t). (7) We now prove (6) from (7) by using mathematical induction for r under the assumption that all solutions α are distinct. Because [t n ] P r (t) n, we need to show [t n ] P r (t) [t n t r 1 r ] r (1 α t) α n r k1,k (α, n r. (8) α k ) (8) is obviously true for r 1. Assume (8) holds for r m. We find [t n ] P m+1 (t) [t n ] 4 t m m+1 (1 α t),

5 which implies [t n ](1 α m+1 t) P m+1 (t) [t n 1 t m 1 m ] m (1 α t) α n 1 m k1,k (α α k ) because of the induction assumption. Noting [t n ] P m+1 (t) one may write m α (m+1) m+1 n 1 which has the solution Indeed, we have (m+1) n α m+1 (m+1) n 1 + m+1 α m+1 m m+1 m+1 n (m+1) m (m+1) n, from the last equations α n 1 m k1,k (α α k ), (9) α n m+1 k1,k (α α k ). (10) α n 1 m k1,k (α α k ) α n 1 m+1 k1,k (α α k ) + α n 1 m α n 1 m k1,k (α α k ) m+1 k1,k (α α k ) (α m+1 +α α m+1 )+ α n m+1 k1,k (α α k ) (m+1) n, α n m+1 m k1 (α m+1 α k ) and (10) is proved. To prove (5), it is sufficient to show it holds for the case of one multiple zero, say α i with the multiplicity m i. The formula of n in this case can be derived from (6) by using the limit process. More precisely, let m i zeros of the characteristic polynomial, denoted by α i, 1 m i, approach α i. Then (6) will be reduced to the corresponding (5) with respect to α i. Indeed, taking the limit as α i, 1 m i, approaches α i on both sides of (6), yields lim (α i1,...,α imi ) (α i,...,α i ) n 1 r mi +1 k1,k i (α lim i α k ) (α i1,...,α imi ) (α i,...,α i ) ) + + α n i mi mi k1,k m i (α imi α ik ) ( α n i 1 mi k (α i 1 α ik ) + αn i mi k1,k (α i α ik ) r α n + r k1,k (α α k ). (11), i 5

6 It is obvious that the summation in the parentheses of equation (11) is the expansion of the divided difference of the function f(t) t n, denoted by f[α i1,α i,...,α imi ], at nodes α i1,α i,...,α imi. Usingthemeanvaluetheorem(see, forexample, [13])fordivideddifference (see, for example, [, Thm. 3.6]), one may obtain lim f[α i1,α i,...,α imi ] f(mi 1) (α i ) (α i1,...,α imi ) (α i,...,α i ) (m i 1)! ( ) n α n m i+1 i. m i 1 Therefore, by taking the limit as α i α i for 1 m i on both sides of (11), we obtain ( n ) lim m n i 1 α n m 1 +1 r i r mi +1 (α i1,...,α imi ) (α i,...,α i ) k1,k i (α i α k ) + α n r k1,k (α α k ),, i which implies the correctness of (5) for the multiple zero α i of multiplicity m i. Therefore, (5) follows by taking the same argument for each multiple zero of the characteristic polynomial P r (t) shown in Proposition 1. Finally, we prove sufficiency. For n r, r i1 l l p i n i ( n m 1 ) α m +1 r i1 p i( n m 1 ) α n i m +1 l k1,k (α α k ) r i1 p iα n i l k1,k (α α k ) ( l n ) n m m 1 α +1 l k1,k (α α k ) n, where we have used α n r i1 p iα n i because of p(α ) 0. Therefore, the sequence { n } satisfies the linear recurrence relation (1) and is the IRS with respect to {p }. The IRS of a set of linear recurring sequences is a kind of basis so that every sequence in the set can be represented in terms of the IRS. For this purpose, we need to extend the indices of IRS n of A r to the negative indices down to till n r+1, where A r is the set of all linear recurring sequences defined by the homogeneous linear recurrence relation (1) with coefficient set E r {p 1,p,...,p r }, p r 0. For instance, 1 is defined by which implies use 1 1/p r because r 1 p 1 r +p r 3 + +p r 1, 0 r 0 and r+ p 1 r+ 1 +p r+ + +p r r 1 1. Similarly, we may todefine,,..., r+1successively. Thefollowingtheoremshowshowtorepresent linear recurring sequences in terms of the extended IRS { n } n r+1. 6

7 Theorem 3. Let A r be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation (1) with coefficient set E r {p 1,p,...,p r }, p r 0, and let { n } n r+1 be the (extended) IRS of A r. Then, for any a n A r, one may write a n as r a n a r 1 n + r 0 k a k p r+ k n 1. (1) Proof. Considering the sequences on the both sides of (1), we find that both sequences satisfy the same recurrence relation (1). Hence, to prove the equivalence of two sequences, we only need to show that they have the same initial vector (a 0,a 1,...,a r 1 ) because of the uniqueness of the linear recurring sequence defined by (1) with fixed initial conditions. First, for n r 1, the conditions 0 r 0 and r 1 1 are applied on the right-hand side of (1) to obtain its value as a r 1, which shows (1) holds for n r 1. Secondly, for all 0 n r, one may write the right-hand side RHS of (1) as r k RHS 0+ k( k0a 0 p r+ k n 1 ) r a k δ k,n a n, where the Kronecker delta symbol δ k,n is 1 when k n and zero otherwise, and the following formula is used: k p r+ k n 1 δ k,n. (13) 0 (13) can be proved by splitting it into three cases: n,>, and < k, respectively. If n k, we have k 0 p r+ k n 1 k 0 k0 p r+ k k 1 p r 1 +p r 1 0 +p r 1 + +p r k k 1 p r 1 1, where we use 1 1/p r and the fact 0 for all 0,1,...,k 1 due to k 1 r 3 and the initial values of n being zero for all 0 n r. If n > k, we have k 0 p r+ k n 1 p r n k 1 +p r 1 n k + +p r k n 1 0 because 0 for all n k 1,n k,...,n 1 due to the definition of condition of 1 < n k 1 n 1 r 3. n and the 7

8 Finally, for 0 n < k, we may find k 0 p r+ k n 1 p r n k 1 +p r 1 n k + +p r k n 1 p r n k 1 + +p r k n+r k 1 0, n 1 +p r k 1 n +p r k n+1 + +p 1 n+r k where the inserted r k 1 terms, n,n+1,...n+r k, are zero due to the definition of n and 0 n n+r k < r. The last two steps of the above equations are from the recurrence relation (1) and the assumptions of 0 k r and 1 n+1 n+r k 1 < r 1, respectively. This completes the proof of theorem. By using Theorems and 3, we immediately obtain Corollary 4. Let A r be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation (1) with coefficient set E r {p 1,p,...,p r }, p r 0, and let α 1,α,...,α l be the zeros of characteristic polynomial, p(t) t r p 1 t r 1 p r, of the recurrence relation (1) with the multiplicities m 1,m,...,m l, respectively, and m 1 + m + +m l r. Then, for any {a n } A r, we have ( l n ) n m m a n a 1 α +1 r 1 l k1,k (α α k ) + r 0 k ( l n 1 ) n m m a k p 1 α r+ k l k1,k (α α k ) (14) We now establish the relationship between a sequence in A r and the IRS of A r by using Toeplitz matrices. A Toeplitz matrix may be defined as a n n matrix A with entries a i, c i, for constants c 1 n,...,c n 1. A m n block Toeplitz matrix is a matrix that can be partitioned into m n blocks and every block is a Toeplitz matrix. Theorem 5. Let A r be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation (1) with coefficient set E r {p 1,p,...,p r }, and let n be the IRS of A r. Then, for even r, we may write n r 1 r/ c a n+r + (3r/) 1 r where the coefficients c, r/ (3r/) 1, satisfy c a n+r 1, (15) 8

9 A e c : a r/ a (r/) 1 a 1 a 1 a a r/ a (r/)+1 a r/ a a 0 a 1 a (r/) a r 1 a r a r/ a (r/) a (r/) 3 a 1 c e r a r a r 1 a (r/)+1 a (r/) 1 a (r/) a a (3r/) 1 a (3r/) a r a r a r 3 a (r/) 1 (16) with c (c r/,c (r/) 1,...,c r,c r+1,c r+,...,c (3r/) 1 ) T and e r (0,0,...,0,1) T R r. For odd r, we have n 3(r 1)/ (r 1)/ where c, (r 1)/ 3(r 1)/, satisfy c a n+r 1, (17) A o c : a (r 1)/ a ((r 1)/) 1 a 1 a 0 a 1 a (r 1)/ a ((r 1)/)+1 a (r 1)/ a a 1 a 0 a (r 1)/ c e r a (3(r 1)/) a (3(r 1)/) 1 a r a r 1 a r a (r 1)/ (18) with c (c (r 1)/,c ((r 1)/) 1,...,c r 1,c r,c r+1,...,c (3(r 1)/) ) T and e r (0,0,...,0,1) T R r. If the block Toeplitz matrix A e and Toeplitz matrix A o are invertible, then we have unique expressions (15) and (17), respectively. Proof. Since the operator L : Z Z Z, L(a n 1,a n,...,a n r ) : p 1 a n 1 + p a n + +p r a n r a n, is linear, sequence {a n } is uniquely determined by L from a given initial vector (a 0, a 1,...,a r 1 ). We may extend {a n } n 0 to a sequence with negative indices using the technique we applied in Theorem 3. For instance, by defining a 1 from a r 1 p r a 1 + p r 1 a 0 +p r a 1 + +p 1 a r ), we obtain a 1 1 p r (a r 1 p r 1 a 0 p r a 1 p 1 a r ), p r 0. Thus, the initial vector (a 1,a 0,...,a r ) generates {a n } n 1 or {a n 1 } n 0 by using the operator L. Since both sequences {a n } and { n } satisfy the recurrence relation (1), they are generated by the same operator L. Hence, for even r, we may write as a linear 9 n

10 combination of a n+r, r/ r 1, and a n+r 1, r (3r/) 1, as shown in (15), provided the initial vectors of the sequences on the two sides are the same. Here, we mention again that the initial values with negative indices have been defined by using (1). Therefore, we substitute the initial values of {a n } with their negative extensions to the right-hand side of (15) and enforce the resulting linear combination equal to the initials of the left-hand side of (15), i.e., e r, which derives (16). Under the assumption of the invertibility of A e, Toeplitz system(16)hasauniquesolutionofc (c r/,c (r/) 1,...,c r,c r+1,c r+,...,c (3r/) 1 ) T, which shows the uniqueness of the expression of (15). (17) can be proved similarly. Remark 6. It can be seen that not every Toeplitz matrix generated from the linear recurring sequence (1) is invertible. For example, consider the sequence {a n } generated by the linear recurrence relation a n a n 1 + 3a n + a n 3 with a 0 1, a 1 0, a 1, there hold a 3 a 0 +3a 1 +a and a 1 a a 1 3a 0. From (17) : A o which is not invertible. If p 1, 1 r, and r >, the corresponding IRS is a higherorder Fibonacci sequence. For instance, (3) n, (4) n, (5) n, (6) n, (7) n, etc. are Tribonacci numbers (A000073), Tetranacci numbers (A000078), Pentanacci numbers (A001591), Heptanacci numbers (A1189), Octanacci numbers (A0796), etc., respectively. 3 Examples We now give some examples of IRS for r and 3. Let A be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation () with coefficient set E {p 1,p }. Hence, from Theorems and 3 and the definition of the impulse response sequence of A with respect to E, we obtain Corollary 7. Let A be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation () with coefficient set E {p 1,p }, and let { n () } be the IRS (i.e., the Lucas sequence) of A. Suppose α and β are two zeros of the characteristic polynomial of A, which do not need to be distinct. Then, () n { α n β n α β, if α β; nα n 1, if α β. (19) In addition, every {a n } A can be written as and a n reduces to a 1 () n α a 0 () n 1 when α β. a n a () 1 n αβa () 0 n 1, (0) 10

11 where Conversely, there is an expression for c 1 () n in terms of {a n } as () n c 1 a n+1 +c a n 1, (1) a 1 a 0 p 1 p 1 (a 1 a 0 a 1 p 1 a 0p ), c a 1 p p 1 (a 1 a 0 a 1 p 1 a 0p ), () provided that p 1 0, and a 1 a 0 a 1 p 1 a 0p 0. Proof. (19) is a special case of (5) and (6) and can be found from (5) and (6) by using the substitution r, a 0 0 and a 1 1. Again from (1) and (14) a n a 1 n +a 0 p n 1 a 1 n αβa 0 n 1 α {a n β n 1 α β αβa 0 αn 1 β n 1, if α β; α β a 1 (nα n 1 ) α ((n 1)α n ), if α β. (3) Byusing(15)and(16),wenowprove(1)and(),respectively. DenotebyL : Z Z Z the operator L(a n 1,a n ) : p 1 a n 1 + p a n a n. As presented before, L is linear, and the sequence {a n } is uniquely determined by L from a given initial vector (a 0, a 1 ). Define a 1 (a 1 p 1 a 0 )/p, then (a 1,a 0 ) is the initial vector that generates {a n 1 } n 0 by L. Similarly, the vector (a 1,p 1 a 1 + p a 0 ) generates sequence {a n+1 } n by using L. Note the initial vector of () n is (0,1). Thus (1) holds if and only if the initial vectors on the two sides are equal: ( ) a1 p 1 a 0 (0,1) c 1 (a 1,p 1 a 1 +p a 0 )+c,a 0, (4) p which yields the solutions () for c 1 and c and completes the proof of the corollary. Recall that [8] presented the following result. Proposition 8. [8] Let {a n } be a sequence of order satisfying linear recurrence relation (), and let α and β be two zeros of the characteristic polynomial x p 1 x p 0 of the relation (). Then {( ) ( ) a 1 βa 0 α n a 1 αa 0 β n, if α β; α β α β a n (5) na 1 α n 1 (n 1)a 0 α n, if α β. The paper [8] also gives a method for finding the expressions of the linear recurring sequences of order and the interrelationship among those sequences. However, [8] also pointed out the method presented in this paper (i.e., Proposition 8) cannot be extended to the higher order setting. In Section, we have shown our method based on the IRS can be extended to the higher setting. Here, Proposition 8 is an application of our method in a particular case. In addition, it can be easily seen that (5) can be derived from (3). Corollary 7 presents the interrelationship between a linear recurring sequence with respect to E {p 1,p ) and its IRS, which can be used to transfer the identities of one sequence to the identities of other sequences in the same set. 11

12 Example 9. Let us consider A, the set of all linear recurring sequences defined by the homogeneous linear recurrence relation () with coefficient set E {p 1,p }. If E {1,1}, then the corresponding characteristic polynomial has zeros α (1 + 5)/ and β (1 5)/, and (1) gives the expression of the IRS of A, which is the Fibonacci sequence {F n }: F n 1 5 {( 1+ ) n 5 ( 1 ) n } 5. The sequence in A with the initial vector (,1) is the Lucas number sequence (which is different from the Lucas sequence) {L n }. From (0) and (1) and noting αβ 1, we have the well-known formulas (see, for example, [9]): L n F n +F n 1 F n+1 +F n 1, F n 1 5 L n L n+1. By using the above formulas, one may transfer identities of Fibonacci number sequence to those of Lucas number sequence and vice versa. For instance, the above relationship can be used to prove that the following two identities are equivalent: F n+1 F n+ F n 1 F n F n+1 L n+1 +L n L n +L n+. It is clear that both of the identities are equivalent to the Carlitz identity, F n+1 L n+ F n+ L n F n+1, shown in [3]. Example 10. Let us consider A, the set of all linear recurring sequences defined by the homogeneous linear recurrence relation () with coefficient set E {p 1 p,p 1}. Then (5) tell us that {a n } A satisfies where α is defined by a n a 1 (p 4+p )a 0 α n a 1 (p+ 4+p )a 0 4+p 4+p ( 1 α) n, (6) α p+ 4+p Similarly, let E {1,q}. Then and β 1 α p 4+p. (7) a n { a1 (1 1+4q)a 0 α n 1+4q 1 a 1 (1+ 1+4q)a 0 α n 1+4q, if q 1; 4 1 (na n 1 (n 1)a 0 ), if q 1, 4 where α 1 (1+ 1+4q) and β 1 (1 1+4q) are solutions of equation x x q 0. The first special case (6) was studied by Falbo [5]. If p 1, the sequence is clearly the 1

13 Fibonacci sequence. If p (resp., q 1), the corresponding sequence is the sequence of numerators (when two initial conditions are 1 and 3) or denominators (when two initial conditions are 1 and ) of the convergent of a continued fraction to : { 1, 3, 7, 17, 41...}, called the closest rational approximation sequence to. The second special case is for the case of q (resp., p 1), the resulting {a n } is the Jacobsthal type sequences (See Bergum, Bennett, Horadam, and Moore [1]). By using Corollary 7, for E {p,1}, the IRS of A with respect to E is {( n () 1 p+ ) n 4+p p ( ) n } 4+p. 4+p 4 In particular, the IRS for E {,1} is the well-known Pell number sequence {P n } {0,1,,5,1,9,...} with the expression P n 1 {(1+ ) n (1 } ) n. Similarly, for E {1,q}, the IRS of A with respect to E is {( ) n ( ) n } n () q 1 1+4q. 1+4q In particular, the IRS for E {1,} is the well-known Jacobsthal number sequence {J n } {0,1,1,3,5,11,1,...} with the expression J n 1 3 (n ( 1) n ). The Jacobsthal-Lucas numbers { n } in A with respect to E {1,} satisfying 0 and 1 1 has the first few elements as {,1,5,7,17,31,...}. From (0), one may deduce n J n +4J n 1 n +( 1) n. In addition, the above formula can transform all identities of the Jacobsthal-Lucas number sequence to those of Jacobsthal number sequence and vice-versa. For example, we have from J n +4J n 1 J n J n, J m J n 1 J n J m 1 ( 1) n n 1 J m n, J m J n +J m J n 1 +J n J m 1 J m+n n J n J n, J m n J n m ( 1) n n+1 J m n, J m n J n m J m+n, respectively. Similarly, we can show that the following two identities are equivalent: n J n+1 +J n 1, J n+1 J n +J n 1. 13

14 Remark 11. Corollary 7 can be extended to the linear nonhomogeneous recurrence relations of order with the form: a n pa n 1 +qa n +l, l 0, for p+q 1. It can be seen that the above recurrence relation is equivalent to the homogeneous form () b n pb n 1 +qb n, where b n a n k and k l 1 p q. Example 1. An obvious example of Remark 11 is the Mersenne number M n n 1, n 0, which satisfies the linear recurrence relation of order : M n 3M n 1 M n ( with M 0 0 and M 1 1) and the non-homogeneous recurrence relation of order 1: M n M n 1 +1 (with M 0 0). It is easy to check that sequence M n (k n 1)/(k 1) satisfies both the homogeneous recurrence relation of order, M n (k +1)M n 1 km n, and the non-homogeneous recurrence relation of order 1, M n km n 1 +1, where M 0 0 and M 1 1. Here, M n is the IRS with respect to E {3, }. Another example is Pell number sequence that satisfies both homogeneous recurrence relation P n P n 1 + P n and the non-homogeneous relation P n P n 1 + P n +1, where P n P n +1/. Remark 13. Niven, Zuckerman, and Montgomery [1] studied some properties of two sequences, {G n } n 0 and {H n } n 0, defined respectively by the linear recurrence relations of order : G n pg n 1 +qg n and H n ph n 1 +qh n with initial conditions G 0 0 and G 1 1 and H 0 and H 1 p, respectively. Clearly, () G n n, the IRS of A with respect to E {p 1 p,p q}. Using Corollary 7, we may rebuild the relationship between the sequences {G n } and {H n }: H n pg n +qg n 1, q G n p +4q H 1 n 1 + p +4q H n+1. We now give more examples of higher order IRS in A 3. Example 14. Consider set A 3 of all linear recurring sequences defined by (1) with coefficient set E 3 {p 1 1,p 1,p 3 1}. The IRS of A 3 is the tribonacci number sequence {0,0,1,1,,4,7,13,4,44,...} (A000073). From Proposition 1, the generating function of the tribonacci sequence is t P 3 (t) 1 t t t 3. Theorem gives the expression of the general term of tribonacci sequence as where (3) n α n 1 (α 1 α )(α 1 α 3 ) + α n (α α 1 )(α α 3 ) + α n 3 (α 3 α 1 )(α 3 α ), α C 3 + 3C, α C 6 (1+i 3) 1 3C (1 i 3), α C 6 (1 i 3) 1 3C (1+i 3), 14

15 inwhichc (( )/) 1/3.Thetribonacci-likesequence{a n } {,1,1,4,6,11,1,...} (A141036) is in the set A 3. From Theorem 3 and its corollary 4, a n can be presented as a n a (3) n k a k (3) n 1. a (3) n +(a 0 +a 1 ) (3) n 1 +a 1 (3) n. (8) Using (17) and (18) in Theorem 5, we may obtain the expression (3) n 6 19 a n a n 1 19 a n 1, n 0, (9) where a 1 : due to the linear recurrence relation a n a n 1 +a n +a n 3. It is easy to see the identity (3) n (3) n 1 (3) n 4, n 4, because the right-hand side is equal to (3) (3) n + n 3 + (3) n 1 + (3) n + (3) n 3 after substituting into (3) n 4. Thus, by using (9) we have an identity for {a n }: 6a n+1 16a n +7a n 1 +a n +6a n 3 4a n 4 a n 5 0 (3) n 1 for all n 5. Similarly from the identity a n a n 1 a n 4, n 4, and relation (8), we have the identity for all n 6. (3) n + (3) n 1 5 (3) n (3) n 3 + (3) n 4 +3 (3) n 5 + (3) n 6 0 It is easy to see there exists r number sequences, denoted by {a () n }, 1,,...,r, in A r with respect to E r such that for any {a n } A r we have a n r c a () n, n 0, where c (c 1,c,...,c r ) can be found from the system consisting of the above equations for n 0,1,...,r 1. In this sense, we may call {a () n }, 1,,...,r, a basis of A r with respect to E r. For instance, for r, let {a (1) n } be {F n }, the IRS of A with respect to E {1,1} (i.e., the Fibonacci sequence), and let {a () n } be the Lucas number sequence {L n }. Then {{F n },{L n }} is a basis of A because. Thus, for any {a n } A with respect to E {1,1} and the initial vector (a 0,a 1 ), we have ( a n a 1 1 ) a 0 F n + 1 a 0L n, 15

16 where the coefficients of the above linear combination are found from the system [ ][ ] [ ] 0 c1 a c Obviously, if {L n } is replaced by any sequence {a () n } in A with respect to E {1,1}, provided the initial vector (a () 0,a () 1 ) of {a () n } satisfies [ ] 0 a () 0 det 1 a () 0, (30) 1 then {{F n },{a () n }} is a basis of A with respect to E {1,1}. However, if a () 0 1 and a () 1 0, then the corresponding a () n F n 1, and the corresponding basis {{F n },{a () n F n 1 }} is called a trivial basis. Thus, for A with respect to E {p 1,p }, {{ n () },{a () n }} forms a non-trivial basis, if the initial vector (a () 0,a () 1 ) of {a () n } satisfies (30) and a () 0 1/p when a () 1 0. The second condition guarantees that the basis is not trivial, otherwise a () n () n 1. 4 Applications In this section, we will give the applications of IRS to the Stirling numbers of the second kind, Wythoff array, and the Boustrophedon transform. Let A r be the set of all linear recurring sequences defined by the homogeneous linear recurrence relation(1) with coefficient n set E r {p 1,p,...,p r }, and let be the IRS of A r. Theorems 3 gives expressions of {a n } A r and the IRS of A r in terms of each other. We will give a different expression of {a n } A r in terms of n using the generating functions of {a n } and { n } shown in Proposition 1. More precisely, let {a n } A r. Then Proposition 1 shows that its generating function P(t) can be written as (3). In particular, the generating function for the IRS with respect to {p } is presented as (4). Comparing the coefficients of P(t) and P(t), one finds Proposition 15. Let {a n } be an element of A r, and let { n } be the IRS of A r. Then Proof. Eq. (31) comes from a 1 n [t n r+1 1 ] p t (31) r 1 k a n a 0 n+r 1 + (a k p a k ) n+r k 1. (3) k1 n [t n ] P(t) [t n r+1 1 ] p t. 16

17 Hence, a n [t n ]P(t) a 0 [t n 1 ] p t +[tn ] ( ) r 1 k a 0 n+r 1 + a k p a k [t n k ] which implies (31). Proposition 16. If k1 g(t) ( r 1 k1 a k ) k p a k t k p t 1 p t, t r 1 p t (33) is the generating function of a sequence {b n }, then {b n } satisfies 0, if n 0,1,...,r ; b n 1, if n r 1; r p b n, if n r. Proof. It is clear that For n r, which completes the proof. b [t ]g(t) [t r+1 1 ] p t [t r+1 ] ( r ) k { p t 0 if 0 r, 1 if r 1. k 0 r p b n r p [t n ]g(t) r [t n ] t r 1 p t i1 p it i ( ) [t n ]t r 1 1 i1 p it 1 i [t n ]g(t) [t n ]t r 1 b n, The Stirling numbers of the second kind S(n,k) count the number of ways to partition a set of n labelled obects into k unlabeled subsets. We now prove that the sequence {b n S(n+1,k)} n 0 is the IRS of A k for any k N 0 by using the structure of the triangle of the Stirling numbers of the second kind shown in Table 1. 17

18 n\k Table 1. Triangle of the Stirling numbers of the second kind Theorem 17. Let S(n,k) be the Stirling numbers of the second kind. Then, for any fixed integer k > 0, S(n+1,k) is the IRS of A k, i.e., S(n+1,k) where A k is the set of linear recurring sequences {a n } generated by (k) n, (34) a n k p a n, (35) which has characteristic polynomial k k P(t) t k p t k (1 t). (36) Thus, the generating function of {S(n+1,k)} is P(t) t k k (37) (1 t). Proof. {S(n,k)} 0 k n obeys the recurrence relation S(n+1,k) ks(n,k)+s(n,k 1), 0 k n, (38) which can be explained by using the definition of S(n,k) shown above. A partition of the n+1 obects into k nonempty subsets either contains the subset {n+1} or it does not. The number of ways that {n + 1} is one of the subsets is given by S(n,k 1), since we must partition the remaining n obects into the available k 1 subsets. The number of ways that {n+1} is not one of the subsets is given by ks(n,k) because we partition all elements other than n+1 into k subsets with k choices for inserting the element n+1. Summing these two values yields (38). Write (38) as S(n,k 1) S(n+1,k) ks(n,k), 0 < k n. 18

19 Denote by g k 1 (t) and g k (t) the generating functions of k 1st and kth columns of the Stirling triangle shown in Table. Then the above equation means [t n ]g k 1 (t) S(n,k 1) S(n+1,k) ks(n,k) [t n+1 ]g k (t) k[t n ]g k (t) [t n+1 ](1 kt)g k (t) for 0 < k n. Therefore, we have g k (t) t 1 kt g k 1(t). Since g 1 (t) t/(1 t), we obtain the generating function of {S(n,k)} for k > 0 g k (t) t k k (39) (1 t). Denote b n S(n+1,k) and the generating function of {b n } by B(t). Then b n [t n ]B(t) S(n+1,k) [t n+1 ]g k (t), 0 < k n, which implies the generating function of {b n } is B(t) 1 t g k(t) t k 1 k (40) (1 t). Using Proposition 16, we immediately know that {S(n+1,k)} is the IRS of A k that has the generating function (37), where A k is the set of linear recurring sequences {a n } generated by recurrence relation (35) with the characteristic polynomial (36). Example 18. Since the second column of the Stirling triangle, {S(n+1,)}, has the generating function t P (t) (1 t)(1 t) 1 1 3t+t, we have the recurrence relation S(n+1,) 3S(n,) S(n 1,) with the initial S(1,) 0 and S(,) 1. Thus we have S(3,) 3, S(4,) 7, S(5,) 15, S(6,) 31, etc. Similarly, {S(n + 1, 3)} is a linear recurring sequence generated by S(n+1,3) 6S(n,3) 11S(n,)+6S(n,1) with initials S(1,3) S(,3) 0 and S(3,3) 1. From the recurrence relation one may find S(4,3) 6, S(5,3) 5, S(6,3) 90, etc. 19

20 If we transfer the entries in the array shown in Table 1 to d n,k k!s(n+1,k+1)/n!, then thearray(d n,k ) 0 k n isariordanarray. ThestructureandcomputationoftheRiordanarray of the generalized Stirling numbers can be found from[6], while the sequence characterization of the Riordan array is shown in [7]. WenowconsidertheWythoffarrayshowninTable, inwhichthetwocolumnstotheleft of the vertical line consist respectively of the nonnegative integers n, and the lower Wythoff sequence (A01), whose nth term is [(n+1)α], where α (1+ 5)/. The rows are linear recurring sequences generated by the Fibonacci rule that each term is the sum of the two previous terms Table. the Wythoff array. The first row sequence in Table is Fibonacci sequence, i.e., the IRS in A with respect to E {1,1}. The th row sequence {a () n } n 0 are linear recurring sequences in A with respect to E {1,1} with initials (,[( +1)α]), 1,,..., where α (1+ 5)/. From (3) or (5), the expression of a () n can be written as a () n [ ] ( +1) [ ] ( +1) ( 1+ 5 ( 1+ ) n 5 1 ) n 5. (41) SinceeveryintegerinN 0 appearsexactlyonce(see[14]), thecollection{a () n } n 0, 0,1,..., gives a partition of N 0. [( +1)α], α (1+ 5)/, can be considered as the representative of the equivalence class {a () n : n 0,1,...}. 0

21 Table 3. the Pell-Wythoff array a 0 b 0 a 1 b 1 a 0 +a 1 b a 1 +a +b 1 a +b 1 a a 3 a 3 +b a +a 3 +b 1 +b b 3 a +a 3 +b 1 +b Table 4. Triangle of the boustrophedon transform The Wythoff array can be extended to similar arrays associated with different linear recurrence relations. For instance, one may define the Pell-Wythoff array shown in Table 3. The first row sequence in Table is the Pell sequence, which is also the IRS in A with respect to E {,1}. The th row sequence {b () n } n 0 are linear recurring sequences in A with respect to E {,1} with initials (,[( +1)r] 1), 1,,..., where r 1+. Noting (3) or (5), we obtain the expression of b () n as b () n ([ ( +1)(1+ ) ] (1 ) 1 ([ ] ( +1)(1+ ) (1+ ) 1 ) ( 1+ ) n ) ( 1 ) n. (4) Millar, Sloane, and Young [11] showed that the boustrophedon transform of a given sequence {a n } n 0 is the sequence produced by the triangle shown in Table 4. Formally, the entries T n,k, n k 0, of the above triangle are defined by T n,0 a n n 0, T n+1,k+1 T n+1,k +T n,n k n k 0, (43) and then b n T n,n for all n 0. Denote the (exponential) generating functions of sequences {a n } and {b n } by A(t) and B(t). Then Theorem 1 of [11] gives B(t) (sect+tant)a(t), 1

22 which implies a relationship of the generating function P r (t) of the IRS { n } of A r and the generating function (BP r )(t) of the boustrophedon transform sequence of { n }: (BP r )(t) (sect+tant)p r (t) (sect+tant) 5 Acknowledgements t r 1 p t. (44) The author would like to express his gratitude to the editor and the anonymous referee for their helpful comments and remarks. The author also thank Michael Dancs for his proofreading of the manuscript. References [1] G. E. Bergum, L. Bennett, A. F. Horadam, and S. D. Moore, Jacobsthal polynomials and a conecture concerning Fibonacci-like matrices, Fibonacci Quart. 3 (1985), [] R. L. Burden and J. D. Faires, Numerical Methods, 8th Edition, Brooks/Cole Publishing Co., 005. [3] L. Carlitz, Problem B-110, Fibonacci Quart. 5 (1967), [4] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Comp., [5] C. Falbo, The golden ratio a contrary viewpoint, College Math. J. 36(005), [6] T. X. He, Expression and computation of generalized Stirling numbers, J. Combin. Math. Combin. Comput., 86 (013), [7] T. X. He and R. Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math., 309 (009), [8] T. X. He and P. J.-S. Shiue, On sequences of numbers and polynomials defined by second order recurrence relations, Int. J. Math. Math. Sci., 009 (009), Article ID , 1 1. [9] T. Koshy, Fibonacci and Lucas Numbers with Applications, Pure and Applied Mathematics, Wiley-Interscience, 001. [10] R. Lidl and H. Niederreiter, Finite Fields, With a foreword by P. M. Cohn, nd edition, Encyclopedia of Mathematics and its Applications, Vol. 0, Cambridge University Press, Cambridge, 1997.

23 [11] J. Millar, N. J. A. Sloane, and N. E. Young, A new operation on sequences: the boustrophedon transform, J. Combin. Theory, Series A, 76 (1996), [1] I. Niven, H. S. Zuckerman, and H. L. Montgomery, An Introduction to the Theory of Numbers, 5th edition, John Wiley & Sons, Inc., [13] W. Rudin, Principles of Mathematical Analysis, 3rd edition, International Series in Pure and Applied Mathematics. McGraw-Hill Book Co., [14] N. J. A. Sloane, My favorite integer sequences, AT&T Lab Research Preprint, [15] G. Strang, Linear Algebra and Its Applications, nd edition, Academic Press, [16] H. S. Wilf, Generatingfunctionology, nd edition, Academic Press, Mathematics Subect Classification: Primary 11B39; Secondary 05A15, 11B73, 11B75, 05A19, 11Y55. Keywords: number sequence, linear recurrence relation, Stirling number of the second kind, generating function, Wythoff array, Boustrophedon transform, Fibonacci number, Pell number, Jacobsthal number, Lucas number, Lucas sequence, Riordan array. (Concerned with sequences A000073, A000078, A001591, A0796, A1189, and A ) Received March 7 013; revised versions received March 9 013; June 1 013; August Published in Journal of Integer Sequences, October Return to Journal of Integer Sequences home page. 3