PURELY PERIODIC SECOND ORDER LINEAR RECURRENCES

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1 THOMAS MCKENZIE AND SHANNON OVERBAY Abstract. Second order linear homogeneous recurrence relations with coefficients in a finite field or in the integers modulo of an ideal have been the subject of much study see for example [1,, 4, 5, 6, 7, 8, 9]). This paper extends many of these results to finite rings. In the first part of this paper we develop polynomials which generate purely periodic sequences over any finite ring, R. We then use these polynomials with coefficients in R to establish bounds on the period of these sequences. 1. Introduction Throughout this paper R is a finite commutative ring with identity. We assume that a ring and its subrings share the same identity. Let Z denote the integers, N denote the nonnegative integers, and Z + denote the positive integers. A sequence s = {s 0, s 1,...} of elements in R is purely periodic with period n Z +, if n is the smallest positive integer with s n+i = s i for all i N. If s is purely periodic with period n then s is said to be uniformly distributed if for every r R, the cardinality of {i N s i = r, 0 i n 1} equals n divided by the order of R. For n Z + we call s an nth order linear homogeneous recurrence relation if there exist a 1,..., a n R with a n 0 such that s i+n = a n s i + a n 1 s i a 1 s i+n 1 for all i N. The aim of this section is to establish a relationship between purely periodic nth order linear homogeneous recurrence relations and the coefficients of certain polynomials. In the next theorem we use the division algorithm to create an nth order linear recurrence. Theorem 1.1. Let fx) = a n x n + + a 1 x 1 R[x] where n Z + and a n 0. Suppose that there exists hx) R[x] with degree at most n 1 and m Z + such that m n and fx) divides hx) x m 1) in R[x]. Now let c jx j R[x] such that hx) x m 1) = c jx j ) fx) and define s = {s 0, s 1,...} as the sequence given by s i = c j where i = j modulo m. Then s is a purely periodic nth order linear recurrence given by for all i N. s i+n = a n s i + a n 1 s i a 1 s i+n 1 Proof. Let i N and let gx) = c jx j. Note that fx) gx) = hx) x m 1) and gx)x mk = s mk x mk + + s mk+1) 1 x mk+1) 1. Use the division algorithm to write i + n as mq + r where q, r Z and 0 r < m. 160 VOLUME 46/47, NUMBER

2 Now fx) s qm x qm + s q x q + + s 1 x + s 0 ) = fx) x qm gx) + x q 1)m gx) + + gx)) = x qm fx)gx) + x q 1)m fx)gx) + + fx)gx) = x qm hx)x m 1) + x q 1)m hx)x m 1) + + hx)x m 1) = x q+1)m hx) x qm hx) + x qm hx) hx) = x q+1)m hx) hx). Since 0 n 1 < i + n < q + 1)m, the coefficient of x i+n in the expansion above equals zero. In other words, a n s i + a n 1 s i a 1 s i+n 1 s i+n )x i+n = 0. Hence, s i+n = a n s i + a n 1 s i a 1 s i+n 1 for all i N. This completes the proof. Example 1.. Assume for the moment that R = Z/5Z). Let fx) = x + 4x 1 and hx) = x. Note that fx) divides xx 0 1) and hx) x 0 1) fx) = 0 1 c j x j = 0 + 1x + 4x + x 3 + x 4 + 0x 5 + x 6 + 3x 7 + 4x 8 + 4x 9 + 0x x x 1 + 3x x x x 16 + x x x 19. If we let s 0 = 0, s 1 = 1, and s i+ = s i + 4s i+1, then for all i N, s i = c j where i is congruent to j modulo 0. Note that this purely periodic sequence is uniformly distributed. In Theorem 1.1 we took polynomials and produced a purely periodic sequence. In the next theorem we reverse that process. The reader should note the similarity between this next theorem and Theorem 6.5 on page 197 of Lidl and Niederreiter [4]. Although this result in Lidl and Niederreiter is stated for finite fields, the proof actually works over more general rings. Theorem 1.3. Suppose s = {s 0, s 1,...} is a purely periodic nth order recurrence in R with period m n. That is, there exist a 1,..., a n R with a n 0 such that, for all i N, Let and s i+n = a n s i + a n 1 s i a 1 s i+n 1. fx) = a n x n + + a 1 x 1 n 1 hx) = a j s 0 + a j 1 s a s j + a 1 s j 1 s j )x j. Then hx) x m 1) = s jx j ) fx). MAY 008/

3 THE FIBONACCI QUARTERLY Proof. Clearly the polynomials fx) s jx j and hx) x m 1) both have degree m+n 1. It suffices to check that these polynomials are equal term by term. Let i {0,..., m+n 1}. Case 1 i < n): In this case the coefficient of x i in fx) s jx j equals a i s 0 +a i 1 s a s i + a 1 s i 1 s i which equals the coefficient of x i in hx). Since i < n < m, the coefficient of x i in hx) x m 1) is equal to the coefficient of x i in hx). Case n i m 1): In this case the coefficient of x i in hx) x m 1) is equal to zero, since the degree of hx) is n 1. The coefficient of x i in fx) s jx j equals a n s i 1 +a n 1 s i n+1 +a n s i n+ + +a 1 s i 1 s i. This is also equal to zero by the recurrence relation. Case 3 m i m+n 1): In this case the coefficient of x i in fx) s jx j equals a n s i n + a n 1 s i n+1 + a n s i n+ + + a i m+1 s equals a i m s m + a i s m a 1 s i 1 s i ) by the recurrence relation) equals a i m s 0 + a i s a 1 s i 1 m s i m ) since s is periodic) equals the coefficient of x i in x m hx) equals the coefficient of x i in x m 1) hx) since n 1 m i).. Purely Periodic Second Order Linear Recurrences Throughout this section let s = {s 0, s 1,...} be the sequence in R generated by second order linear recurrence relation s i+ = a s i + a 1 s i+1 for all i N where a 1 R and a is a unit in R. The polynomial gx) = x a 1 x a is the characteristic polynomial associated to this recurrence. In this section we establish a formula for s n in terms of the roots of the characteristic polynomial. We then use this formula to give an upper bound for the period of s. Theorem.1. There exists a ring S which contains R as a subring, an element r 1 S, and an element r R[r 1 ] such that gx) = x a 1 x a = x r 1 )x r ). Proof. Let S = R[x]/gx)) and let r 1 be the element x + gx)), S. Note that R is a subring of S and gr 1 ) = 0. By the division algorithm on page 158 of [3], there exists a polynomial hx) with coefficients in R[r 1 ] and element c in R[r 1 ] such that gx) = x r 1 )hx) + c in R[r 1 ][x]. Plugging r 1 into both sides of this equation and recalling that gr 1 ) = 0 yields c = 0 and gx) = x r 1 )hx). Since gx) is a monic polynomial of degree two and x r 1 ) is a monic polynomial of degree one, it follows that there exists r R[r 1 ] such that hx) = x r ) and gx) = x r 1 )x r ). Consider a ring S which contains R as a subring with elements r 1, r S such that gx) = x a 1 x a = x r 1 )x r ). Since r 1 r = a, and a was assumed to be a unit in R, it follows that r 1 and r are units in S. Now r 1 is integral over R and thus by Theorem 5.3 on page 395 of [3], R[r 1 ] is a finitely generated R module. This, along with the fact that R is a finite ring, implies that the units of R[r 1 ] form a finite group. We use r 1, r Z + to denote the orders of r 1 and r in this group. Hence, r r 1 1 = r r = 1 R. Recall that gx) = x a 1 x a is the characteristic polynomial corresponding to the recurrence s i+ = a s i + a 1 s i VOLUME 46/47, NUMBER

4 Theorem.. Let S be a ring which contains R as a subring and let r 1, r S such that gx) = x r 1 )x r ). Then for n 1, n 1 s n = s 0 r1 n + s 1 s 0 r 1 ) Proof. We proceed by induction on n. If n = 1 then When n = we have s 0 r s 1 s 0 r 1 ) 0 r j 1r n j 1. r j 1r n j 1 = s 0 r s 1 s 0 r 1 ) = s 1. 1 s 0 r1 + s 1 s 0 r 1 ) r1r j j 1 = s 0 r1 + s 1 s 0 r 1 )r 1 + r ) = r 1 + r )s 1 + s 0 r 1 r ) = a 1 s 1 + a s 0 = s. Now let k be an integer greater than and assume that the equality holds for all integers n greater than zero and less than k. We show the equality holds for n = k. Note that s k = a 1 s k 1 + a s k ) = a 1 s 0 r1 k 1 + s 1 s 0 r 1 ) r1r j k 1 j 1 k 1 ) + a s 0 r1 k + s 1 s 0 r 1 ) r1r j k j 1 ) = r 1 + r ) s 0 r1 k 1 + s 1 s 0 r 1 ) r1r j k 1 j 1 k 1 ) + r 1 r ) s 0 r1 k + s 1 s 0 r 1 ) r1r j k j 1 ) = r 1 s 0 r1 k 1 + s 1 s 0 r 1 ) r1r j k 1 j 1 ) + r s 0 r1 k 1 + s 1 s 0 r 1 ) r1r j k 1 j 1 k 1 ) + r 1 r ) s 0 r1 k + s 1 s 0 r 1 ) r1r j k j 1 MAY 008/

5 THE FIBONACCI QUARTERLY = s 0 r1 k + s 1 s 0 r 1 ) r j+1 1 r k 1 j 1 + s 0 r1 k 1 r + s 1 s 0 r 1 ) r1r j k j 1 k 1 s 0 r1 k 1 r s 1 s 0 r 1 ) r j+1 1 r k j k ) = s 0 r1 k + s 1 s 0 r 1 ) r1 k 1 + r1r j k j 1 k 1 = s 0 r1 k + s 1 s 0 r 1 ) r j 1r k j 1. We did not use the fact that R is finite or the fact that a is a unit in the proof of the last theorem. The result even holds when r 1 r is a zero divisor. If we set r 1 equal to r in the last theorem we get our next result. Corollary.3. Let S be a ring which contains R as a subring and let r S such that gx) = x r). Then for k 1, s k = s 0 1 k)r k + s 1 kr k 1. Write n R for the characteristic of the finite ring R. Theorem.4. Let S be a ring which contains R as a subring and let r 1, r S such that gx) = x r 1 )x r ). Let λ be the least common multiple or r 1 and r. Then s is purely periodic with period dividing λ n R. Proof. Let m = λ n R and let k N. Then m+k 1 s m+k = s 0 r1 m+k + s 1 s 0 r 1 ) r1r j m+k j 1 = s 0 r1 m+k + s 1 s 0 r 1 ) r1r j m+k j 1 m+k 1 + s 1 s 0 r 1 ) r1r j m+k j 1 j=m m+k 1 where r1r j m+k j 1 = 0 if k = 0) j=m = s 0 r m 1 r k 1 + s 1 s 0 r 1 )r m 1 k 1 r1r j k j VOLUME 46/47, NUMBER

6 + s 1 s 0 r 1 ) r1r j m+k j 1 k 1 = s 0 r1 k + s 1 s 0 r 1 ) r1r j k j 1 + s 1 s 0 r 1 ) r1r j m+k j 1 since r1 λ = 1) k 1 = s 0 r1 k + s 1 s 0 r 1 ) + λ 1 j=λ r j 1r m+k j λ 1 r1r j k j 1 + s 1 s 0 r 1 ) j=λn R 1) k 1 = s 0 r1 k + s 1 s 0 r 1 ) r1r j k j 1 λ 1 ) + s 1 s 0 r 1 ) n R r1r j m+k j 1 k 1 = s 0 r1 k + s 1 s 0 r 1 ) r1r j k j 1 = s k. r j 1r m+k j 1 ) r j 1r m+k j 1 since r λ 1 = r λ = 1) Acknowledgment We would like to thank the referee for a very careful reading of this paper and for many valuable suggestions. References [1] R. T. Bumby, A Distribution Property for Linear Recurrences of the Second Order, Proc. American Mathematical Society, ), [] J. R. Burke, Some Remarks on the Distribution of Second Order Recurrences and a Related Group Structure, Applications of Fibonacci Numbers, ), [3] T. Hungerford, Algebra, Springer, New York, [4] R. Lidl and H. Niederreiter, Introduction to Finite Fields and Their Applications, Cambridge University Press, Cambridge, [5] W. Narkiewicz, Uniform Distribution of Sequences of Integers in Residue Classes, Lecture Notes in Math. 1087, Springer, Berlin, [6] M. Nathanson, Linear Recurrences and Uniform Distribution, Proc. American Mathematical Society, ), [7] H. Niederreiter and J.-S. Shiue, Equidistribution of Linear Recurring Sequences in Finite Fields, Indag. Math., ), [8] W. Valez, Uniform Distribution of Two-Term Recurrence Sequences, Trans. American Mathematical Society, ), [9] W. A. Webb and C. T. Long, Distribution Modulo p of the General Second Order Recurrence, Atti Accad. Lincei, ), MAY 008/

7 THE FIBONACCI QUARTERLY MSC000: 11B37, 11B50 Department of Mathematics, Gonzaga University, Spokane, WA address: Department of Mathematics, Gonzaga University, Spokane, WA address: 166 VOLUME 46/47, NUMBER