ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES

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1 ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES BROTHER ALFRED BROUSSEAU St. Mary's College, Moraga, California 947 Given a third-order recursion relation (D T n + 1 = a l T n " a 2 T n _ i + a T n Let the auxiliary equation (2) x - 2L t x 2 + a 2 x - a = 0 have three distinct roots r l r 2, r. Then any term of a sequence governed by this recursion relation can be expressed in the form () T n = A i r f + A 2 r? + A r f. THE SEQUENCE S = r? Since the individual elements of these sums are powers of the roots, the sums obey the given recursion relation. Hence it is possible to determine a few terms of S by means of symmetric functions and thereafter generate additional terms of the S sequence. Since this sequence is basic to all the algorithms, its generation constitutes the first algorithm. (Note. This use of the S sequence is exemplified in [ 1 ].) ALGOEITHM FOE FINDING THE TERMS OF S Three consecutive terms of the sequence are: Si = a x S 2 = a 2 x - 2a 2! S = &\ - a A a 2 + a Then use the recursion relation to obtain positive and negative subscript terms of the sequence, The algorithm will be illustrated for two recursion relations which will be used to check other algorithms numerically. 167

2 168 ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES [April E X A M P L E 1: x - x EXAMPLE 2: X x^ 2 + x + 4 = S n <-4) n RECURSION RELATIONS FOR SPACED TERMS OF A SEQUENCE Given a sequence T satisfying the given recursion relation. It is desired to find the recursion relation for a spacing of k among the terms, namely, for the sequence T.. Since () nk+a A nk+a, A nk+a, A nk+a A t r t + A 2 r 2 + A r

3 1974] ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES 169 and since there is a change of r, from one term to the next, the recursion relation is that k x whose roots correspond to r.. Let the coefficients be given in the relation Then x - B t x 2 + B 2 x - B = 0. -r, v ^ k k k - k k_ B 2 = E r i r j = a9e r = i a s _k B = 4 Hence the recursion relation is given by (6) x - S k x 2 + a!f S_ k - a!f = 0. EXAMPLE FOR x - x 2 - x - 1 = 0 with k =. Using the sequence S with n = 20, T _,_-= -21T + T _ + T 1A. n+ n n- n0 T 2 = 21* = EXAMPLE FOR x - 7x 2 + x + 4 = 0 using the terms of the S sequence. T = (-9 T Ti - T 4 )/64 T_ = (-9*81/ *7-19)/64 = -4446/1024. SECOND-DEGREE HOMOGENEOUS SEQUENCE FUNCTIONS If there are several sequences satisfying the given recursion relation, a sum of terms of the form T T would form a homogeneous sequence function of the second degree. mi m 2 Such terms if expanded using the roots of the auxiliary equation would yield terms of the form r m i m 2 a n (j others of the form C r. 1 r. 2. The first type obey the recursion relation for i i 1 1 j J P J r 2 since there is a change of 2 in the power in going from one term in the product to the next as the m f s change by 1. The second type obey the recursion relation for the quantities r.r.. 1 J given by: ALGORITHM FOR THE SECOND-DEGREE FUNCTIONS The recursion relation governing the quantities r? has already been obtained and is (7) x - S 2 x 2 + ass_ 2 x - 4 = 0.

4 170 ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES [April For the second we need to find the symmetric functions of the roots r. r.. B i = X) r i r j = a 2 B = r? r? r? = a. Hence the recursion relation is (8) x - a 2 x 2 + a ^ x - a = 0. The total recursion relation is the product of (7) and (8): (9) (x - S 2 x 2 + a S_ 2 x - a )(x - a 2 x 2 + a ^ x - a ) = 0. EXAMPLE FOR x - 7x 2 + x + 4 = 0. Si = 44 s sf - 6 Si S^ S 2, - 26 S 2^ = 44* * * * * *(/4) 2 = = THIRD-DEGREE HOMOGENEOUS SEQUENCE FUNCTIONS An expression of the form gives rise to terms of the form T(1) T (2) T () m x m 2 m m-i+mo nii+mo m«mi m? mo i i j * i j k The first type corresponds to the recursion relation for r?, the second to the recursion r e - lation for r 2 r. and the third to the recursion relation for ao. The first relation is: I j ' * (10) x - Sgx 2 + a S_ x - a = 0 The last relation is: () x - a = 0. For the second we have a relation of the sixth degree with coefficients symmetric functions of the roots R x = r ^ r 2, R 2 = r f r ^ R = rfr, R 4 = r r l 9 R = r 2 r, R 6 = r r 2.

5 1974] ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES 171 B i = E R i = < 21 ) = " a + a 2 ai, where the notation (21) = J^r? r. taken as a symmetric function. B a - S R j R. = (41 2 ) + ( 2 ) + (21) + (222) B 2 = 6a - a a2a x + a a + a 2 B = L R. R. R k = (1) + 2(42) + 2( ) (12) B = -7a + 6a a 2 a t - 2a aj - 2a a 2 + a a 2 a 2 B 4 = (6 2 ) + ( 2 2) + (4) + (444) = a () + a ( 2 ) + a (21) + (4 ) B 4 = 6a - a a 2 a x + aga^ + a a 2 B = (64) = af (21) = -a + a f a ^ B 6 = ag The product of (10), () and the polynomial whose coefficients are given by (12) is the r e - quired recursion relation for the third degree. APPLIED TO x - x 2 - x - l = 0, we have or (x - 7x 2 + x - l)(x - l)(x 6 + 4x + llx X + l l x 2 + 4x + 1) = 0 x 10-4x 9-9x 8-4x x 6-2x + 40x 4-14x - x 2-2x + 1 = 0. Starting with S 9 = 241 we have: 4* * + 4*71-24*9 + 2*21-40* + 14* *1 - = = 44 - S Z 1Q. FOURTH-DEGREE HOMOGENEOUS SEQUENCE FUNCTIONS We proceed as before but without going through the preliminary details we arrive at the conclusion that the symmetric functions of the roots are given by the partitions (4), (1), (22), (2) of four into three parts or less. We determine the recursion relations or equivalently the coefficients for each of these..(4) (1) x - S 4 x 2 + afs_ 4 x - af = 0. (2) Since this symmetric function is equivalent to a rj in its terms, we have the relation (14) x - a ^ x 2 + a a 2 x - af = 0. (1) A t = (1) = -SL Z SL t - 2a 2 + a 2 a 2

6 172 ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES [April A 2 = (6) + (44) + (41) + (2) = a () + (44) + a (2) + a a 2 A 2 = -a a 2 + Sa^a^ + 2a a 2 a 1 - a a 2 a + a a^ + a 2 A = (741) + (642) +. (4) + 2(444) = a (6) + a (42) + ajj (21) + 2af ( 1 ) A = 2a - I a a 2 a 1 + a a^ + a a + 10a a 2 ai - a a 2 a^ - S a g a ^ i + a a 2 af A 4 = af( 2 ) + a (4) + af (1) + a (1) A 4 = -afa-i + a a 2 + 2a a 2 a^ - a a 2 a x + a af + a af A = aj(2) = - a a 2-2afa *1 A 6 = a (22) B x = (2 2 ) = -2a ai + a 2 (16) B 2 = (422) = a (2) = -2a a 2 + a af B = 4 The product of the polynomials given by (1), (14), (1), and (16) gives the required recursion relation for the fourth degree. APPLICATION TO x - x 2 - x - 1 = 0. (x - llx 2 - x - l)(x 6 + 4x + 1x 4-24X + 7x 2 + l)(x + x 2 + x - 1) or X (x - x 2 - x - 1) = 0 X1 _ 7x14 _ x i _ 22x x + 41 x x 9-41 x 8-7x 7-67x x + x X - x 2 - x - 1 = 0. REMARKS The determination of the coefficients of the polynomials for higher degrees in terms of the coefficients of the original recursion relation leads to expressions of ever greater complexity which make calculations tedious and present a greater possibility of error. A simpler approach is to use symmetric functions of the roots which in turn can be calculated by means of the S sequence of the given recursion relation. For three roots all such symmetric functions can be reduced to one of the forms (ab), (a 2 ) or (a). The last is simply S while the others are given by:

7 1974] ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES 17 (17) (ab) = S a S b - S a + b (18) (a 2 ) = (S 2 a - S 2 a ) / 2. FIFTH-DEGREE HOMOGENEOUS SEQUENCE FUNCTIONS On the basis of partitions we consider symmetric functions of the roots of the forms (), (41), (2), (), (221). IS : - S x 2 + afs_ 6 x () Bi = = a s (2) B 2 = a (2 8 ) B (221) (41) Ci = c 2 = c = a a 2 4*1 4 Di = ( 4 1 ) D 2 = (81 2 ) + ( 2 ) + (41) + (442) = a (7) + ( 2 ) + a (4) + a 2 (2 2 ) D = (91) + (82) + (64) + 2( ) = a (84) + a (6) + a (21) + 2ajj D 4 = (992) + (10,) + (96) + (866) = al(7 2 ) + a () + a (41) + a 6 (2) D = (10,96) = a 6 (4) D 6 = aj (2) E t = (2) E 2 = (622) + () + (2) + (4) = a (4) + ( 2 ) + a (1) + 4*1 E = (82) + (64) + 2() + (7) = af (6) + a (21) + 2a + ajj (42) E 4 = (884) + (10,) + (87) + (776) = af(4 2 ) + a () + af(2) + a 6 a 2 E = (10,87) = a '(1) E 6 = aj

8 174 ALGORITHMS FOR THIRD-ORDER RECURSION SEQUENCES April 1974 APPLICATION TO x - x 2 - x - 1 = 0. (x - 21x 2 - x - l)(x - x 2 - x - imx + x 2 + x - l)(x 6 + Ox + 7x 4-24x + 1x 2 + 4x + 1) X (x x + 7X x - x 2-6x + 1) = 0. The product is x 21-1x 20-0x x 18 ± 242x x x 1-976x x x x x x x x 7-892x x.- 7x x - 10x 2 + x - 1 = 0. CONCLUDING NOTES 1. That the symmetric functions of the roots can always be expressed in terms of the quantities S is an elementary proposition in combinatorial analysis. (See [2; p. 7].) 2. For the n degree, the recursion relation has degree / n + 2 \ 2 V This follows from the fact that the number of terms involving the roots is equivalent to the solution of x + y + z = n in positive integers and zero.. For the sixth-degree relations, the coefficients A A and A, A 2 and A 4, are complementary, the respective quantities in the symmetric functions adding up to 2n. 4. Each term in a coefficient has a weight. The coefficient A, would have its terms of weight kn where n is the degree being considered for the terms of the original recursion relation. Thus for n = 8, E 4 has a term a (7 2 ) which has a weight 6 X + 2 X 7 = 2== 4 X 8.. If a = 1, all the factors for the n degree are found for degree n With some modifications on the symmetric functions involved, this approach could be used to produce algorithms relating to recursion relations of higher order. 7. The algorithms were checked numerically by using a relation with roots 1, 2, and 4, finding the symmetric functions directly and comparing the result with that given by the algorithms. REFERENCES 1. Trudy Y. H. Tong, "Some Properties of the Tribonacci Sequence and the Special Lucas Sequence," Master T s Thesis, San Jose State University, August P. A. MacMahon, Combinatory Analysis, Cambridge University P r e s s, 191, Reprinted by Chelsea Publishing Company, I ' Editor's Note: There are an additional twelve pages on this subject, going through the tenth degree. If you would like a Xerox copy of the additional material at four cents a page (which includes postage, materials and labor), send your request to: Brother Alfred Brousseau, St. Mary! s College, Moraga, California 947.