RECENT RESULTS ON LINEAR RECURRENCE SEQUENCES

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1 RECENT RESULTS ON LINEAR RECURRENCE SEQUENCES Jan-Hendrik Evertse (Leiden) General Mathematics Colloquium, Delft May 19,

2 INTRODUCTION A linear recurrence sequence U = {u n } n=0 (in C) is a sequence given by a linear recurrence (1) u n = c 1 u n 1 + c 2 u n c k u n k (n k) with coefficients c i C and initial values u 0,..., u k 1 C. The smallest k such that U satisfies a recurrence of type (1) is called the order of U. If k is the order of U, then the coefficients c 1,..., c k are uniquely determined. In that case the companion polynomial of U is given by F U (X) := X k c 1 X k 1 c 2 X k 2 c k. 2

3 FACT. Let U = {u n } n=0 be a linear recurrence sequence. Assume that its companion polynomial can be factored as (2) F U (X) = (X α 1 ) e 1 (X α r ) e r with distinct α 1,..., α r and e i > 0. Then u n can be expressed as a polynomialexponential sum, (3) u n = r i=1 f i (n)α n i for n 0 where f i is a polynomial of degree e i 1 (i = 1,..., r). Conversely, if {u n } n=0 is given by (3) then it is a linear recurrence sequence with companion polynomial given by (2). 3

4 Proof. Let U = {u n } n=0 be a linear recurrence sequence of order k with companion polynomial F U (X) = X k c 1 X k 1 c k = (X α 1 ) e 1 (X α r ) er. Then for some polynomial A of degree < k and for certain constants c ij, n=0 u n X n = = = A(X) 1 c 1 X c 2 X 2 c k X k r e i i=1 j=1 c ij (1 α i X) j r e i c ij i=1 j=1 n=0 ( ) n+j 1 j 1 α n i X n. 4

5 ZERO MULTIPLICITY The zero multiplicity of a linear recurrence sequence U = {u n } n=0 is given by N(U) := #{n Z 0 : u n = 0}. Assume that U has companion polynomial F U (X) = (X α 1 ) e 1 (X α r ) e r with α i distinct, e i > 0. U is called non-degenerate if none of the quotients α i /α j (1 i < j r) is a root of unity. THEOREM (Skolem-Mahler-Lech, ) Let U be a non-degenerate linear recurrence sequence. Then N(U) is finite. 5

6 Example. Let U = {u n } n=0 be given by u n = 3 n +( 3) n +n(2 n (2e 2πi/3 ) n ) (n 0). Then U has companion polynomial F U (X) = (X 3)(X+3)(X 2) 2 (X 2e 2πi/3 ) 2 and u n = 0 for n = 3,9,15,.... Problem. Suppose that U is non-degenerate. Find a good upper bound for N(U). 6

7 Example. Let U = {u n } n=0 be a linear recurrence sequence of order k with terms in R. Suppose that its companion polynomial is F U (X) = (X α 1 ) e 1 (X α r ) e r with 0 < α 1 < < α r. Then U is nondegenerate and u n = r i=1 f i (n)α n i (n 0) where the f i are polynomials with real coefficients. FACT (Follows from Rolle s Theorem) The function u(x) := r i=1 f i (x)α x i has at most r i=1 deg f i k 1 zeros in R. Hence N(U) k 1. Old conjecture: N(U) C(k) for every nondegenerate linear recurrence sequence U of order k with terms in C. 7

8 Linear recurrence sequences of order 3 THEOREM (Beukers, 1991) Let U = {u n } n=0 be a non-degenerate linear recurrence sequence of order 3 with terms in Q. Then N(U) 6. Example (Berstel, 1974) u n+3 = 2u n+2 4u n+1 + 4u n (n 3), u 0 = u 1 = 0, u 2 = 1. Then u 0 = u 1 = u 4 = u 6 = u 13 = u 52 = 0. THEOREM (Beukers, Schlickewei, 1996) Let U = {u n } n=0 be a non-degenerate linear recurrence sequence of order 3 with terms in C. Then N(U) 61. 8

9 Linear recurrence sequences of arbitrary order Earlier results in the 1990 s: Schlickewei, van der Poorten and Schlickewei, Schlickewei and Schmidt: upper bounds for N(U) valid for linear recurrence sequences with algebraic terms and depending on the order k of U and other parameters. THEOREM (Schmidt, 2000). Let U = {u n } n=0 be a non-degenerate linear recurrence sequence of order k with terms in C. Then N(U) expexpexp(20k). 9

10 Steps in the proof. 1) Reduce to the case that all terms of U are algebraic numbers, using a specialization argument from algebraic geometry. 2) Apply techniques from Diophantine approximation, the Quantitative p-adic Subspace Theorem. 3) Write u n = r i=1 f i (n)α n i, where the f i are polynomials. The proof is by induction on ri=1 deg f i. 4) Special case (Schlickewei, Schmidt, Ev.) Suppose that u n = k i=1 c i α n i where the c i are non-zero constants. Then N(U) e (6k)3k. 10

11 THE QUOTIENT OF TWO LINEAR RECURRENCE SEQUENCES If {u n } n=0, {v n} n=0 are linear recurrence sequences, then so are {λu n + µv n } n=0 (λ, µ C) and {u n v n } n=0. What about {u n/v n } n=0? If this is a linear recurrence sequence then u n /v n = r i=1 h i (n)γ n i for certain polynomials h i and certain γ i. Hence all terms u n /v n lie in a finitely generated subring of C, namely the ring generated by the γ i and the coefficients of the h i. 11

12 THEOREM (Pourchet, 1979, van der Poorten, 1988) Let U = {u n } n=0, V = {v n} n=0 be two linear recurrence sequences with terms in C. Suppose that there is a finitely generated subring R of C such that u n /v n R for all but finitely many n. Then there is n 0 0 such that {u n /v n } n=n 0 is a linear recurrence sequence. Can we weaken the condition u n /v n R for all but finitely many n to u n /v n R for infinitely many n? 12

13 THEOREM (Corvaja, Zannier, 2002) Let U = {u n } n=0, V = {v n} n=0 be two linear recurrence sequences with terms in C. Assume that there is a finitely generated subring R of C such that u n /v n R for infinitely many n. Then there are a polynomial g(x) and positive integers a, b such that { g(an + b) u an+b v an+b } n=0 are linear recurrence sequences., { van+b g(an + b) } Proof. 1) Reduce to the case that U, V have algebraic terms by a specialization argument. 2) Apply the p-adic Subspace Theorem. 13

14 Example. Let u n = 4 n 1 ( 1) n 1, v n = n 2 n 1 + n ( 1) n 1 (n 0). For every prime number n 3 we have u n = 4n 1 1 v n n(2 n 1 + 1) = 2n 1 1 n (using Fermat s little theorem). Z Hence u n /v n Z for infinitely many n. Verify that (2n+1) u2n+1 v 2n+1 = 2 2n 1, v 2n+1 2n + 1 = 22n +1 are linear recurrence sequences, but that {u n /v n } n=0 and {nu n/v n } n=0 are not linear recurrence sequences. 14

15 D-TH ROOTS OF LINEAR RECURRENCE SEQUENCES Let d be a positive integer, let {u n } n=0 be a linear recurrence sequence and suppose that there is a linear recurrence sequence {v n } n=0 such that u n = vn d for all n. Write v n = r i=1 h i (n)γ n i where γ i C and h i is a polynomial. Let R be the ring generated by the γ i and the coefficients of the h i. Then for every n there is y R with y d = u n. 15

16 THEOREM (Zannier, 2000) Let d be a positive integer, let {u n } n=0 be a linear recurrence sequence with terms in C and suppose that there is a finitely generated subring of C such that for every n 0 there is y R with y d = u n. Then there is a linear recurrence sequence {v n } n=0 such that vd n = u n for every n 0. Proof. Specialization, algebraic number theory, arithmetic geometry. What if there is y R with y d = u n for infinitely many n? 16

17 THEOREM (Corvaja, Zannier, 1998) Let d be a positive integer. Let {u n } n=0 be a linear recurrence sequence with terms in Q, satisfying the following condition: u n = r i=1 c i α n i for n 0, where the c i are non-zero constants, and where α 1 > max( α 2,..., α r ). Assume that for infinitely many n there is y Q such that y d = u n. Then there are a linear recurrence sequence {v n } n=0 with terms in Q, as well as positive integers a, b, such that v d n = u an+b for every n 0. Proof. p-adic Subspace Theorem. 17

18 THE SUBSPACE THEOREM For x = (x 1,..., x m ) Z m put x := max( x 1,..., x m ) SUBSPACE THEOREM (Schmidt, 1972) Let L i (X) = α i1 X 1 + +α im X m (i = 1,..., m) be m linearly independent linear forms in m variables with algebraic coefficients in C and let δ > 0. Then the set of solutions x Z m of L 1 (x) L m (x) x δ is contained in the union of finitely many proper linear subspaces of Q m. 18

19 Example (m = 3). Consider (4) (x 1 2x 2 )(x 1 + 2x 2 )(x 3 2x 2 ) x 1. 1) (4) has infinitely many solutions x = (x 1, x 2, x 3 ) Z 3 in the subspace x 1 = x 3, which are given by x 1 = x 3, x 2 1 2x2 2 = 1, x 1 x ) (4) has infinitely many solutions x Z 3 in the subspace x 1 = x 3, which are given by x 1 = x 3, x 2 1 2x2 2 = 1, x 1x ) (4) has only finitely many solutions x Z 3 with x 1 ±x 3, given by (±1,0,0), (0,0, ±1). Remark. The Pell equation x 2 1 2x2 2 = 1 has infinitely many solutions in integers x 1, x 2. 19

20 p-adic absolute values Given a prime number p we define the p-adic absolute value p on Q by 0 p = 0; a p = p r if a = p r b/c where b, c are integers not divisible by p. Example: = 2 3 since = Likewise = 3 2. Properties: ab p = a p b p ; a + b p max( a p, b p ); a Z, a divisible by p r a p p r. Product formula: a composed of primes p 1,..., p t a a p1 a pt = 1. 20

21 P-ADIC SUBSPACE THEOREM (Schlickewei, 1977) Let p 1,..., p t be distinct prime numbers. Let L 1 (X),..., L m (X) be m linearly independent linear forms in m variables with coefficients in Q. For each p {p 1,..., p t }, let L 1p (X),..., L mp (X) be m linearly independent linear forms in m variables with coefficients in Q. Let δ > 0. Then the set of solutions x Z m of L 1 (x) L m (x) t i=1 L 1,pi (x) L m,pi (x) pi x δ is contained in the union of finitely many proper linear subspaces of Q m. 21

22 Remark 1. There is a more general result in which the coefficients of the linear forms L i, L ip are algebraic, and the solutions x have their coordinates in a given algebraic number field (Schmidt, Schlickewei). Remark 2. All proofs given up to now for the (p-adic) Subspace Theorem are ineffective, i.e., these proofs do not allow to determine effectively the subspaces containing all solutions. Remark 3. There is however a Quantitative p-adic Subspace Theorem, giving an explicit upper bound for the number of subspaces containing all solutions (Schmidt 1989, Schlickewei 1991,..., Schlickewei& Ev., 2002). This is a crucial tool in the proof of Schmidt s theorem on the zero multiplicity of linear recurrence sequences. 22

23 AN APPLICATION We prove: THEOREM Let p, q be two prime numbers with p > q. Then there are only finitely many positive integers n such that p n 1 q n 1 Z. (This is a special case of the Theorem of Corvaja and Zannier on quotients of linear recurrence sequences). 23

24 Let h be a positive integer. Then (q hn 1) pn ( h 1 1 q n 1 = (pn 1) q ). in i=0 Hence q hnpn h 1 1 q n 1 + i=0 q in h 1 i=0 p n q in = pn 1 q n 1, or x 1 + x x 2h+1 = pn 1 q n 1 where x 1 = q hn p n 1 q n 1, x i = q (i 2)n (i = 2,..., h + 1), x i = p n q (i h 2)n (i = h + 2,...,2h + 1). 24

25 Put x n := (x 1, x 2,..., x 2h+1 ) = (q hn p n 1 q n 1,1,..., q(h 1)n, p n,..., p n q (h 1)n ). LEMMA Let q h+1 > p. Then there is δ > 0 such that for all sufficiently large n with pn 1 q n 1 Z we have (x x 2h+1 )x 2 x 2h+1 x 1 x 2h+1 p x 1 x 2h+1 q x n δ. Hence {x n : pn 1 q n 1 Z} is contained in a finite union of proper linear subspaces of Q 2h+1. 25

26 Proof. Suppose z n := pn 1 q n 1 Z. Recall x n = (x 1, x 2,..., x 2h+1 ) Hence = (q hn p n 1 q n 1,1,..., q(h 1)n, x n = q hn p n 1 q n 1 (pqh 1 ) n. Further, x x 2h+1 = pn 1 q n 1 ; x 1 p = q hn z n p = 1; x 1 q = q hn z n q q hn ; p n,..., p n q (h 1)n ). x i x i p x i q = 1 for i = 2,...,2h + 1 and so the product of these terms is at most q hnpn 1 q n 1 (qh+1 /p) n x n δ where δ = log qh+1 /p pq h 1. QED. 26

27 The set {x n : pn 1 q n 1 Z} is contained in the union of finitely many proper linear subspaces of Q 2h+1. It suffices to show that if T is any proper linear subspace of Q 2h+1, then there are only finitely many n such that x n T. 27

28 W.l.o.g. T is given by an equation a 1 x a 2h+1 x 2h+1 = 0 with a i Q. Substitute x n = (q hn p n 1 q n 1,1,..., q(h 1)n, and multiply with q n 1. p n,..., p n q (h 1)n ) Then we obtain an equation r i=1 c i α n i = 0 where each α i is an integer composed of p and q and each c i is a constant. The left-hand side is a non-degenerate linear recurrence sequence. So by the Skolem-Mahler-Lech Theorem (or Rolle s Theorem), there are only finitely many possibilities for n. QED 28

29 Two integers a, b are called multiplicatively independent if there are no positive integers m, n such that a m = b n. By extending the above argument, the following result can be proved: THEOREM (Bugeaud, Corvaja, Zannier, 2003) Let a, b be two multiplicatively independent integers. Then lim n loggcd(a n 1, b n 1) n = 0. 29

On Schmidt s Subspace Theorem. Jan-Hendrik Evertse

On Schmidt s Subspace Theorem. Jan-Hendrik Evertse On Schmidt s Subspace Theorem Jan-Hendrik Evertse Universiteit Leiden Winter school on Lang s and Vojta s conjectures March 3, 2014, CIRM, Luminy Slides can be downloaded from http://www.math.leidenuniv.nl/