Michel Waldschmidt. Linear recurrence sequences, exponential polynomials and Diophantine approximation. Institut de Mathématiques de Jussieu Paris VI

Transcription

1 8th ICSMEDC December 6, 2015 The Eighth International Conference on Science and Mathematics Education in Developing Countries Yangon University, Yangon, The Republic of the Union of Myanmar. Linear recurrence sequences, exponential polynomials and Diophantine approximation Michel Waldschmidt Institut de Mathématiques de Jussieu Paris VI

2 Abstract Linear recurrence sequences are ubiquitous. They occur in biology, economics, computer science (analysis of algorithms), digital signal processing. We give a survey of this subject, including connections with linear combinations of powers and with exponential polynomials, with an emphasis on arithmetic questions. This lecture will include new results, arising from a joint work with Claude Levesque, involving families of Diophantine equations, with explicit examples related to some units of L. Bernstein and H. Hasse.

3 Leonardo Pisano (Fibonacci) Fibonacci sequence (F n ) n 0 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, is defined by F 0 =0,F 1 =1, F n+2 = F n+1 + F n (n 0). Leonardo Pisano (Fibonacci) ( )

4 Fibonacci rabbits Fibonacci considers the growth of a rabbit population. Anewlybornpairofrabbits, one male, one female, are put in a field. Rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits ; rabbits never die and amatingpairalwaysproduces one new pair (one male, one female) every month from the second month on. The puzzle that Fibonacci posed was : how many pairs will there be in one year? Answer : F 12 =144.

5 Fibonacci squares

6 Geometric construction of the Fibonacci sequence

7 Fibonacci numbers in nature Ammonite (Nautilus shape)

8 Reflections of a ray of light Consider three parallel sheets of glass and a ray of light which crosses the first sheet. Each time it touches one of the sheets, it can cross it or reflect on it. Denote by p n the number of di erent paths with the ray going out of the system after n reflections. p 0 =1, p 1 =2, p 2 =3, p 3 =5. In general, p n = F n+2.

9 Reflections of a ray of light Consider three parallel sheets of glass and a ray of light which crosses the first sheet. Each time it touches one of the sheets, it can cross it or reflect on it. Denote by p n the number of di erent paths with the ray going out of the system after n reflections. p 0 =1, p 1 =2, p 2 =3, p 3 =5. In general, p n = F n+2.

10 Levels of energy of an electron of an atom of hydrogen An atom of hydrogen can have three levels of energy, 0 at the ground level when it does not move, 1 or 2. Ateachstep, alternatively, it gains or losses 1 or 2 levels of energy, without going below 0 nor above 2. Let`n. bethenumberofdi erent possible histories of this electron are there after n steps. We have `0 =1(initial state level 0) `1 =2:state1 or 2, histories 01 or 02. `2 =3:histories010, 021 or 020. `3 =5:histories0101, 0102, 0212, 0201 or In general, `n = F n+2.

11 Rhythmic patterns The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. Several Indian scholars, Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135), and the Jain scholar Hemachandra (c. 1150) studied rhythmic patterns that are formed from one-beat notes (or short syllables, ti in Morse Alphabet) : and two-beat notes (or long syllables, ta ta in Morse) :. 1 beat, 1 pattern : 2 beats, 2 patterns : and 3 beats, 3 patterns :, and 4 beats, 5 patterns :,,,, n beats, F n+1 patterns

12 Rhythmic patterns The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. Several Indian scholars, Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135), and the Jain scholar Hemachandra (c. 1150) studied rhythmic patterns that are formed from one-beat notes (or short syllables, ti in Morse Alphabet) : and two-beat notes (or long syllables, ta ta in Morse) :. 1 beat, 1 pattern : 2 beats, 2 patterns : and 3 beats, 3 patterns :, and 4 beats, 5 patterns :,,,, n beats, F n+1 patterns

13 Rhythmic patterns The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. Several Indian scholars, Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135), and the Jain scholar Hemachandra (c. 1150) studied rhythmic patterns that are formed from one-beat notes (or short syllables, ti in Morse Alphabet) : and two-beat notes (or long syllables, ta ta in Morse) :. 1 beat, 1 pattern : 2 beats, 2 patterns : and 3 beats, 3 patterns :, and 4 beats, 5 patterns :,,,, n beats, F n+1 patterns

14 Rhythmic patterns The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. Several Indian scholars, Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135), and the Jain scholar Hemachandra (c. 1150) studied rhythmic patterns that are formed from one-beat notes (or short syllables, ti in Morse Alphabet) : and two-beat notes (or long syllables, ta ta in Morse) :. 1 beat, 1 pattern : 2 beats, 2 patterns : and 3 beats, 3 patterns :, and 4 beats, 5 patterns :,,,, n beats, F n+1 patterns

15 Rhythmic patterns The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. Several Indian scholars, Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135), and the Jain scholar Hemachandra (c. 1150) studied rhythmic patterns that are formed from one-beat notes (or short syllables, ti in Morse Alphabet) : and two-beat notes (or long syllables, ta ta in Morse) :. 1 beat, 1 pattern : 2 beats, 2 patterns : and 3 beats, 3 patterns :, and 4 beats, 5 patterns :,,,, n beats, F n+1 patterns

16 Rhythmic patterns The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. Several Indian scholars, Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135), and the Jain scholar Hemachandra (c. 1150) studied rhythmic patterns that are formed from one-beat notes (or short syllables, ti in Morse Alphabet) : and two-beat notes (or long syllables, ta ta in Morse) :. 1 beat, 1 pattern : 2 beats, 2 patterns : and 3 beats, 3 patterns :, and 4 beats, 5 patterns :,,,, n beats, F n+1 patterns

17 Fibonacci sequence and the Golden ratio For n 0, thefibonacci number F n is the nearest integer to 1 p 5 n, where is the Golden Ratio : = 1+p 5 2 which satisfies = lim n!1 F n+1 F n = =1+ 1

18 Binet s formula For n 0, F n = n ( ) n p 5 Jacques Philippe Marie Binet (1843) = (1 + p 5) n (1 2 np 5 p 5) n, = 1+p 5, 2 1 = 1 p 5, 2 X 2 X 1=(X )(X + 1 ).

19 The so called Binet Formula Formula of A. De Moivre (1718, 1730), Daniel Bernoulli (1726), L. Euler (1728, 1765), J.P.M. Binet (1843) : for n 0, F n = p 1 1+ p! n p p! n 5. Abraham de Moivre ( ) Daniel Bernoulli ( ) Leonhard Euler ( ) Jacques P.M. Binet ( )

20 Generating series AsingleseriesencodesalltheFibonacci sequence : X F n X n = X + X 2 +2X 3 +3X 4 +5X F n X n + n 0 Fact : this series is the Taylor expansion of a rational fraction : X F n X n X = 1 X X 2 n 0 Proof :theproduct (X + X 2 +2X 3 +3X 4 +5X 5 +8X 6 + )(1 X X 2 ) is a telescoping series X + X 2 +2X 3 +3X 4 +5X 5 +8X 6 + = X. X 2 X 3 2X 4 3X 5 5X 6 X 3 X 4 2X 5 3X 6

21 Generating series AsingleseriesencodesalltheFibonacci sequence : X F n X n = X + X 2 +2X 3 +3X 4 +5X F n X n + n 0 Fact : this series is the Taylor expansion of a rational fraction : X F n X n X = 1 X X 2 n 0 Proof :theproduct (X + X 2 +2X 3 +3X 4 +5X 5 +8X 6 + )(1 X X 2 ) is a telescoping series X + X 2 +2X 3 +3X 4 +5X 5 +8X 6 + = X. X 2 X 3 2X 4 3X 5 5X 6 X 3 X 4 2X 5 3X 6

22 Generating series AsingleseriesencodesalltheFibonacci sequence : X F n X n = X + X 2 +2X 3 +3X 4 +5X F n X n + n 0 Fact : this series is the Taylor expansion of a rational fraction : X F n X n X = 1 X X 2 n 0 Proof :theproduct (X + X 2 +2X 3 +3X 4 +5X 5 +8X 6 + )(1 X X 2 ) is a telescoping series X + X 2 +2X 3 +3X 4 +5X 5 +8X 6 + = X. X 2 X 3 2X 4 3X 5 5X 6 X 3 X 4 2X 5 3X 6

23 Generating series of the Fibonacci sequence Remark. The denominator 1 X X 2 in the right hand side of X + X 2 +2X 3 +3X F n X n + = X 1 X X 2 is X 2 f(x 1 ),wheref(x) =X 2 X 1 is the irreducible polynomial of the Golden ratio.

24 Fibonacci and powers of matrices The Fibonacci linear recurrence relation F n+2 = F n+1 + F n for n 0 can be written Fn Fn =. F n F n+1 By induction one deduces, for n 0, n Fn = F n+1 An equivalent formula is, for n 1, n 0 1 Fn = 1 F n. 1 1 F n F n+1

25 Fibonacci and powers of matrices The Fibonacci linear recurrence relation F n+2 = F n+1 + F n for n 0 can be written Fn Fn =. F n F n+1 By induction one deduces, for n 0, n Fn = F n+1 An equivalent formula is, for n 1, n 0 1 Fn = 1 F n. 1 1 F n F n+1

26 Fibonacci and powers of matrices The Fibonacci linear recurrence relation F n+2 = F n+1 + F n for n 0 can be written Fn Fn =. F n F n+1 By induction one deduces, for n 0, n Fn = F n+1 An equivalent formula is, for n 1, n 0 1 Fn = 1 F n. 1 1 F n F n+1

27 Characteristic polynomial The characteristic polynomial of the matrix 0 1 A = 1 1 is det(xi X 1 A) =det = X 2 X 1, 1 X 1 which is the irreducible polynomial of the Golden ratio.

28 Fibonacci sequence and the Golden ratio (continued) For n 1, n 2 Z[ ] = Z + Z is a linear combination of 1 and with integer coe cients, namely n = F n 1 + F n

29 Fibonacci sequence and the Golden ratio (continued) For n 1, n 2 Z[ ] = Z + Z is a linear combination of 1 and with integer coe cients, namely n = F n 1 + F n

30 Fibonacci sequence and Hilbert s 10thproblem Yuri Matiyasevich (1970) showed that there is a polynomial P in n, m, andanumberofothervariablesx, y, z,... having the property that n = F 2m i there exist integers x, y, z,... such that P (n, m, x, y, z,... )=0. This completed the proof of the impossibility of the tenth of Hilbert s problems (does there exist a general method for solving Diophantine equations?) thankstothe previous work of Hilary Putnam, Julia Robinson and Martin Davis.

31 The Fibonacci Quarterly The Fibonacci sequence satisfies a lot of very interesting properties. Four times a year, the Fibonacci Quarterly publishes an issue with new properties which have been discovered.

32 Lucas sequence The Lucas sequence (L n ) n 0 satisfies the same recurrence relation as the Fibonacci sequence, namely L n+2 = L n+1 + L n (n 0), only the initial values are di erent : L 0 =2,L 1 =1. The sequence of Lucas numbers starts with 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, AclosedforminvolvingtheGoldenratio is L n = n +( ) n, from which it follows that for n to n. 2, L n is the nearest integer

33 Lucas sequence The Lucas sequence (L n ) n 0 satisfies the same recurrence relation as the Fibonacci sequence, namely L n+2 = L n+1 + L n (n 0), only the initial values are di erent : L 0 =2,L 1 =1. The sequence of Lucas numbers starts with 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, AclosedforminvolvingtheGoldenratio is L n = n +( ) n, from which it follows that for n to n. 2, L n is the nearest integer

34 François Édouard Anatole Lucas ( ) Edouard Lucas is best known for his results in number theory. He studied the Fibonacci sequence and devised the test for Mersenne primes still used today. Mathematicians/Lucas.html

35 Generating series of the Lucas sequence The generating series of the Lucas sequence X L n X n =2+X +3X 2 +4X L n X n + n 0 is nothing else than 2 X 1 X X 2

36 The Lucas sequence and power of matrices From the linear recurrence relation L n+2 = L n+1 + L n one deduces, (as we did for the Fibonacci sequence), for n 0, Ln Ln =, 1 1 L n+2 L n+1 hence Ln L n+1 = n 2. 1 Any one of the three sequences (F n ) n 0, (L n ) n 0, ( n ) n 0 can be written as a linear combination of the two others.

37 The Lucas sequence and power of matrices From the linear recurrence relation L n+2 = L n+1 + L n one deduces, (as we did for the Fibonacci sequence), for n 0, Ln Ln =, 1 1 L n+2 L n+1 hence Ln L n+1 = n 2. 1 Any one of the three sequences (F n ) n 0, (L n ) n 0, ( n ) n 0 can be written as a linear combination of the two others.

38 Perrin sequence The Perrin sequence (also called skiponacci sequence) is the linear recurrence sequence defined by with the initial conditions P n+3 = P n+1 + P n for n 0, P 0 =3,P 1 =0,P 2 =2. It starts with 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, François Olivier Raoul Perrin :

39 Perrin sequence The Perrin sequence (also called skiponacci sequence) is the linear recurrence sequence defined by with the initial conditions P n+3 = P n+1 + P n for n 0, P 0 =3,P 1 =0,P 2 =2. It starts with 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, François Olivier Raoul Perrin :

40 Plastic (or silver) constant The ratio of successive terms in the Perrin sequence approaches the plastic number %, whichistheminimal Pisot Vijayaraghavan number, real root of x 3 x 1 which has a value of approximately This constant is equal to % = p p p p 69 6

41 Plastic (or silver) constant The ratio of successive terms in the Perrin sequence approaches the plastic number %, whichistheminimal Pisot Vijayaraghavan number, real root of x 3 x 1 which has a value of approximately This constant is equal to % = p p p p 69 6

42 Perrin sequence and the plastic constant Decompose the polynomial X 3 X 1 into irreducible factors over C and over R X 3 X 1=(X %)(X )(X ). X 3 X 1=(X %)(X 2 + %X + % 1 ). Hence and are the roots of X 2 + %X + % 1.Then,for n 0, P n = % n + n + n, It follows that, for n 0, P n is the nearest integer to % n

43 Perrin sequence and the plastic constant Decompose the polynomial X 3 X 1 into irreducible factors over C and over R X 3 X 1=(X %)(X )(X ). X 3 X 1=(X %)(X 2 + %X + % 1 ). Hence and are the roots of X 2 + %X + % 1.Then,for n 0, P n = % n + n + n, It follows that, for n 0, P n is the nearest integer to % n

44 Perrin sequence and the plastic constant Decompose the polynomial X 3 X 1 into irreducible factors over C and over R X 3 X 1=(X %)(X )(X ). X 3 X 1=(X %)(X 2 + %X + % 1 ). Hence and are the roots of X 2 + %X + % 1.Then,for n 0, P n = % n + n + n, It follows that, for n 0, P n is the nearest integer to % n

45 Perrin sequence and the plastic constant Decompose the polynomial X 3 X 1 into irreducible factors over C and over R X 3 X 1=(X %)(X )(X ). X 3 X 1=(X %)(X 2 + %X + % 1 ). Hence and are the roots of X 2 + %X + % 1.Then,for n 0, P n = % n + n + n, It follows that, for n 0, P n is the nearest integer to % n

46 Generating series of the Perrin sequence The generating series of the Perrin sequence X P n X n =3+2X 2 +3X 3 +2X P n X n + n 0 is nothing else than 3 X 2 1 X 2 X 3 The denominator 1 X 2 X 3 is X 3 f(x 1 ) where f(x) =X 3 X 1 is the irreducible polynomial of %.

47 Generating series of the Perrin sequence The generating series of the Perrin sequence X P n X n =3+2X 2 +3X 3 +2X P n X n + n 0 is nothing else than 3 X 2 1 X 2 X 3 The denominator 1 X 2 X 3 is X 3 f(x 1 ) where f(x) =X 3 X 1 is the irreducible polynomial of %.

48 Perrin sequence and power of matrices From we deduce P n+3 = P n+1 + P n P n P P n+2 A 0 0 P n+1 A P n P n+2 Hence P n 0 1 P n+1 A 0 0 1A P n A 2 n 0

49 Perrin sequence and power of matrices From we deduce P n+3 = P n+1 + P n P n P P n+2 A 0 0 P n+1 A P n P n+2 Hence P n 0 1 P n+1 A 0 0 1A P n A 2 n 0

50 Characteristic polynomial The characteristic polynomial of the matrix A 0 0 1A is det(xi 0 X A) = 0 X 1A = X 3 X 1, 1 1 X which is the irreducible polynomial of the plastic constant %.

51 Perrin pseudoprimes If p is prime, then p divides P p. The smallest composite n such that n divides P n is For n either = or = , the number n divides P n. Jon Grantham has proved that there are infinitely many Perrin pseudoprimes. The number c of decimal digits of P satisfies 10 c = % ,hencec =271441(log%)/(log 10) The website maintained by Richard Turk is devoted to Perrin numbers. See OEISA

52 Perrin pseudoprimes If p is prime, then p divides P p. The smallest composite n such that n divides P n is For n either = or = , the number n divides P n. Jon Grantham has proved that there are infinitely many Perrin pseudoprimes. The number c of decimal digits of P satisfies 10 c = % ,hencec =271441(log%)/(log 10) The website maintained by Richard Turk is devoted to Perrin numbers. See OEISA

53 Perrin pseudoprimes If p is prime, then p divides P p. The smallest composite n such that n divides P n is For n either = or = , the number n divides P n. Jon Grantham has proved that there are infinitely many Perrin pseudoprimes. The number c of decimal digits of P satisfies 10 c = % ,hencec =271441(log%)/(log 10) The website maintained by Richard Turk is devoted to Perrin numbers. See OEISA

54 Perrin pseudoprimes If p is prime, then p divides P p. The smallest composite n such that n divides P n is For n either = or = , the number n divides P n. Jon Grantham has proved that there are infinitely many Perrin pseudoprimes. The number c of decimal digits of P satisfies 10 c = % ,hencec =271441(log%)/(log 10) The website maintained by Richard Turk is devoted to Perrin numbers. See OEISA

55 Padovan sequence The Padovan sequence p n satisfies the same recurrence p n+3 = p n+1 + p n as the Perrin sequence but has di erent initial values : p 0 =1, p 1 = p 2 =0. It starts with 1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, Richard Padovan

56 Generating series and power of matrices 1+X 3 + X p n X n + = 1 X 2 1 X 2 X 3 For n 0, p n 0 1 p n+1 A 0 0 1A p n A 0 n 0

57 Generating series and power of matrices 1+X 3 + X p n X n + = 1 X 2 1 X 2 X 3 For n 0, p n 0 1 p n+1 A 0 0 1A p n A 0 n 0

58 Padovan triangles p n = p n 2 + p n 3 p n 1 = p n 3 + p n 4 Hence p n 2 = p n 4 + p n 5 p n p n 1 = p n 5 p n = p n 1 + p n 5

59 Padovan triangles

60 Padovan triangles vs Fibonacci squares

61 Narayana sequence Narayana sequence is defined by the recurrence relation C n+3 = C n+2 + C n with the initial values C 0 =2, C 1 =3, C 2 =4. It starts with 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595,... Real root of x 3 x 2 1 s p s p =

62 Narayana sequence Narayana sequence is defined by the recurrence relation C n+3 = C n+2 + C n with the initial values C 0 =2, C 1 =3, C 2 =4. It starts with 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595,... Real root of x 3 x 2 1 s p s p =

63 Generating series and power of matrices 2+3X +4X 2 +6X C n X n + = 2+X + X2 1 X X 3 For n 0, C n 0 1 C n+1 A 0 0 1A C n A 4 n 0

64 Generating series and power of matrices 2+3X +4X 2 +6X C n X n + = 2+X + X2 1 X X 3 For n 0, C n 0 1 C n+1 A 0 0 1A C n A 4 n 0

65 Narayana s cows Narayana was an Indian mathematician in the 14th century who proposed the following problem : Acowproducesonecalfeveryyear.Beginninginitsfourth year each calf produces one calf at the beginning of each year. How many calves are there altogether after, for example, 17 years?

66 Music : In working this out, Tom Johnson found a way to translate this into a composition called Narayana s Cows. Music : Tom Johnson Saxophones : Daniel Kientzy

67 Narayana s cows Year Original Cow Second generation Third generation Fourth generation Fifth generation Sixth generation Total

68 Jean-Paul Allouche and Tom Johnson bibliorecente.html

69 Cows, music and morphisms Jean-Paul Allouche and Tom Johnson Narayana s Cows and Delayed Morphisms In 3èmes Journées d Informatique Musicale (JIM 96), Ile de Tatihou, Les Cahiers du GREYC (1996 no. 4), pages 2-7, May Finite automata and morphisms in assisted musical composition, Journal of New Music Research, no. 24 (1995), nl/jnmr/vol24_2.html

70 Linear recurrence sequences : definitions A linear recurrence sequence is a sequence of numbers u =(u 0,u 1,u 2,...) for which there exist a positive integer d together with numbers a 1,...,a d with a d 6=0such that, for n 0, (?) u n+d = a 1 u n+d a d u n. Here, a number means an element of a field K of zero characteristic. Given a =(a 1,...,a d ) 2 K d,thesete a of linear recurrence sequences u =(u n ) n 0 satisfying (?) isak vector subspace of dimension d of the space K N of all sequences. The characteristic (or companion) polynomial of the linear recurrence is f(x) =X d a 1 X d 1 a d.

71 Linear recurrence sequences : definitions A linear recurrence sequence is a sequence of numbers u =(u 0,u 1,u 2,...) for which there exist a positive integer d together with numbers a 1,...,a d with a d 6=0such that, for n 0, (?) u n+d = a 1 u n+d a d u n. Here, a number means an element of a field K of zero characteristic. Given a =(a 1,...,a d ) 2 K d,thesete a of linear recurrence sequences u =(u n ) n 0 satisfying (?) isak vector subspace of dimension d of the space K N of all sequences. The characteristic (or companion) polynomial of the linear recurrence is f(x) =X d a 1 X d 1 a d.

72 Linear recurrence sequences : definitions A linear recurrence sequence is a sequence of numbers u =(u 0,u 1,u 2,...) for which there exist a positive integer d together with numbers a 1,...,a d with a d 6=0such that, for n 0, (?) u n+d = a 1 u n+d a d u n. Here, a number means an element of a field K of zero characteristic. Given a =(a 1,...,a d ) 2 K d,thesete a of linear recurrence sequences u =(u n ) n 0 satisfying (?) isak vector subspace of dimension d of the space K N of all sequences. The characteristic (or companion) polynomial of the linear recurrence is f(x) =X d a 1 X d 1 a d.

73 Linear recurrence sequences : definitions A linear recurrence sequence is a sequence of numbers u =(u 0,u 1,u 2,...) for which there exist a positive integer d together with numbers a 1,...,a d with a d 6=0such that, for n 0, (?) u n+d = a 1 u n+d a d u n. Here, a number means an element of a field K of zero characteristic. Given a =(a 1,...,a d ) 2 K d,thesete a of linear recurrence sequences u =(u n ) n 0 satisfying (?) isak vector subspace of dimension d of the space K N of all sequences. The characteristic (or companion) polynomial of the linear recurrence is f(x) =X d a 1 X d 1 a d.

74 Linear recurrence sequence : examples Constant sequence : u n = u 0. Linear recurrence sequence of order 1 : u n+1 = u n. Characteristic polynomial : f(x) =X 1. Generating series : X n 0 X n = 1 1 X Geometric progression : u n = u n 0. Linear recurrence sequence of order 1 : u n = u n 1. Characteristic polynomial f(x) =X. Generating series : X n 0 u 0 n X n = u 0 1 X

75 Linear recurrence sequence : examples Constant sequence : u n = u 0. Linear recurrence sequence of order 1 : u n+1 = u n. Characteristic polynomial : f(x) =X 1. Generating series : X n 0 X n = 1 1 X Geometric progression : u n = u n 0. Linear recurrence sequence of order 1 : u n = u n 1. Characteristic polynomial f(x) =X. Generating series : X n 0 u 0 n X n = u 0 1 X

76 Linear recurrence sequence : examples u n = n. Linearrecurrencesequenceoforder2 : Characteristic polynomial n +2=2(n +1) n. f(x) =X 2 2X +1=(X 1) 2. Generating series X nx n = n X + X 2 Power of matrices : n 0 1 = 1 2 n +1 n. n n+1

77 Linear recurrence sequence : examples u n = f(n), f polynomial of degree d. Linearrecurrence sequence of order d +1. Proof. The sequences (f(n)) n 0, (f(n +1)) n 0,, (f(n + k)) n 0 are K linearly independent in K N for k = d dependent for k = d. 1 and linearly

78 Linear recurrence sequence : examples u n = f(n), f polynomial of degree d. Linearrecurrence sequence of order d +1. Proof. The sequences (f(n)) n 0, (f(n +1)) n 0,, (f(n + k)) n 0 are K linearly independent in K N for k = d dependent for k = d. 1 and linearly

79 Linear sequences which are ultimately recurrent The sequence 1, 0, 0,... is not a linear recurrence sequence. The condition is satisfied only for n 1. u n+1 = u n The relation u n+2 = u n+1 +0u n with d =2, a d =0does not fulfill the requirement a d 6=0.

80 Linear sequences which are ultimately recurrent The sequence 1, 0, 0,... is not a linear recurrence sequence. The condition is satisfied only for n 1. u n+1 = u n The relation u n+2 = u n+1 +0u n with d =2, a d =0does not fulfill the requirement a d 6=0.

81 Linear sequences which are ultimately recurrent The sequence 1, 0, 0,... is not a linear recurrence sequence. The condition is satisfied only for n 1. u n+1 = u n The relation u n+2 = u n+1 +0u n with d =2, a d =0does not fulfill the requirement a d 6=0.

82 Order of a linear recurrence sequence If u =(u n ) n 0 satisfies the linear recurrence, the characteristic polynomial of which is f, then,foranymonicpolynomial g 2 K[X], thissequenceualso satisfies the linear recurrence, the characteristic polynomial of which is fg. Example :forg(x) =X,from (?) u n+d a 1 u n+d 1 a d u n =0 we deduce u n+d+1 a 1 u n+d a d u n+1 + (u n+d a 1 u n+d 1 a d u n )=0. The order of a linear recurrence sequence is the smallest d such that (?) holdsforalln 0.

83 Order of a linear recurrence sequence If u =(u n ) n 0 satisfies the linear recurrence, the characteristic polynomial of which is f, then,foranymonicpolynomial g 2 K[X], thissequenceualso satisfies the linear recurrence, the characteristic polynomial of which is fg. Example :forg(x) =X,from (?) u n+d a 1 u n+d 1 a d u n =0 we deduce u n+d+1 a 1 u n+d a d u n+1 + (u n+d a 1 u n+d 1 a d u n )=0. The order of a linear recurrence sequence is the smallest d such that (?) holdsforalln 0.

84 Polynomial combinations of powers The sum of any two linear recurrence sequences is a linear recurrence sequence. The set [ a E a of all linear recurrence sequences with coe cients in K is a sub K algebra of K N. Given polynomials p 1,...,p` in K[X] and elements in K,thesequence 1,..., ` is a linear recurrence sequence. p 1 (n) n p`(n) ǹ n 0 Fact : any linear recurrence sequence is of this form.

85 Polynomial combinations of powers The sum of any two linear recurrence sequences is a linear recurrence sequence. The set [ a E a of all linear recurrence sequences with coe cients in K is a sub K algebra of K N. Given polynomials p 1,...,p` in K[X] and elements in K,thesequence 1,..., ` is a linear recurrence sequence. p 1 (n) n p`(n) ǹ n 0 Fact : any linear recurrence sequence is of this form.

86 Polynomial combinations of powers The sum of any two linear recurrence sequences is a linear recurrence sequence. The set [ a E a of all linear recurrence sequences with coe cients in K is a sub K algebra of K N. Given polynomials p 1,...,p` in K[X] and elements in K,thesequence 1,..., ` is a linear recurrence sequence. p 1 (n) n p`(n) ǹ n 0 Fact : any linear recurrence sequence is of this form.

87 Polynomial combinations of powers The sum of any two linear recurrence sequences is a linear recurrence sequence. The set [ a E a of all linear recurrence sequences with coe cients in K is a sub K algebra of K N. Given polynomials p 1,...,p` in K[X] and elements in K,thesequence 1,..., ` is a linear recurrence sequence. p 1 (n) n p`(n) ǹ n 0 Fact : any linear recurrence sequence is of this form.

88 Linear recurrence sequence and Brahmagupta Pell Fermat Equation Let d be a positive integer, not a square. The solutions (x, y) 2 Z Z of the Brahmagupta Pell Fermat Equation x 2 dy 2 = ±1 form a sequence (x n,y n ) n2z defined by From x n + p dy n =(x 1 + p dy 1 ) n. 2x n =(x 1 + p dy 1 ) n +(x 1 p dy1 ) n we deduce that (x n ) n 0 is a linear recurrence sequence. Same for y n,andalsoforn 0.

89 Linear recurrence sequence and Brahmagupta Pell Fermat Equation Let d be a positive integer, not a square. The solutions (x, y) 2 Z Z of the Brahmagupta Pell Fermat Equation x 2 dy 2 = ±1 form a sequence (x n,y n ) n2z defined by From x n + p dy n =(x 1 + p dy 1 ) n. 2x n =(x 1 + p dy 1 ) n +(x 1 p dy1 ) n we deduce that (x n ) n 0 is a linear recurrence sequence. Same for y n,andalsoforn 0.

90 Doubly infinite linear recurrence sequence Asequence(u n ) n2z indexed by Z is a linear recurrence sequence if it satisfies (?) u n+d = a 1 u n+d a d u n. for all n 2 Z. Recall a d 6=0. Such a sequence is determined by d consecutive values.

91 Doubly infinite linear recurrence sequence Asequence(u n ) n2z indexed by Z is a linear recurrence sequence if it satisfies (?) u n+d = a 1 u n+d a d u n. for all n 2 Z. Recall a d 6=0. Such a sequence is determined by d consecutive values.

92 Doubly infinite linear recurrence sequence Asequence(u n ) n2z indexed by Z is a linear recurrence sequence if it satisfies (?) u n+d = a 1 u n+d a d u n. for all n 2 Z. Recall a d 6=0. Such a sequence is determined by d consecutive values.

93 Linear recurrence sequence : simple roots AbasisofE a over K is obtained by attributing to the initial values u 0,...,u d 1 the values given by the canonical basis of K d. Given in K,anecessaryandsu cientconditionfora sequence ( n ) n 0 to satisfy (?) isthat is a root of the characteristic polynomial f(x) =X d a 1 X d 1 a d. If this polynomial has d distinct roots 1,..., d in K, f(x) =(X 1) (X d), i 6= j, then a basis of E a over K is given by the d sequences ( n i ) n 0, i =1,...,d.

94 Linear recurrence sequence : simple roots AbasisofE a over K is obtained by attributing to the initial values u 0,...,u d 1 the values given by the canonical basis of K d. Given in K,anecessaryandsu cientconditionfora sequence ( n ) n 0 to satisfy (?) isthat is a root of the characteristic polynomial f(x) =X d a 1 X d 1 a d. If this polynomial has d distinct roots 1,..., d in K, f(x) =(X 1) (X d), i 6= j, then a basis of E a over K is given by the d sequences ( n i ) n 0, i =1,...,d.

95 Linear recurrence sequence : simple roots AbasisofE a over K is obtained by attributing to the initial values u 0,...,u d 1 the values given by the canonical basis of K d. Given in K,anecessaryandsu cientconditionfora sequence ( n ) n 0 to satisfy (?) isthat is a root of the characteristic polynomial f(x) =X d a 1 X d 1 a d. If this polynomial has d distinct roots 1,..., d in K, f(x) =(X 1) (X d), i 6= j, then a basis of E a over K is given by the d sequences ( n i ) n 0, i =1,...,d.

96 Linear recurrence sequence : double roots The characteristic polynomial of the linear recurrence u n =2 u n 1 2 u n 2 is X 2 2 X + 2 =(X ) 2 with a double root. The sequence (n n ) n 0 satisfies n n =2 (n 1)n n 1 2 (n 2) n 2. AbasisofE a for a 1 =2, a 2 = sequences ( n ) n 0, (n n ) n 0. 2 is given by the two Given 2 K,anecessaryandsu cientconditionforthe sequence n n to satisfy the linear recurrence relation (?) is that is a root of multiplicity 2 of f(x).

97 Linear recurrence sequence : double roots The characteristic polynomial of the linear recurrence u n =2 u n 1 2 u n 2 is X 2 2 X + 2 =(X ) 2 with a double root. The sequence (n n ) n 0 satisfies n n =2 (n 1)n n 1 2 (n 2) n 2. AbasisofE a for a 1 =2, a 2 = sequences ( n ) n 0, (n n ) n 0. 2 is given by the two Given 2 K,anecessaryandsu cientconditionforthe sequence n n to satisfy the linear recurrence relation (?) is that is a root of multiplicity 2 of f(x).

98 Linear recurrence sequence : double roots The characteristic polynomial of the linear recurrence u n =2 u n 1 2 u n 2 is X 2 2 X + 2 =(X ) 2 with a double root. The sequence (n n ) n 0 satisfies n n =2 (n 1)n n 1 2 (n 2) n 2. AbasisofE a for a 1 =2, a 2 = sequences ( n ) n 0, (n n ) n 0. 2 is given by the two Given 2 K,anecessaryandsu cientconditionforthe sequence n n to satisfy the linear recurrence relation (?) is that is a root of multiplicity 2 of f(x).

99 Linear recurrence sequence : double roots The characteristic polynomial of the linear recurrence u n =2 u n 1 2 u n 2 is X 2 2 X + 2 =(X ) 2 with a double root. The sequence (n n ) n 0 satisfies n n =2 (n 1)n n 1 2 (n 2) n 2. AbasisofE a for a 1 =2, a 2 = sequences ( n ) n 0, (n n ) n 0. 2 is given by the two Given 2 K,anecessaryandsu cientconditionforthe sequence n n to satisfy the linear recurrence relation (?) is that is a root of multiplicity 2 of f(x).

100 Linear recurrence sequence : multiple roots In general, when the characteristic polynomial splits as Ỳ X d a 1 X d 1 a d = (X i) t i, i=1 abasisofe a is given by the d sequences (n k i n ) n 0, 0 apple k apple t i 1, 1 apple i apple `.

101 Generating series of a linear recurrence sequence Let u =(u n ) n 0 be a linear recurrence sequence (?) u n+d = a 1 u n+d a d u n for n 0 with characteristic polynomial f(x) =X d a 1 X d 1 a d. Denote by f the reciprocal polynomial of f : Then f (X) =X d f(x 1 )=1 a 1 X a 0 X d. 1X n=0 u n X n = r(x) f (X), where r is a polynomial of degree less than d determined by the initial values of u.

102 Generating series of a linear recurrence sequence Assume u n+d = a 1 u n+d a d u n for n 0. Then 1X n=0 u n X n = r(x) f (X) Proof. Comparingthecoe cientsofx n for n d shows that 1X f (X) u n X n n=0 is a polynomial of degree less than d

103 Generating series of a linear recurrence sequence Assume u n+d = a 1 u n+d a d u n for n 0. Then 1X n=0 u n X n = r(x) f (X) Proof. Comparingthecoe cientsofx n for n d shows that 1X f (X) u n X n n=0 is a polynomial of degree less than d

104 Taylor coe cients of rational functions Conversely, the coe cients the Taylor expansion of any rational fraction a(x)/b(x) with deg a<deg b satisfies the recurrence relation with characteristic polynomial f 2 K[X] given by f(x) =b (X). Therefore a sequence u =(u n ) n 0 satisfies the recurrence relation (?) withcharacteristicpolynomialf 2 K[X] if and only if 1X u n X n = r(x) f (X), n=0 where r is a polynomial of degree less than d determined by the initial values of u.

105 Taylor coe cients of rational functions Conversely, the coe cients the Taylor expansion of any rational fraction a(x)/b(x) with deg a<deg b satisfies the recurrence relation with characteristic polynomial f 2 K[X] given by f(x) =b (X). Therefore a sequence u =(u n ) n 0 satisfies the recurrence relation (?) withcharacteristicpolynomialf 2 K[X] if and only if 1X u n X n = r(x) f (X), n=0 where r is a polynomial of degree less than d determined by the initial values of u.

106 Matrix notation for a linear recurrence sequence The linear recurrence sequence (?) u n+d = a 1 u n+d a d u n for n 0 can be written u n+1 u n B A =. B A u n+d a d a d 1 a d 2 a 1 u n u n+1. u n+d 1 1 C A.

107 Matrix notation for a linear recurrence sequence with 0 U n = u n u n+1. u n+d 1 The determinant of I d X A) isnothingbut U n+1 = AU n C A, A =. B A a d a d 1 a d 2 a 1 A (the characteristic polynomial of f(x) =X d a 1 X d 1 a d, the characteristic polynomial of the linear recurrence sequence. By induction U n = A n U 0.

108 Matrix notation for a linear recurrence sequence with 0 U n = u n u n+1. u n+d 1 The determinant of I d X A) isnothingbut U n+1 = AU n C A, A =. B A a d a d 1 a d 2 a 1 A (the characteristic polynomial of f(x) =X d a 1 X d 1 a d, the characteristic polynomial of the linear recurrence sequence. By induction U n = A n U 0.

109 Matrix notation for a linear recurrence sequence with 0 U n = u n u n+1. u n+d 1 The determinant of I d X A) isnothingbut U n+1 = AU n C A, A =. B A a d a d 1 a d 2 a 1 A (the characteristic polynomial of f(x) =X d a 1 X d 1 a d, the characteristic polynomial of the linear recurrence sequence. By induction U n = A n U 0.

110 Powers of matrices Let A =(a ij ) 1applei,jappled 2 GL d d (K) be a d d matrix with coe cients in K and nonzero determinant. For n 0, define A n = a ij (n) 1applei,jappled. Then each of the d 2 sequences a ij (n) n 0, (1 apple i, j apple d) is a linear recurrence sequence. The roots of the characteristic polynomial of these linear recurrences are the eigenvalues of A. In particular the sequence Tr(A n ) n 0 satisfies the linear recurrence, the characteristic polynomial of which is the characteristic polynomial of the matrix A.

111 Powers of matrices Let A =(a ij ) 1applei,jappled 2 GL d d (K) be a d d matrix with coe cients in K and nonzero determinant. For n 0, define A n = a ij (n) 1applei,jappled. Then each of the d 2 sequences a ij (n) n 0, (1 apple i, j apple d) is a linear recurrence sequence. The roots of the characteristic polynomial of these linear recurrences are the eigenvalues of A. In particular the sequence Tr(A n ) n 0 satisfies the linear recurrence, the characteristic polynomial of which is the characteristic polynomial of the matrix A.

112 Powers of matrices Let A =(a ij ) 1applei,jappled 2 GL d d (K) be a d d matrix with coe cients in K and nonzero determinant. For n 0, define A n = a ij (n) 1applei,jappled. Then each of the d 2 sequences a ij (n) n 0, (1 apple i, j apple d) is a linear recurrence sequence. The roots of the characteristic polynomial of these linear recurrences are the eigenvalues of A. In particular the sequence Tr(A n ) n 0 satisfies the linear recurrence, the characteristic polynomial of which is the characteristic polynomial of the matrix A.

113 Conversely : Given a linear recurrence sequence u 2 K N,thereexistan integer d 1 and a matrix A 2 GL d (K) such that, for each n 0, u n = a 11 (n). The characteristic polynomial of A is the characteristic polynomial of the linear recurrence sequence. Everest G., van der Poorten A., Shparlinski I., Ward T. Recurrence sequences, Mathematical Surveys and Monographs (AMS, 2003), volume 104.

114 Conversely : Given a linear recurrence sequence u 2 K N,thereexistan integer d 1 and a matrix A 2 GL d (K) such that, for each n 0, u n = a 11 (n). The characteristic polynomial of A is the characteristic polynomial of the linear recurrence sequence. Everest G., van der Poorten A., Shparlinski I., Ward T. Recurrence sequences, Mathematical Surveys and Monographs (AMS, 2003), volume 104.

115 Discrete version of linear di erential equations Asequenceu 2 K N can be viewed as a linear map N! K. Define the discrete derivative D by Du : N! K n 7! u n+1 u n. Asequenceu 2 K N is a linear recurrence sequence if and only if there exists Q 2 K[T ] with Q(1) 6= 1such that Q(D)u =0. Linear recurrence sequences are a discrete version of linear di erential equations with constant coe cients. The condition Q(1) 6= 0reflects a d 6=0 otherwise one gets ultimately recurrent sequences.

116 Joint work with Claude Levesque Linear recurrence sequences and twisted binary forms. Proceedings of the International Conference on Pure and Applied Mathematics ICPAM-GOROKA South Pacific Journal of Pure and Applied Mathematics. pdf/procconfpng2014.pdf

117 Families of binary forms Consider a binary form F 0 (X, Y ) 2 C[X, Y ] which satisfies F 0 (1, 0) = 1. Wewriteitas F 0 (X, Y )=X d + a 1 X d 1 Y + + a d Y d = dy (X i Y ). Let 1,..., d be d nonzero complex numbers not necessarily distinct. Twisting F 0 by the powers n 1,..., n d (n 2 Z), we obtain the family of binary forms dy F n (X, Y )= (X i n i Y ), which we write as Therefore i=1 i=1 X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d. U h (0) = ( 1) h a h (1 apple h apple d).

118 Families of binary forms Consider a binary form F 0 (X, Y ) 2 C[X, Y ] which satisfies F 0 (1, 0) = 1. Wewriteitas F 0 (X, Y )=X d + a 1 X d 1 Y + + a d Y d = dy (X i Y ). Let 1,..., d be d nonzero complex numbers not necessarily distinct. Twisting F 0 by the powers n 1,..., n d (n 2 Z), we obtain the family of binary forms dy F n (X, Y )= (X i n i Y ), which we write as Therefore i=1 i=1 X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d. U h (0) = ( 1) h a h (1 apple h apple d).

119 Families of binary forms Consider a binary form F 0 (X, Y ) 2 C[X, Y ] which satisfies F 0 (1, 0) = 1. Wewriteitas F 0 (X, Y )=X d + a 1 X d 1 Y + + a d Y d = dy (X i Y ). Let 1,..., d be d nonzero complex numbers not necessarily distinct. Twisting F 0 by the powers n 1,..., n d (n 2 Z), we obtain the family of binary forms dy F n (X, Y )= (X i n i Y ), which we write as Therefore i=1 i=1 X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d. U h (0) = ( 1) h a h (1 apple h apple d).

120 Families of binary forms Consider a binary form F 0 (X, Y ) 2 C[X, Y ] which satisfies F 0 (1, 0) = 1. Wewriteitas F 0 (X, Y )=X d + a 1 X d 1 Y + + a d Y d = dy (X i Y ). Let 1,..., d be d nonzero complex numbers not necessarily distinct. Twisting F 0 by the powers n 1,..., n d (n 2 Z), we obtain the family of binary forms dy F n (X, Y )= (X i n i Y ), which we write as Therefore i=1 i=1 X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d. U h (0) = ( 1) h a h (1 apple h apple d).

121 Families of Diophantine equations With Claude Levesque, we consider some families of diophantine equations F n (x, y) =m obtained in the same way from a given irreducible form F (X, Y ) with coe cients in Z, when 1,..., d are algebraic units and when the algebraic numbers 1 1,..., d d are Galois conjugates with d 3. Theorem. Let K be a number field of degree d 3, S afinite set of places of K containing the places at infinity. Denote by O S the ring of S integers of K and by O S the group of S units of K. Assume 1,..., d, 1,..., d belong to K Then there are only finitely many (x, y, n) in O S O S Z satisfying F n (x, y) 2O S, xy 6= 0 and Card{ 1 n 1,..., d n d} 3.

122 Families of Diophantine equations With Claude Levesque, we consider some families of diophantine equations F n (x, y) =m obtained in the same way from a given irreducible form F (X, Y ) with coe cients in Z, when 1,..., d are algebraic units and when the algebraic numbers 1 1,..., d d are Galois conjugates with d 3. Theorem. Let K be a number field of degree d 3, S afinite set of places of K containing the places at infinity. Denote by O S the ring of S integers of K and by O S the group of S units of K. Assume 1,..., d, 1,..., d belong to K Then there are only finitely many (x, y, n) in O S O S Z satisfying F n (x, y) 2O S, xy 6= 0 and Card{ 1 n 1,..., d n d} 3.

123 Families of Diophantine equations Each of the sequences U h (n) n2z coming from the coe cients of the relation F n (X, Y )=X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d is a linear recurrence sequence. For example, for n 2 Z, U 1 (n) = dx i n i, U d (n) = i=1 dy i n i. i=1 For 1 apple h apple d, thesequence combination of the sequences U h (n) n2z is a linear ( i1 ih ) n n2z, (1 apple i 1 < <i h apple d).

124 Families of Diophantine equations Each of the sequences U h (n) n2z coming from the coe cients of the relation F n (X, Y )=X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d is a linear recurrence sequence. For example, for n 2 Z, U 1 (n) = dx i n i, U d (n) = i=1 dy i n i. i=1 For 1 apple h apple d, thesequence combination of the sequences U h (n) n2z is a linear ( i1 ih ) n n2z, (1 apple i 1 < <i h apple d).

125 Families of Diophantine equations Each of the sequences U h (n) n2z coming from the coe cients of the relation F n (X, Y )=X d U 1 (n)x d 1 Y + +( 1) d U d (n)y d is a linear recurrence sequence. For example, for n 2 Z, U 1 (n) = dx i n i, U d (n) = i=1 dy i n i. i=1 For 1 apple h apple d, thesequence combination of the sequences U h (n) n2z is a linear ( i1 ih ) n n2z, (1 apple i 1 < <i h apple d).

126 Units of Bernstein and Hasse Let t and s be two positive integers, D an integer c 2 { 1, +1}. Let!>1 satisfy 1, and! st = D st + c, where it is assumed that Q(!) is of degree st. Consider = D!, = D t! t. L. Bernstein and H. Hasse noticed that and are units of degree st and s respectively, and showed that these units can be obtained from the Jacobi Perron algorithm. H.-J. Stender proved that for s = t =2, {, } is a fundamental system of units of the quartic field Q(!).

127 Units of Bernstein and Hasse Let t and s be two positive integers, D an integer c 2 { 1, +1}. Let!>1 satisfy 1, and! st = D st + c, where it is assumed that Q(!) is of degree st. Consider = D!, = D t! t. L. Bernstein and H. Hasse noticed that and are units of degree st and s respectively, and showed that these units can be obtained from the Jacobi Perron algorithm. H.-J. Stender proved that for s = t =2, {, } is a fundamental system of units of the quartic field Q(!).

128 Helmut Hasse ( ) D>0, s 1, t 1, c 2 { 1, +1},!>0,! st = D st + c, = D!, = D t! t. ( D) st =( 1) st (D st + c).

129 Diophantine equations associated with some units of Bernstein and Hasse The irreducible polynomial of is F 0 (X, 1), with F 0 (X, Y )=(X DY ) st ( 1) st (D st + c)y st. For n 2 Z, thebinaryformf n (X, Y ), obtainedbytwisting F 0 (X, Y ) with the powers n of, isthehomogeneousversion of the irreducible polynomial F n (X, 1) of n.sof n depends of the parameters n, D, s, t and c. Theorem (with Claude Levesque). Suppose st 3. There exists an e ectively computable constant apple, dependingonlyon D, s and t, withthefollowingproperty.letm, a, x, y be rational integers satisfying m 2, xy 6= 0, [Q( a ):Q] =st and F n (x, y) applem. Then max{log x, log y, n } apple apple log m.

130 Diophantine equations associated with some units of Bernstein and Hasse The irreducible polynomial of is F 0 (X, 1), with F 0 (X, Y )=(X DY ) st ( 1) st (D st + c)y st. For n 2 Z, thebinaryformf n (X, Y ), obtainedbytwisting F 0 (X, Y ) with the powers n of, isthehomogeneousversion of the irreducible polynomial F n (X, 1) of n.sof n depends of the parameters n, D, s, t and c. Theorem (with Claude Levesque). Suppose st 3. There exists an e ectively computable constant apple, dependingonlyon D, s and t, withthefollowingproperty.letm, a, x, y be rational integers satisfying m 2, xy 6= 0, [Q( a ):Q] =st and F n (x, y) applem. Then max{log x, log y, n } apple apple log m.

131 Diophantine equations associated with some units of Bernstein and Hasse The irreducible polynomial of is F 0 (X, 1), with F 0 (X, Y )=(X DY ) st ( 1) st (D st + c)y st. For n 2 Z, thebinaryformf n (X, Y ), obtainedbytwisting F 0 (X, Y ) with the powers n of, isthehomogeneousversion of the irreducible polynomial F n (X, 1) of n.sof n depends of the parameters n, D, s, t and c. Theorem (with Claude Levesque). Suppose st 3. There exists an e ectively computable constant apple, dependingonlyon D, s and t, withthefollowingproperty.letm, a, x, y be rational integers satisfying m 2, xy 6= 0, [Q( a ):Q] =st and F n (x, y) applem. Then max{log x, log y, n } apple apple log m.

132 8th ICSMEDC December 6, 2015 The Eighth International Conference on Science and Mathematics Education in Developing Countries Yangon University, Yangon, The Republic of the Union of Myanmar. Linear recurrence sequences, exponential polynomials and Diophantine approximation Michel Waldschmidt Institut de Mathématiques de Jussieu Paris VI