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Abstract December 2016 Linear recurrence sequences are ubiquitous. They occur in biology, economics, computer science (analysis of algorithms), digital signal processing. Linear recurrence sequences, exponential polynomials and Diophantine approximation We give a survey of this subject, together with connections with linear combinations of powers and with exponential polynomials, with an emphasis on arithmetic questions. Michel Waldschmidt Institut de Mathématiques de Jussieu — Paris VI 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. http://webusers.imj-prg.fr/~michel.waldschmidt/ 1 / 119 Applications of linear recurrence sequences 2 / 119 Leonardo Pisano (Fibonacci) Combinatorics Elimination Fibonacci sequence (Fn )n 0 , Symmetric functions Leonardo Pisano (Fibonacci) (1170–1250) 0, 1, 1, 2, 3, 5, 8, 13, 21, Hypergeometric series 34, 55, 89, 144, 233, . . . Language is defined by Communication, shift registers F0 = 0, F1 = 1, Finite di↵erence equations Fn+2 = Fn+1 + Fn Logic for n 0. http://oeis.org/A000045 Approximation Pseudo–random sequences 3 / 119 4 / 119 Fibonacci rabbits Fibonacci squares Fibonacci considered the growth of a rabbit population. A newly born pair of rabbits, a male and a 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 a mating pair always produces 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 : F12 = 144. http://mathforum.org/dr.math/faq/faq.golden.ratio.html 5 / 119 6 / 119 Fibonacci numbers in nature Geometric construction of the Fibonacci sequence 8 5 3 2 1 1 Ammonite (Nautilus shape) 2 7 / 119 8 / 119 Reflections of a ray of light Levels of energy of an electron of an atom of hydrogen 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. An atom of hydrogen can have three levels of energy, 0 at the ground level when it does not move, 1 or 2. At each step, it alternatively gains and looses some level of energy, either 1 or 2, without going sub 0 nor above 2. Let `n be the number of di↵erent possible scenarios for this electron after n steps. Denote by pn the number of di↵erent paths with the ray going out of the system after n reflections. p0 = 1, We have `0 = 1 (initial state level 0) `1 = 2 : state 1 or 2, scenarios 01 or 02. `2 = 3 : scenarios 010, 021 or 020. `3 = 5 : scenarios 0101, 0102, 0212, 0201 or 0202. In general, `n = Fn+2 . p1 = 2, p2 = 3, p3 = 5. In general, pn = Fn+2 . 9 / 119 Rhythmic patterns 10 / 119 Fibonacci sequence and the Golden ratio 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 : For n 0, the Fibonacci number Fn is the nearest integer to 1 p 5 where n is the Golden Ratio : http://oeis.org/A001622 p 1+ 5 Fn+1 = = lim = 1.6180339887499 . . . n!1 Fn 2 which satisfies =1+ • • ••, ⌅⌅ • •, •⌅⌅•, • • ⌅⌅, ⌅⌅⌅⌅ , 1 · n beats, Fn+1 patterns 11 / 119 12 / 119 Binet’s formula For n n (1 + p ( p ) Formula of A. De Moivre (1718, 1730), Daniel Bernoulli (1726), L. Euler (1728, 1765), J.P.M. Binet (1843) : for n p !n p !n 1 1+ 5 1 1 5 p Fn = p . 2 2 5 5 Jacques Philippe Marie Binet (1843) 0, Fn = = The so–called Binet Formula n 5 5)n (1 p 2n 5 p 5)n , p 1+ 5 = , 2 X 2 X 1 = (X Abraham de Moivre (1667–1754) 1 = 1 p 2 )(X + 5 Daniel Bernoulli (1700–1782) Leonhard Euler (1707–1783) 0, Jacques P.M. Binet (1786–1856) , 1 ). 13 / 119 14 / 119 Fn is the nearest integer to Generating series Generating series of the Fibonacci A single series encodes all the Fibonacci sequence : X Fn X n = X + X 2 + 2X 3 + 3X 4 + 5X 5 + · · · + Fn X n + · · · 1 p n. sequence 5 n 0 Fact : this series is the Taylor expansion of a rational fraction : X X Fn X n = · 1 X X2 n 0 X X X 2 in the right hand side X + X 2 + 2X 3 + 3X 4 + · · · + Fn X n + · · · = Proof : the product (X + X 2 + 2X 3 + 3X 4 + 5X 5 + 8X 6 + · · · )(1 Remark. The denominator 1 of X 2) is X 2 f (X 1 ), where f (X) = X 2 polynomial of the Golden ratio . is a telescoping series X 1 X X X2 1 is the irreducible X + X 2 + 2X 3 + 3X 4 + 5X 5 + 8X 6 + · · · X2 X 3 2X 4 3X 5 5X 6 · · · X3 X 4 2X 5 3X 6 · · · = X. 15 / 119 16 / 119 Fibonacci and powers of matrices Characteristic polynomial The Fibonacci linear recurrence relation Fn+2 = Fn+1 + Fn for n 0 can be written ✓ ◆ ✓ ◆✓ ◆ Fn+1 0 1 Fn = . Fn+2 1 1 Fn+1 The characteristic polynomial of the matrix ✓ ◆ 0 1 A= 1 1 By induction one deduces, for n 0, ✓ ◆ ✓ ◆n ✓ ◆ Fn 0 1 0 = . Fn+1 1 1 1 is det(XI An equivalent formula is, for n 1, ✓ ◆n ✓ ◆ 0 1 Fn 1 Fn = . 1 1 Fn Fn+1 A) = det ✓ X 1 1 X 1 ◆ = X2 X which is the irreducible polynomial of the Golden ratio 1, . 17 / 119 Fibonacci sequence and the Golden ratio (continued) Fibonacci sequence and Hilbert’s 10th problem Yuri Matiyasevich (1970) showed that there is a polynomial P in n, m, and a number of other variables x, y, z, . . . having the property that n = F2m i↵ there exist integers x, y, z, . . . such that P (n, m, x, y, z, . . . ) = 0. For n 1, n 2 Z[ ] = Z + Z is a linear combination of 1 and with integer coefficients, namely n = Fn 1 18 / 119 This completed the proof of the impossibility of the tenth of Hilbert’s problems (does there exist a general method for solving Diophantine equations ?) thanks to the previous work of Hilary Putnam, Julia Robinson and Martin Davis. + Fn 19 / 119 20 / 119 The Fibonacci Quarterly Lucas sequence http://oeis.org/000032 The Lucas sequence (Ln )n 0 satisfies the same recurrence relation as the Fibonacci sequence, namely for n Ln+2 = Ln+1 + Ln 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. 0, only the initial values are di↵erent : L0 = 2, L1 = 1. The sequence of Lucas numbers starts with 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, . . . A closed form involving the Golden ratio Ln = n +( from which it follows that for n to n . ) n is , 2, Ln is the nearest integer 21 / 119 François Édouard Anatole Lucas (1842 - 1891) 22 / 119 Generating series of the Lucas sequence 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. The generating series of the Lucas sequence X Ln X n = 2 + X + 3X 2 + 4X 3 + · · · + Ln X n + · · · n 0 is nothing else than 2 X · 1 X X2 http://www-history.mcs.st-andrews.ac.uk/history/ Mathematicians/Lucas.html 23 / 119 24 / 119 The Lucas sequence and power of matrices Perrin sequence The Perrin sequence (also called skiponacci sequence) is the linear recurrence sequence (Pn )n 0 defined by From the linear recurrence relation Ln+2 = Ln+1 + Ln one deduces, (as we did for the Fibonacci sequence), for n 0, ✓ ◆ ✓ ◆✓ ◆ Ln+1 0 1 Ln = , Ln+2 1 1 Ln+1 hence ✓ Ln Ln+1 ◆ = ✓ 0 1 1 1 http://oeis.org/A001608 for n Pn+3 = Pn+1 + Pn 0, with the initial conditions P0 = 3, P1 = 0, P2 = 2. ◆n ✓ ◆ 2 . 1 It starts with Any one of the three sequences (Fn )n 0 , (Ln )n 0 , ( n )n can be written as a linear combination of the two others. 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, . . . 0 François Olivier Raoul Perrin (1841-1910) : https://en.wikipedia.org/wiki/Perrin_number 25 / 119 Plastic (or silver) constant https://oeis.org/A060006 The ratio of successive terms in the Perrin sequence approaches the plastic number %, which is the minimal Pisot–Vijayaraghavan number, real root of x 3 x 26 / 119 Perrin sequence and the plastic constant Decompose the polynomial X 3 factors over C X3 1 1 = (X %)(X 1 into irreducible ⇢)(X ⇢) and over R which has a value of approximately 1.324718. X3 This constant is equal to p p p 3 3 108 + 12 69 + 108 %= 6 X X p 12 69 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 27 / 119 0, Pn is the nearest integer to %n . 28 / 119 Generating series of the Perrin sequence Perrin sequence and power of matrices The generating series of the Perrin sequence X Pn X n = 3 + 2X 2 + 3X 3 + 2X 4 + · · · + Pn X n + · · · From Pn+3 = Pn+1 + Pn we deduce n 0 0 is nothing else than 3 X2 · 1 X2 X3 Hence 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 %. 1 0 10 1 Pn+1 0 1 0 Pn @Pn+2 A = @0 0 1A @Pn+1 A . Pn+3 1 1 0 Pn+2 0 1 0 1n 0 1 Pn 0 1 0 3 @Pn+1 A = @0 0 1A @0A . Pn+2 1 1 0 2 29 / 119 Characteristic polynomial The characteristic polynomial of 0 0 A = @0 1 is det(XI 0 X @ A) = det 0 1 30 / 119 Perrin’s remark the matrix 1 1 0 0 1A 1 0 1 X 1 1 0 1A = X 3 X X R. Perrin L’intermédiaire des mathématiciens, Query 1484, v.6, 76–77 (1899). 1, The website www.Perrin088.org maintained by Richard Turk is devoted to Perrin numbers. See OEISA113788. which is the irreducible polynomial of the plastic constant %. 31 / 119 32 / 119 Perrin pseudoprimes https://oeis.org/A013998 Padovan sequence https://oeis.org/A000931 The Padovan sequence (pn )n If p is prime, then p divides Pp . 0 satisfies the same recurrence pn+3 = pn+1 + pn The smallest composite n such that n divides Pn is 5212 = 271441. The number P271441 has 33 150 decimal digits (the number c which satisfies 10c = %271441 is c = 271441(log %)/(log 10)). as the Perrin sequence but has di↵erent initial values : p0 = 1, p1 = p2 = 0. It starts with Also for the composite number n = 904631 = 7 ⇥ 13 ⇥ 9941, the number n divides Pn . 1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, . . . Jon Grantham has proved that there are infinitely many Perrin pseudoprimes. Richard Padovan http://mathworld.wolfram.com/LinearRecurrenceEquation.html 33 / 119 Generating series and power of matrices 1 + X 3 + X 5 + · · · + pn X n + · · · = For n 34 / 119 Padovan triangles 1 X2 · 1 X2 X3 pn = pn 0, 0 1n 0 1 0 1 0 pn pn+2 pn+1 @0 0 1A = @pn+1 pn+3 pn+2 A . 1 1 0 pn+2 pn+4 pn+3 2 + pn 3 pn 1 = pn 3 + pn 4 pn 2 = pn 4 + pn 5 Hence pn pn = pn 35 / 119 1 + pn pn 1 = pn 5 5 36 / 119 Padovan triangles Padovan triangles vs Fibonacci squares C 2 curve, not C 3 C 1 curve, not C 2 37 / 119 Padovan, Euler, Zagier and Brown 38 / 119 Narayana sequence https://oeis.org/A000930 Narayana sequence is defined by the recurrence relation For n 0, the number of compositions s = (s1 , . . . , sk ) with si 2 {2, 3} and s1 + · · · + sk = n is pn+3 . This is (an upper bound for) the dimension of the space spanned by the multiple zeta values of weight n of Euler and Zagier. Cn+3 = Cn+2 + Cn with the initial values C0 = 2, C1 = 3, C2 = 4. It starts with 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, . . . Leonhard Euler Don Zagier Real root of x3 x2 1 s s p p 93 3 93 3 29 + 3 3 29 + +1 2 2 = 1.465571231876768 . . . 3 Francis Brown 39 / 119 40 / 119 Generating series and power of matrices 2 + 3X + 4X 2 + 6X 3 + · · · + Cn X n + · · · = For n Narayana’s cows 2 + X + X2 · 1 X X3 Narayana was an Indian mathematician in the 14th century who proposed the following problem : A cow produces one calf every year. Beginning in its fourth year each calf produces one calf at the beginning of each year. How many calves are there altogether after, for example, 17 years ? 7, 0 1n 0 0 1 0 Cn @0 0 1 A = @C n 1 0 1 Cn 6 5 4 Cn Cn Cn 7 6 5 1 Cn 5 Cn 4 A . Cn 3 41 / 119 Music : 42 / 119 Narayana’s cows http://www.pogus.com/21033.html 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 http://webusers.imj-prg.fr/~michel.waldschmidt/ 43 / 119 Year 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Original Cow 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Second generation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Third generation 0 0 0 1 3 6 10 15 21 28 36 45 55 66 Fourth generation 0 0 0 0 0 0 1 4 10 20 35 56 84 120 Fifth generation 0 0 0 0 0 0 0 0 0 1 5 15 35 70 Sixth generation 0 0 0 0 0 0 0 0 0 0 0 0 1 6 Total 2 3 4 6 9 13 19 28 41 60 88 129 189 277 44 / 119 Jean-Paul Allouche and Tom Johnson 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 1996. http://kalvos.org/johness1.html • Finite automata and morphisms in assisted musical composition, Journal of New Music Research, no. 24 (1995), 97 – 108. http://www.tandfonline.com/doi/abs/10.1080/ 09298219508570676 http://web.archive.org/web/19990128092059/www.swets. nl/jnmr/vol24_2.html http://webusers.imj-prg.fr/~jean-paul.allouche/ bibliorecente.html http://www.math.jussieu.fr/~allouche/johnson1.pdf 45 / 119 Linear recurrence sequences : definitions Linear recurrence sequences : examples A linear recurrence sequence is a sequence of numbers u = (u0 , u1 , u2 , . . . ) for which there exist a positive integer d together with numbers a1 , . . . , ad with ad 6= 0 such that, for n 0, (?) un+d = a1 un+d 1 • Constant sequence : un = u0 . Linear recurrence sequence of order 1 : un+1 = un . Characteristic polynomial : f (X) = X 1. Generating series : X + · · · + ad u n . Here, a number means an element of a field K of zero characteristic. Given a = (a1 , . . . , ad ) 2 Kd , the set Ea of linear recurrence sequences u = (un )n 0 satisfying (?) is a K–vector subspace of dimension d of the space KN of all sequences . The characteristic (or companion) polynomial of the linear recurrence is f (X) = X d a1 X d 1 ··· 46 / 119 n 0 Xn = 1 1 X · • Geometric progression : un = u0 n . Linear recurrence sequence of order 1 : un = un 1 . Characteristic polynomial f (X) = X . Generating series : X u0 u0 n X n = · 1 X n 0 ad . 47 / 119 48 / 119 Linear recurrence sequences : examples Linear recurrence sequences : examples • un = n. Linear recurrence sequence of order 2 : n + 2 = 2(n + 1) • un = f (n), f polynomial of degree d. Linear recurrence sequence of order d + 1. n. Characteristic polynomial Proof. The sequences f (X) = X 2 2X + 1 = (X 1)2 . (f (n))n 0 , Generating series X nX n = n 0 1 ··· , (f (n + 1))n 0 , (f (n + k))n are K–linearly independent in KN for k = d dependent for k = d . 1 · 2X + X 2 0 1 and linearly A basis of the space of polynomials of degree d is given by the d + 1 polynomials Power of matrices : ✓ ◆n ✓ ◆ 0 1 n+1 n = . 1 2 n n+1 f (X), f (X + 1), . . . , f (X + d). 49 / 119 Linear sequences which are ultimately recurrent Order of a linear recurrence sequence If u = (un )n 0 satisfies the linear recurrence, the characteristic polynomial of which is f , then, for any monic polynomial g 2 K[X] with g(0) 6= 0, this sequence u also satisfies the linear recurrence, the characteristic polynomial of which is f g. Example : for g(X) = X with 6= 0, from The sequence (1, 0, 0, . . . ) is not a linear recurrence sequence. The condition (?) un+1 = un is satisfied only for n 50 / 119 un+d a1 un+d 1 ··· ad u n = 0 we deduce 1. un+d+1 The relation un+2 = un+1 + 0un with d = 2, ad = 0 does not fulfill the requirement ad 6= 0. a1 un+d · · · ad un+1 (un+d a1 un+d 1 · · · ad un ) = 0. The order of a linear recurrence sequence is the smallest d such that (?) holds for all n 0. 51 / 119 52 / 119 Generating series of a linear recurrence sequence Let u = (un )n (?) 0 be a linear recurrence sequence un+d = a1 un+d 1 + · · · + ad u n Assume for n 0 un+d = a1 un+d with characteristic polynomial f (X) = X d Denote by f Then 1 ··· 1 X ad . 1 X 1 )=1 a1 X ··· 1 + · · · + ad u n un X n = n=0 the reciprocal polynomial of f : f (X) = X d f (X Then a1 X d Generating series of a linear recurrence sequence for n r(X) · f (X) Proof. Comparing the coefficients of X n for n ad X d . f (X) r(X) un X = , f (X) n=0 n 1 X 0. d shows that un X n n=0 is a polynomial of degree less than d where r is a polynomial of degree less than d determined by the initial values of u. 53 / 119 Taylor coefficients of rational functions 54 / 119 Matrix notation for a linear recurrence sequence Conversely, the coefficients the Taylor expansion of any rational fraction a(X)/b(X) with deg a < deg b and b(0) 6= 0 satisfies the recurrence relation with characteristic polynomial f 2 K[X] given by f (X) = b (X). The linear recurrence sequence (?) un+d = a1 un+d 1 + · · · + ad u n for n 0 can be written Therefore a sequence u = (un )n 0 satisfies the recurrence relation (?) with characteristic polynomial f 2 K[X] if and only if 1 X r(X) un X n = , f (X) n=0 0 1 0 0 1 0 un+1 B0 0 1 Bun+2 C B B C B .. . .. .. B .. C = B . . @ . A B @0 0 0 un+d ad ad 1 ad 2 where r is a polynomial of degree less than d determined by the initial values of u. 55 / 119 ··· ··· .. . ··· ··· 1 1 0 0 un C 0CB C .. C B un+1 C . B .. C .C C@ . A 1A un+d 1 a1 56 / 119 Matrix notation for a linear recurrence sequence Powers of matrices Un+1 = AUn with 0 B B Un = B @ 1 un un+1 .. . un+d 1 C C C, A The determinant of Id X A) is nothing but 0 0 1 0 B0 0 1 B B .. . .. .. A=B. . B @0 0 0 ad ad 1 ad 2 f (X) = X d ··· ··· .. . ··· ··· Let A = (aij )1i,jd 2 GLd⇥d (K) be a d ⇥ d matrix with coefficients in K and nonzero determinant. For n 0, define 1 0 0C C .. C . .C C 1A a1 (n) An = aij 1i,jd . (n) Then each of the d2 sequences aij n 0 , (1 i, j d) is a linear recurrence sequence. The roots of the characteristic polynomial of these linear recurrences are the eigenvalues of A. A (the characteristic polynomial of a1 X d 1 ··· In particular the sequence Tr(An ) n 0 satisfies the linear recurrence, the characteristic polynomial of which is the characteristic polynomial of the matrix A. ad , the characteristic polynomial of the linear recurrence sequence. By induction Un = An U0 . 57 / 119 Conversely : 58 / 119 Linear recurrence sequences : simple roots A basis of Ea over K is obtained by attributing to the initial values u0 , . . . , ud 1 the values given by the canonical basis of Kd . Given in K⇥ , a necessary and sufficient condition for a sequence ( n )n 0 to satisfy (?) is that is a root of the characteristic polynomial Given a linear recurrence sequence u 2 KN , there exist an integer d 1 and a matrix A 2 GLd (K) such that, for each n 0, (n) un = a11 . f (X) = X d The characteristic polynomial of A is the characteristic polynomial of the linear recurrence sequence. a1 X d 1 If this polynomial has d distinct roots Everest G., van der Poorten A., Shparlinski I., Ward T. – Recurrence sequences, Mathematical Surveys and Monographs (AMS, 2003), volume 104. f (X) = (X 1 ) · · · (X ··· ad . 1, . . . , d ), d i in K, 6= j, then a basis of Ea over K is given by the d sequences ( i = 1, . . . , d. 59 / 119 n i )n 0 , 60 / 119 Linear recurrence sequences : double roots Linear recurrence sequences : multiple roots The characteristic polynomial of the linear recurrence 2 un = 2 u n 1 un 2 is X 2 2 X + 2 = (X )2 with a double root . The sequence (n n n n )n 0 In general, when the characteristic polynomial splits as satisfies = 2 (n 1)n n 1 2 (n 2) n 2 Xd a1 X d . 1 ··· ad = Ỳ (X i) ti , i=1 a basis of Ea is given by the d sequences A basis of Ea for a1 = 2 , a2 = sequences ( n )n 0 , (n n )n 0 . 2 is given by the two (nk n i )n 0 , 0 k ti 1, 1 i `. Given 2 K⇥ , a necessary and sufficient condition for the sequence n n to satisfy the linear recurrence relation (?) is that is a root of multiplicity 2 of f (X). 61 / 119 Polynomial combinations of powers 62 / 119 Sum of polynomial combinations of powers The sum and the product of any two linear recurrence sequences are linear recurrence sequences. If u1 and u2 are two linear recurrence sequences of characteristic polynomials f1 and f2 respectively, then u1 + u2 satisfies the linear recurrence, the characteristic polynomial of which is f1 f2 · gcd(f1 , f2 ) The set [a Ea of all linear recurrence sequences with coefficients in K is a sub–K–algebra of KN . Given polynomials p1 , . . . , p` in K[X] and elements in K⇥ , the sequence p1 (n) n 1 + · · · + p` (n) 1, . . . , ` n ` n 0 is a linear recurrence sequence. Fact : any linear recurrence sequence is of this form. 63 / 119 64 / 119 Product of polynomial combinations of powers Linear recurrence sequences and Brahmagupta–Pell–Fermat Equation If the characteristic polynomials of the two linear recurrence sequences u1 and u2 are respectively f1 (T ) = Ỳ 0 (T j) tj and f2 (T ) = j=1 ` Y (T Let d be a positive integer, not a square. The solutions (x, y) 2 Z ⇥ Z of the Brahmagupta–Pell–Fermat Equation 0 t0k k) , x2 k=1 form a sequence (xn , yn )n2Z defined by p p xn + dyn = (x1 + dy1 )n . then u1 u2 satisfies the linear recurrence, the characteristic polynomial of which is 0 ` Ỳ Y (T dy 2 = ±1 From 0 tj +t0k 1 . j k) 2xn = (x1 + j=1 k=1 p dy1 )n + (x1 p dy1 )n we deduce that (xn )n 0 is a linear recurrence sequence. Same for yn , and also for n 0. 65 / 119 Doubly infinite linear recurrence sequences Discrete version of linear di↵erential equations A sequence u 2 KN can be viewed as a linear map N ! K. Define the discrete derivative D by A sequence (un )n2Z indexed by Z is a linear recurrence sequence if it satisfies (?) un+d = a1 un+d 1 66 / 119 Du : N ! K n 7 ! un+1 un . + · · · + ad u n . for all n 2 Z. A sequence u 2 KN is a linear recurrence sequence if and only if there exists Q 2 K[T ] with Q(1) 6= 1 such that Recall ad 6= 0. Q(D)u = 0. Linear recurrence sequences are a discrete version of linear di↵erential equations with constant coefficients. Such a sequence is determined by d consecutive values. The condition Q(1) 6= 0 reflects ad 6= 0 – otherwise one gets ultimately recurrent sequences. 67 / 119 68 / 119 97th Indian Science Congress, 2010 A. P. J. Abdul Kalam (1931-2015) Public Lecture during the 97th Indian Science Congress, Thiruvananthapuram – 4 January 2010 Thiruvananthapuram A.K. Agarwal Invited by Ashok Agrawal to the 97th Indian Science Congress in Thiruvananthapuram (Trivandrum, Kerala), January 3-7, 2010. • Lecture on Number Theory Challenges of 21st Century Basic research is vital for enhancing national and international competitiveness http://www.abdulkalam.com/kalam/theme/jsp/guest/ content-display.jsp 69 / 119 Kerala 2010 Sudhir Ghorpade 70 / 119 KSOM 2010 Jugal K. Verma January 8, 2010, Calicut = Kozhikode (Kerala) The Kerala School of Mathematics (KSoM) Ambar Vijayatkumar January 9-10, 2010, Cochin = Kochi (Kerala) Department of Mathematics, Cochin University of Science and Technology CUSAT A. J. Parameswaran, Director of the Kerala School of Mathematical Science (KSOM) in Kozikhode (Calicut) 71 / 119 72 / 119 KSOM 2010 A dynamical system Work on dynamical systems by A. J. Parameswaran and S.G. Dani Let V be a finite dimensional vector space over a field of zero characteristic, H an hyperplane of V , f : V ! V an endomorphism (linear map) and x an element in V . Theorem. If there exist infinitely many n 1 such that f n (x) 2 H, then there is an (infinite) arithmetic progression of n for which it is so. A. J. Parameswaran S.G. Dani 73 / 119 Skolem – Mahler – Lech Theorem Wolfgang M. Schmidt Theorem (Skolem 1934 – Mahler 1935 – Lech 1953). Given a linear recurrence sequence, the set of indices n 0 such that un = 0 is a finite union of arithmetic progressions. Linear recurrence sequence : un+d = a1 un+d 1 + · · · + ad u n , n 74 / 119 0 Thue – Siegel – Roth – Schmidt, Schmidt’s Subspace Theorem. The generalized S–unit Theorem Let K be a field of characteristic zero, let G be a finitely multiplicative subgroup of the multiplicative group K⇥ = K \ {0} and let n 2. Then the equation (ad 6= 0). Characteristic polynomial : Xd a1 X d 1 un = ··· ad = j 1 X̀ tX Ỳ u1 + u2 + · · · + un = 1, (X j) where the values of the unknowns u1 , u2 , · · · , un are in G for which no nontrivial subsum X ui ;= 6 I ⇢ {1, . . . , n} tj j=1 cij ni n j. i2I j=1 i=0 vanishes, has only finitely many solutions. 75 / 119 76 / 119 Schmidt’s subspace Theorem Balu’s 60’s Birthday, 2011 December 15 - 20, 2011 : HRI : International Meeting on Number Theory 2011 celebrating the 60th Birthday of Professor R. Balasubramanian. Pietro Corvaja M. Manickam, director of KSOM. Wolfgang M. Schmidt Pietro Corvaja December 16, 2011 : lecture on Families of Thue-Mahler equations. Umberto Zannier 77 / 119 Joint work with Claude Levesque 78 / 119 http://www.ramanujanmathsociety.org/publications/rms-lecture-notes-series http://arxiv.org/abs/1505.06653 Solving simultaneously Thue Diophantine equations : almost totally imaginary case Proceedings of the International Meeting on Number Theory HRI 2011, in honor of R. Balasubramanian. Ramanujan Mathematical Society, Lecture Notes Series 23, Highly composite : papers in number theory, (2016), 137–156. Editors Kumar Murty, Ravindranathan Thangadurai. http://www.ramanujanmathsociety.org/publications/ rms-lecture-notes-series 79 / 119 80 / 119 KSOM 2013 Reference Workshop number theory and dynamical systems in KSOM (Director M. Manickam) in February 2013 M. Waldschmidt. Diophantine approximation with applications to dynamical systems. Proceedings of the International Conference on Pure and Applied Mathematics ICPAM–LAE 2013, South Pacific Journal of Pure and Applied Mathematics, vol. 1, No 2 (2014), 1–18. Yann Bugeaud Pietro Corvaja S.G. Dani 81 / 119 Skolem – Mahler – Lech Theorem A dynamical system Theorem (Skolem 1934 – Mahler 1935 – Lech 1953). Given a linear recurrence sequence, the set of indices n 0 such that un = 0 is a finite union of arithmetic progressions. Thoralf Albert Skolem (1887 – 1963) Kurt Mahler (1903 – 1988) 82 / 119 Let V be a finite dimensional vector space over a field of zero characteristic, W a subspace of V , f : V ! V an endomorphism (linear map) and x an element in V . Christer Lech Corollary of the Skolem – Mahler – Lech Theorem. The set of n 0 such that f n (x) 2 W is a finite union of arithmetic progressions. By induction, it suffices to consider the case where W = H is an hyperplane of V . An arithmetic progression is a set of positive integers of the form {n0 , n0 + k, n0 + 2k, . . .}. Here, we allow k = 0. 83 / 119 84 / 119 Idea of the proof of the corollary Theorem of Cayley – Hamilton Arthur Cayley (1821 – 1895) Choose a basis of V . The endomorphism f is given by a square d ⇥ d matrix A, where d is the dimension of V . Consider the characteristic polynomial of A, say Xd ad 1 X d 1 ··· a1 X Sir William Rowan Hamilton (1805 – 1865) a0 . By the Theorem of Cayley – Hamilton, Ad = ad 1 Ad 1 Hence, for n + · · · + a1 A + a0 I d 0, An+d = ad 1 An+d where Id is the identity d ⇥ d matrix. 1 + · · · + a1 An+1 + a0 An . It follows that each entry aij (n), 1 i, j d, satisfies a linear recurrence relation, the same for all i, j. 85 / 119 Hyperplane membership 86 / 119 Remark on the theorem of Skolem–Mahler–Lech Let b1 x1 + · · · + bd xd = 0 be an equation of the hyperplane H in the selected basis of V . Let t b denote the 1 ⇥ d matrix (b1 , . . . , bd ) (transpose of a column matrix b). Using the notation v for the d ⇥ 1 (column) matrix given by the coordinates of an element v in V , the condition v 2 H can be written t b v = 0. T.A. Skolem treated the case K = Q of in 1934 K. Mahler the case K = Q, the algebraic closure of Q, in 1935 Let x be an element in V and x the d ⇥ 1 (column) matrix given by its coordinates. The condition f n (x) 2 H can now be written t bAn x = 0. The general case was settled by C. Lech in 1953. The entry un of the 1 ⇥ 1 matrix t bAn x satisfies a linear recurrence relation, hence, the Skolem – Mahler – Lech Theorem applies. 87 / 119 88 / 119 Finite characteristic Polynomial-linear recurrence relation C. Lech pointed out in 1953 that such a result may not hold if the characteristic of K is positive : he gave as an example the sequence un = (1 + x)n xn 1, a third-order linear recurrence over the field of rational functions in one variable over the field Fp with p elements, where un = 0 for n 2 {1, p, p2 , p3 , . . .}. A substitute is provided by a result of Harm Derksen (2007), who proved that the zero set in characteristic p is a p–automatic sequence. Further results by Boris Adamczewski and Jason Bell. A generalization of the Theorem of Skolem–Mahler–Lech has been achieved by Jason P. Bell, Stanley Burris and Karen Yeats who prove that the same conclusion holds if the sequence (un )n 0 satisfies a polynomial-linear recurrence relation un = d X Pi (n)un i i=1 where d is a positive integer and P1 , . . . , Pd are polynomials with coefficient in the field K of zero characteristic, provided that Pd (x) is a nonzero constant. Harm Derksen Boris Adamczewski Jason Bell 89 / 119 Algebraic maps, algebraic groups Open problem One main open problem related with Theorem of Skolem–Mahler–Lech is that it is not e↵ective : explicit upper bounds for the number of arithmetic progressions, depending only on the order d of the linear recurrence sequence, are known (W.M. Schmidt, U. Zannier), but no upper bound for the arithmetic progressions themselves is known. A related open problem raised by T.A. Skolem and C. Pisot is : Given an integer linear recurrence sequence, is the truth of the statement “xn 6= 0 for all n” decidable in finite time ? There are also analogues of the Theorem of Skolem–Mahler–Lech for algebraic maps on varieties (Jason Bell). A version of the Skolem–Mahler–Lech Theorem for any algebraic group is due to Umberto Zannier. Jason Bell 90 / 119 T. Tao, E↵ective Skolem Mahler Lech theorem. In “Structure and Randomness : pages from year one of a mathematical blog”, American Mathematical Society (2008), 298 pages. Umberto Zannier http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/ 91 / 119 92 / 119 Zeros of linear recurrence sequences Zeros of linear recurrence sequences Jean Berstel et Maurice Mignotte. – Deux propriétés décidables des suites récurrentes linéaires Bulletin de la S.M.F., tome 104 (1976), p. 175-184. Maurice Mignotte Propriétés arithmétiques des suites récurrentes linéaires. Besançon, 1989 http://pmb.univ-fcomte.fr/1989/Mignotte.pdf http://www.numdam.org/item?id=BSMF_1976__104__175_0 Given a linear recurrence sequence with integer coefficients ; are there finitely or infinitely many zeroes ? E. Bavenco↵e and J-P. Bézivin Une famille remarquable de suites recurrentes lineaires. – Monatshefte für Mathematik, (1995) 120 3, 189–203 Philippe Robba. – Zéros de suites récurrentes linéaires. Groupe de travail d’analyse ultramétrique (1977-1978) Volume : 5, page 1-5. Karim Samake. – Suites récurrentes linéaires, problème d’e↵ectivité. Inst. de Recherche Math. Avancée, 1996 - 62 pages L. Cerlienco, M. Mignotte, F. Piras. Suites récurrentes linéaires. Propriétés algébriques et arithmétiques. L’Enseignement Mathématique 33 (1987). 93 / 119 Reference 94 / 119 Berstel’s sequence Everest, Graham ; van der Poorten, Alf ; Shparlinski, Igor ; Ward, Tom – Recurrence sequences, Mathematical Surveys and Monographs (AMS, 2003), volume 104. 1290 references. 0, 0, 1, 2, 0, 4, 0, 16, 16, http://oeis.org/A007420 32, 64, 64, 256, 0, 768, . . . b0 = b1 = 0, b2 = 1, bn+3 = 2bn+2 4bn+1 + 4bn for n 0. Graham Everest Linear recurrence sequence of order 3 with exactly 6 zeros : n = 0, 1, 4, 6, 13, 52. Alf van der Poorten Jean Berstel Igor Shparlinski http://www-igm.univ-mlv.fr/~berstel/ Tom Ward 95 / 119 96 / 119 Ternary linear recurrences Edgard Bavenco↵e and Jean-Paul Bézivin Berstel’s sequence is a linear recurrence sequence of order 3 with 6 zeroes. Let n 2. The sequence with initial values u0 = 1, u1 = · · · = un Frits Beukers (1991) : up to trivial transformation, any other linear recurrence of order 3 with finitely many zeroes has at most 5 zeros. 1 =0 satisfying the recurrence relation of order n with characteristic polynomial X n+1 ( 2)n 1 X + ( 2)n X +2 has at least n(n + 1) 1 2 zeroes. Frits Beukers 97 / 119 Edgard Bavenco↵e and Jean-Paul Bézivin Berstel’s sequence For n = 3 one obtains Berstel’s sequence which happens to have an extra zero. X 4 + 4X X +2 8 = X3 2X 2 + 4X 98 / 119 0, 0, 1, 2, 0, 4, 0, 16, 16, 32, b0 = b1 = 0, b2 = 1, bn+3 = 2bn+2 4. 64, 64, 256, 0, 4bn+1 + 4bn for n 768, . . . 0. The equation bm = ±bn has exactly 21 solutions (m, n) with m 6= n. Edgard Bavenco↵e Jean-Paul Bézivin Maurice Mignotte 99 / 119 The equation bn = ±2r 3s has exactly 44 solutions (n, r, s). 100 / 119 Joint work with Claude Levesque Families of binary forms Consider a binary form F0 (X, Y ) 2 C[X, Y ] which satisfies F0 (1, 0) = 1. We write it as Linear recurrence sequences and twisted binary forms. Proceedings of the International Conference on Pure and Applied Mathematics ICPAM-GOROKA 2014. South Pacific Journal of Pure and Applied Mathematics. d F0 (X, Y ) = X + a1 X d 1 Fn (X, Y ) = (X ↵i Y ). i=1 d Y ↵i ✏ni Y ), (X i=1 Xd http://webusers.imj-prg.fr/~michel.waldschmidt//articles/ pdf/ProcConfPNG2014.pdf Therefore 101 / 119 Families of Diophantine equations U1 (n)X d 1 Y + · · · + ( 1)d Ud (n)Y d . Uh (0) = ( 1)h ah (1 h d). 102 / 119 Families of Diophantine equations With Claude Levesque, we considered some families of diophantine equations Each of the sequences Uh (n) coefficients of the relation Fn (x, y) = m obtained in the same way from a given irreducible form F (X, Y ) with coefficients 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 a finite set of places of K containing the places at infinity. Denote by OS the ring of S–integers of K and by OS⇥ 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 OS ⇥ OS ⇥ Z satisfying xy 6= 0 and Card{↵1 ✏n1 , . . . , ↵d ✏nd } = d Y Let ✏1 , . . . , ✏d be d nonzero complex numbers not necessarily distinct. Twisting F0 by the powers ✏n1 , . . . , ✏nd (n 2 Z) boils down to considering the family of binary forms which we write as Fn (x, y) 2 OS⇥ , Y + · · · + ad Y d Fn (X, Y ) = X d 103 / 119 coming from the U1 (n)X d 1 Y + · · · + ( 1)d Ud (n)Y d is a linear recurrence sequence. For example, for n 2 Z, U1 (n) = d X ↵i ✏ni , Ud (n) = i=1 n2Z d Y ↵i ✏ni . i=1 For 1 h d, the sequence Uh (n) combination of the sequences (✏i1 · · · ✏ih )n 3. n2Z , n2Z is a linear (1 i1 < · · · < ih d). 104 / 119 Some units of Bernstein and Hasse Let t and s be two positive integers, D an integer c 2 { 1, +1}. Let ! > 1 satisfy st Helmut Hasse (1898-1979) 1, and D > 0, s 1, t 1, c 2 { 1, +1}, ! > 0, st ! = D + c, ! st = Dst + c, where it is assumed that Q(!) is of degree st. Consider ↵ = D !, ✏ = Dt ! 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(!). ↵=D !, ✏ = Dt !t. (↵ D)st = ( 1)st (Dst + c). 105 / 119 Diophantine equations associated with some units of Bernstein and Hasse The irreducible polynomial of ↵ is F0 (X, 1), with F0 (X, Y ) = (X DY )st ( 1)st (Dst + c)Y st . For n 2 Z, the binary form Fn (X, Y ), obtained by twisting F0 (X, Y ) with the powers ✏n of ✏, is the homogeneous version of the irreducible polynomial Fn (X, 1) of ↵✏n . So Fn depends of the parameters n, D, s, t and c. Theorem (LW). Suppose st 3. There exists an e↵ectively computable constant , depending only on D, s and t, with the following property. Let m, a, x, y be rational integers satisfying m 2, xy 6= 0, [Q(↵✏a ) : Q] = st and Then |Fn (x, y)| m. max{log |x|, log |y|, |n|} log m. 106 / 119 Hankel determinants To test an arbitrary sequence u = (un )n 0 of elements of a field K for the property of being a linear recurrence sequence, consider the Hankel determinants N,d (u) The sum f (z) = 1 X un z n n=0 Hermann Hankel (1839–1873) 107 / 119 = det (ud+i+j )0i,jN . represents a rational function if and only if for some d, N,d (u) = 0 for all sufficiently large N 108 / 119 Hankel determinants Perfect powers in the Fibonacci sequence Alan Haynes, Wadim Zudilin. – Hankel determinants of zeta values (Submitted on 7 Oct 2015) Yann Bugeaud, Maurice Mignotte, Samir Siksek (2004) : The only perfect powers (squares, cubes, etc.) in the Fibonacci sequence are 1, 8 and 144. Abstract: We study the asymptotics of Hankel determinants constructed using the values ⇣(an + b) of the Riemann zeta function at positive integers in an arithmetic progression. Our principal result is a Diophantine application of the asymptotics. Y. Bugeaud Alan Haynes M. Mignotte S. Siksek Wadim Zudilin 109 / 119 Powers in recurrence sequences 110 / 119 Bases of the space of linear recurrence sequences Given a1 , . . . , ad with ad 6= 0, consider the vector space of linear recurrence sequences satisfying, for n 0, (?) M. A. Bennett, Powers in recurrence sequences : Pell equations, Trans. Amer. Math. Soc. 357 (2005), 1675-1691. un+d = a1 un+d 1 + · · · + ad u n . Assuming the characteristic polynomial f (X) = X d a1 X d 1 ··· ad of the recurrence splits completely in K, Mike Bennett f (X) = http://www.math.ubc.ca/~bennett/paper31.pdf Ỳ (X j) ti j=1 we have two bases. The first one given by the initial conditions (u0 , . . . , ud 1 ), and the second one is given by the sequences (ni 111 / 119 n j )n 0 , 0 i tj 1, 1 j `. 112 / 119 Change of basis Exponential polynomials The sequence of derivatives of an exponential polynomial evaluated at one point satisfies a linear recurrence relation. Let p1 (z), . . . , p` (z) be nonzero polynomials of C[z] of degrees smaller than t1 , . . . , t` respectively. Let 1 , . . . , ` be distinct complex numbers. Suppose that the function The matrix of change of bases is 0 1 M1 B .. C M =@ . A M` F (z) = p1 (z)e where 0 1 B B B0 B B Mj = B B0 B. B .. @ 0 2 j j 1 2 1 ··· j tj 1 j tj j ··· d 1 j C C C C C 3 C. C C C A ... tj 1 1 tj 2 j tj 1 tj 1 j ... d 1 1 d 2 j tj 1 2 tj 3 j tj 2 tj 2 j ... .. . d 1 2 d j ··· d 1 tj 1 0 .. . 1 .. . ... .. . 0 0 ... .. . 1 .. . tj tj 1 j .. . 1 d tj j 1z + · · · + p` (z)e `z is not identically 0. Then its vanishing order at a point z0 is smaller than or equal to t1 + · · · + t` 1. In other terms, when the complex numbers j are distinct, the determinant ✓ ◆a d z i e j z z=0 0itj 1, 1j` dz 0ad 1 is di↵erent from 0. This is no surprise that we come across the determinant of the matrix M . 113 / 119 The matrix M 114 / 119 Interpolation The determinant of M is det M = Y ( j i) ti tj . 1i<j` For 1 j `, 0 i tj 1, 0 k d entry of the matrix M is ✓ ◆i ✓ ◆ 1 d k k T = i! dT i T= 1, the (sj + i, k) k i j . di f ( j ) = ⌘ij , dz i j The matrix M is associated with the linear system of d equations in d unknowns which amounts to finding a polynomial f 2 K[z] of degree < d for which the d numbers di f ( j ), (1 j `, 0 i tj dz i take prescribed values. Let j (1 j `) be distinct elements in K, tj (1 j `) be positive integers, ⌘ij (1 j `, 0 i tj 1) be elements in K. Set d = t1 + · · · + t` . There exists a unique polynomial f 2 K[z] of degree < d satisfying (1 j `, 0 i tj 1). 1) 115 / 119 116 / 119 Truncated Taylor expansion Explicit solution to the interpolation problem Let g 2 K(z), let z0 2 K and let t pole of g. We set Tg,z0 ,t (z) = t 1 i X dg i=0 dz i For j = 1, . . . , `, define 1. Assume z0 is not a (z0 ) (z hj (z) = z0 )i · i! Y ✓z 1k` k6=j j k k ◆t k tj 1 and pj (z) = X ⌘ij (z j) i! i=0 i · Then the solution f of the interpolation problem In other words, Tg,z0 ,t is the unique polynomial in K[z] of degree < t such that there exists r(z) 2 K(z) having no pole at z0 with g(z) = Tg,z0 ,t (z) + (z z0 )t r(z). di f ( j ) = ⌘ij , dz i (1 j `, 0 i tj 1). is given by Notice that if g is a polynomial of degree < t, then g = Tg,z0 ,t for any z0 2 K. f= X̀ j=1 117 / 119 December 2016 Linear recurrence sequences, exponential polynomials and Diophantine approximation Michel Waldschmidt Institut de Mathématiques de Jussieu — Paris VI http://webusers.imj-prg.fr/~michel.waldschmidt/ 119 / 119 h j T pj , hj j ,tj . 118 / 119