On primary pseudo-polynomials (Around Ruzsa's Conjecture)
aa r X i v : . [ m a t h . N T ] F e b On primary pseudo-polynomials (Around Ruzsa’sConjecture) ´E. Delaygue and T. RivoalFebruary 3, 2021
Abstract
Every polynomial P ( X ) ∈ Z [ X ] satisfies the congruences P ( n + m ) ≡ P ( n )mod m for all integers n, m ≥
0. An integer valued sequence ( a n ) n ≥ is called apseudo-polynomial when it satisfies these congruences. Hall characterized pseudo-polynomials and proved that they are not necessarily polynomials. A long standingconjecture of Ruzsa says that a pseudo-polynomial a n is a polynomial as soon aslim sup n | a n | /n < e . Under this growth assumption, Perelli and Zannier proved thatthe generating series P ∞ n =0 a n z n is a G -function. A primary pseudo-polynomial is aninteger valued sequence ( a n ) n ≥ such that a n + p ≡ a n mod p for all integers n ≥ p . The same conjecture has been formulated for them, whichimplies Ruzsa’s, and this paper revolves around this conjecture. We obtain a Halltype characterization of primary pseudo-polynomials and draw various consequencesfrom it. We prove that any primary pseudo-polynomial with an algebraic generatingseries is a polynomial. We make the Perelli-Zannier Theorem effective and show thatits conclusion does not necessarily hold if lim sup n | a n | /n ≤ e . We prove a P´olya typeresult: if there exists a function F analytic in a right-half plane with not too largeexponential growth (in a precise sense) and such that for all large n the primarypseudo-polynomial a n = F ( n ), then a n is a polynomial. Finally, we show how toconstruct a non-polynomial primary pseudo-polynomial starting from any primarypseudo-polynomial generated by a G -function different of 1 / (1 − x ). Given a sequence ( a n ) n ≥ ∈ C N , we define its binomial transform ( b n ) n ≥ ∈ C N as ∀ n ≥ , b n := n X k =0 ( − n − k (cid:18) nk (cid:19) a k . (1.1)It is well-known that ( a n ) n ≥ can be recovered from ( b n ) n ≥ by ∀ n ≥ , a n = n X k =0 (cid:18) nk (cid:19) b k , (1.2)1e, the binomial transform is “almost” involutive. An important property of the binomialtransform is that “there exists P ( X ) ∈ C [ X ] such that a n = P ( n ) for all n large enough” ifand only if “ b n = 0 for all n large enough”. In the sequel, we will say that a sequence a n is“eventually in A [ n ]” (where A is a given ring) when there exist a polynomial P ( X ) ∈ A [ X ]and an integer N such that a n = P ( n ) for all n ≥ N .A sequence ( a n ) n ≥ ∈ Z N is said to be a pseudo-polynomial when the following propertyholds: for any integers n, k ≥ a n + k ≡ a n mod k . The terminology comes from thefact that for any P ( X ) ∈ Z [ X ], the sequence ( P ( n )) n ≥ is a pseudo-polynomial. Notethat if P ( X ) ∈ C [ X ] is such that the sequence ( P ( n )) n ≥ is a pseudo-polynomial, then P ( X ) ∈ Q [ X ], but P ( X ) does not necessarily belong to Z [ X ] as P ( X ) = X ( X + 1)shows. Pseudo-polynomials have long been studied for themselves, but they have alsofound recent applications in analytic number theory [8, 9].We now set d n := lcm { , , . . . , n } for n ≥ d := 1. We recall that d n = Q p ≤ n p ⌊ log p ( n ) ⌋ ≤ n for all n and that d /nn → e as n → + ∞ (by the Prime NumberTheorem). Hall [5] proved the following fundamental property: a sequence ( a n ) n ≥ ∈ Z N is a pseudo-polynomial if and only if its binomial transform ( b n ) n ≥ ∈ Z N satisfies ∀ n ≥ , d n | b n . (1.3)Using this characterization with b n := n ! in (1.3), we see from (1.2) that the resultingpseudo-polynomial a n is simply equal to ⌊ n ! e ⌋ for n ≥ n = 0), which isobviously not a polynomial. With b n := d n in (1.3), we obtain from (1.2) another pseudo-polynomial of slower growth a n := P nk =0 (cid:0) nk (cid:1) d n ≤ n ; since a n ≥ n for all n ≥
0, this isnot a polynomial either.The search of minimal growth conditions that can be attained by non-polynomialpseudo-polynomials has been the subject of many papers. Hall [5] and Ruzsa [16] in-dependently proved that if lim sup n → + ∞ | a n | /n < e − a n is eventually in Q [ n ]. Ruzsa proposed the following Conjecture 1 (Ruzsa) . Let ( a n ) n ≥ ∈ Z N be a pseudo-polynomial such that lim sup n → + ∞ | a n | /n < e. (1.5) Then a n is eventually in Q [ n ] . Using his characterization (1.3), Hall [5, p. 76] sketched an inductive construction ofa non-polynomial pseudo-polynomial ( a n ) n ≥ such that lim sup n → + ∞ | a n | /n ≤ e , show-ing that the upper bound < e is best possible in Ruzsa’s conjecture. Perelli and Zan-nier [13] then proved a highly non-trivial property: under the growth condition (1.5) inRuzsa’s Conjecture, the pseudo-polynomial sequence ( a n ) n ≥ satisfies a linear recurrencewith polynomials coefficients; see (1.9) below. In other words, the generating function f a ( x ) := P ∞ n =0 a n x n ∈ Z [[ x ]] satisfies a linear differential equation with coefficients in Z [ x ].2ence, f a is a G -function ( ). Perelli and Zannier also proved a form of Ruzsa’s Conjecture1 under a stronger assumption than (1.5), ie with e replaced by e . .In fact, many results towards Ruzsa’s conjecture have been proven for sequences weshall call primary pseudo-polynomial (for lack of better terminology). Definition 1.
A sequence ( a n ) n ≥ ∈ Z N is said to be a primary pseudo-polynomial whenthe following property holds: for any integer n ≥ and any prime number p , a n + p ≡ a n mod p . The set of primary pseudo-polynomials is a ring for the term by term sum and productof sequences in Z N , with the null and unit sequences defined with all terms equal to 0 andall terms equal to 1 respectively. A pseudo-polynomial is a primary pseudo-polynomial butthe converse is false (see the comments following Theorem 1 below). Many authors deltwith the following conjecture, the truth of which would imply that of Ruzsa. Conjecture 2.
Let ( a n ) n ≥ ∈ Z N be a primary pseudo-polynomial such that lim sup n → + ∞ | a n | /n < e. (1.6) Then a n is eventually in Q [ n ] . As in the case of pseudo-polynomials, < e cannot be replaced by ≤ e , and we referagain to the comments following Theorem 1 for a proof of this. In fact, the above quotedresults of Perelli and Zannier hold more generally for primary pseudo-polynomials, andZannier [19] was even able to replace e . by e . .In this paper, we are interested in the properties of primary pseudo-polynomials and oftheir generating functions. We now present our four main results, make comments abouttheir significance and mention further open problems. • A Hall type characterization of primary pseudo-polynomials.
We shall firstprove an analogue (ie Eq. (1.7) below) of Hall’s characterization for pseudo-polynomialsand deduce some consequences of it. We set P n := Q p ≤ n p for n ≥ P = P := 1,where the product is over prime numbers. By the Prime Number Theorem, P /nn → e as n → + ∞ . Theorem 1. ( i ) A sequence ( a n ) n ≥ ∈ Z N is primary pseudo-polynomial if and only if itsbinomial transform ( b n ) n ≥ ∈ Z N is such that ∀ n ≥ , P n | b n . (1.7) A power series P ∞ n =0 a n x n ∈ Q [[ x ]] is said to be a G -function when it is solution of a non-zerolinear differential equation over Q ( x ) ( D -finiteness), and the maximum of the modulus of all the Galoisianconjuguates of a , . . . , a n as well as the positive denominator of a , . . . , a n are both bounded for all n ≥ C n +1 , for some C ≥
1. For instance, any D -finite series in Z [[ x ]] with positive radius of convergenceis a G -function. A power series P ∞ n =0 a n n ! z x ∈ Q [[ x ]] is said to be an E -function when P ∞ n =0 a n x n is a G -function. See [1, 17] for the properties satisfied by these functions. ii ) Given a primary pseudo-polynomial ( a n ) n ≥ , if there is no Q ( X ) ∈ Q [ X ] such that a n = Q ( n ) for all n large enough, then lim inf n → + ∞ | b n | /n ≥ e. ( iii ) If a primary pseudo-polynomial ( a n ) n ≥ is such that lim sup n → + ∞ | a n | /n < e − ,then a n is eventually a polynomial. ( iv ) Given any function ϕ : N → R with ϕ (0) = 1 , there exists a non-polynomialprimary pseudo-polynomial ( A n ) n ≥ such that ϕ ( n ) ≤ A n ≤ ϕ ( n ) + 2 P n for all n ∈ N . We recall that d n = Q p ≤ n p ⌊ log p ( n ) ⌋ . Hence, for all n ≥ P n divides d n , but obviously d n divides P n for no n ≥
4. Choosing b n := P n in (1.7), the resulting sequence in (1.2) a n := P nk =0 (cid:0) nk (cid:1) P k is a primary pseudo-polynomial, but not a pseudo-polynomial because itis does not satisfy Hall’s criterion (1.3). Under the assumption in ( ii ), if we also assume that b n is eventually of the same sign, then lim inf n → + ∞ | a n | /n ≥ e +1 because a n = P nk =0 (cid:0) nk (cid:1) b k .Consequently, any putative counter-example ( a n ) n ≥ to Conjecture 2 must be such thatits binomial transform ( b n ) n ≥ changes sign infinitely often. A similar remark applies toRuzsa’s Conjecture 1.Assertion ( iii ) is the analogue of the Hall-Ruzsa result recalled at Eq. (1.4).In ( iv ), given ϕ , the existence of sequence ( A n ) n ≥ is proved constructively by aninductive process. An important consequence of ( iv ) is the existence of a non-polynomialprimary pseudo-polynomial of any growth ϕ ( n ) > n ϕ ( n ) /n ≥ e . Inparticular, with ϕ ( n ) = P n , we deduce that < e cannot be replaced by ≤ e on the right-hand side of (1.6) in Conjecture 2. • Primary pseudo-polynomials with an algebraic generating series.
The gen-erating functions f a and f b of the sequences ( a n ) n ≥ and its binomial transform ( b n ) n ≥ satisfy the relations f b ( x ) = 11 + x f a (cid:16) x x (cid:17) , f a ( x ) = 11 − x f b (cid:16) x − x (cid:17) . (1.8)In particular f a ( x ) is algebraic over Q ( x ) if and only if f b ( x ) is algebraic over Q ( x ). Theorem 2.
Let ( a n ) n ≥ ∈ Z N be a primary pseudo-polynomial, and let ( b n ) n ≥ ∈ Z N beits binomial transform. ( i ) If f a ( x ) is algebraic over Q ( x ) , then a n is eventually in Q [ n ] (and thus f a ( x ) ∈ Q ( x ) ). ( ii ) If f b ( x ) is algebraic over Q ( x ) , then f b ( x ) is in Z [ x ] . Theorem 2 implies in particular that the Taylor coefficients of an irrational algebraicfunction over Q ( x ) do not define a primary pseudo-polynomial, and a fortiori not a pseudo-polynomial either. It also implies that if there exists a counter example to Conjecture 2 orto Ruzsa’s Conjecture 1, then its generating function f a ( x ) is transcendental over C ( x ). Aslight generalization of Theorem 2( i ) is presented in § P n ,...,n k ≥ u n ,...,n k z n · · · z n k k is the univariate power series P ∞ n =0 u n,n,...,n z n . A classical result of Furstenberg [4] saysthat algebraic series coincide with diagonals of rational functions in two variables. It wouldbe very interesting to know if the conclusions of Theorem 2 also hold when f a and f b areassumed to be diagonals of rational functions of an arbitrary (but fixed) number of variables(which they are, or not, simultaneously). Moreover, diagonal of rational functions areglobally bounded G -functions in the sense of Christol, who conjectured that the converseholds (see [2]). In particular, G -functions with integer coefficients are globally bounded.Therefore given the Perelli-Zannier Theorem quoted just below, Christol’s Conjecture andthe above putative generalization of Theorem 2 would together imply Conjecture 2 andRuzsa’s Conjecture 1. See also related comments in [19, pp. 392–393]. • An effective version of a result of Perelli and Zannier.
In [13], Perelli andZannier sketched the proof of the following result.
Theorem (Perelli-Zannier) . Let ( a n ) n ≥ be a primary pseudo-polynomial such that thereexist c > and < δ < e such that | a n | ≤ cδ n for all n ≥ . Then there exist an integer S ≥ and S + 1 polynomials p ( X ) , . . . , p S ( X ) ∈ Z [ X ] not all zero such that S X j =0 p j ( n ) a n + j = 0 . (1.9) In other words, f a ( x ) is D -finite, and even a G -function. Perelli and Zannier mentioned that it would be possible to provide upper bounds for S and the degree/height of the p j ( X ) in terms of δ , but they did not write them down. Wemake more precise their theorem as follows, where given Q ( X ) = P j q j X j ∈ C [ X ], we set H ( Q ) := max j | q j | . Theorem 3.
In the conditions and notations of the Perelli-Zannier Theorem, for any δ ∈ (1 , e ) , there exists an effectively computable constant H ( δ ) ≥ such that a non-triviallinear recurrence for ( a n ) n ≥ as in (1.9) holds with max j deg( p j ) ≤ max (cid:16) , l δ ) − − log( δ ) m(cid:17) , max j H ( p j ) ≤ H ( δ ) ,S ≤ log (cid:0) H ( δ ) (cid:1) / log( δ ) . (1.10) Moreover, the Perelli-Zannier Theorem is best possible in the sense that its conclusiondoes not necessarily hold if δ = e . To prove the final statement in Theorem 3, we take ϕ ( n ) := P n in Theorem 1( iv ): weobtain a non-polynomial primary pseudo-polynomial A n such that | A n | /n → e . Hence5 A n ) n ≥ does not satisfy a non-zero linear recurrence with coefficients in Q [ n ]. ( )Lower and upper bounds for the function H ( δ ) are given in (4.8) and (4.10) respectivelyin § p j ) in (1.10) is obviously not optimal; in fact, at the cost ofmore complicated computations, Perelli-Zannier [13] and then Zannier [19] obtained betterbounds when δ ≤ e . and δ ≤ e . .The classification of primary pseudo-polynomials with a D -finite generating series is anopen problem. As shown by the above example ( A n ) n ≥ , Theorem 1 rules out the possibilitythat every primary pseudo-polynomial satisfies a linear recurrence with coefficients in Q [ n ].Another example is the primary pseudo-polynomial D n := P nk =0 (cid:0) nk (cid:1) P k : it cannot satisfysuch a linear recurrence because otherwise P n = P nk =0 ( − n − k (cid:0) nk (cid:1) D k would satisfy oneas well, which is not possible because P /nn → e . Since P n ≥ P n = e n + o ( n ) , asimple analytic argument shows that D n = ( e + 1) n + o ( n ) . More specifically, it would beinteresting to know if there is any non-polynomial primary pseudo-polynomial ( a n ) n ≥ suchthat f a ( x ) is a G -function. There exist primary pseudo-polynomials ( a n ) n ≥ such that f a is D -finite but is not a G -function. For instance, e n := ⌊ ( n +1)! e ⌋ = P n +1 k =0 (cid:0) n +1 k (cid:1) k ! ( n ≥
0) is aprimary pseudo-polynomial by Theorem 1, and for all n ≥ e n +2 = ( n + 4) e n +1 − ( n + 2) e n ( e = 2 , e = 5), so that P ∞ n =0 e n x n is D -finite. A method to obtain further examples ispresented at the end of § • A P´olya type result for primary pseudo-polynomials.
Perelli and Zannier alsoproved in [12] that if a (primary) pseudo-polynomial a n = F ( n ) for some entire function F such that lim sup R →∞ R log max | x | = R | F ( x ) | < log( e + 1), then a n is in Q [ n ]. We provehere a result of a similar flavor with a different analyticity condition. Theorem 4.
Let ( a n ) n ≥ ∈ Z N be a primary pseudo-polynomial. Let us assume that thereexists F ( x ) analytic in a right-half plane ℜ ( x ) > u such that a n = F ( n ) for all n > u , and c > , < ρ < log(2 √ e ) such that (cid:12)(cid:12) F ( x ) (cid:12)(cid:12) ≤ c · e ρ ℜ ( x ) (1.11) for ℜ ( x ) > u . Then F ( x ) is in Q [ x ] and a n is eventually in Q [ n ] . We have log(2 √ e ) ≈ .
193 while log( e + 1) ≈ . a n is eventually a polynomial before proving that F is a polynomial. Because of thePerelli-Zannier Theorem recalled before Theorem 3, the assumptions of Theorem 4 arein fact natural in the context of Ruzsa’s Conjecture 1. Indeed, for any given G -function f ( x ) := P ∞ n =0 v n x n ∈ Q [[ x ]], there exists a function λ ( x ) := P pj =1 c j ( x ) · e ρ j x for somefunctions c j ( x ) analytic in ℜ ( x ) > u and of polynomial growth (at most), and such that v n = λ ( n ) for all n > u ; the numbers e − ρ j are the finite singularities of f ( x ) (see [14, § e ρ j ℜ ( x ) is a priori different of a bound involving Indeed, if a solution ( a n ) n ≥ of a linear recurrence with coefficients in Q [ n ] is such that | a n | /n → α finite, then α is an algebraic number. Moreover, if a sequence of rational numbers satisfies a non-zerolinear recurrence of minimal order with coefficients in C [ n ], then these coefficients are necessarily in Q [ n ],up to a common non-zero multiplicative constant. Hence ( A n ) n ≥ , ( d n ) n ≥ and ( P n ) n ≥ do not satisfyany non-zero linear recurrence with coefficients in C [ n ]. e ρ j x | = e ℜ ( ρ j x ) , but they are the same when ρ j ∈ R . In particular, if all the singularities of f ( x ) are positive real numbers, then a bound as in (1.11) holds for λ ( x ) for some ρ ∈ R .In Theorem 1( iv ), take ϕ ( n ) = δ n with e < δ < √ e . This yields of non-polynomialprimary pseudo-polynomial ( a n ) n ≥ such that a n = δ n + o ( n ) as n → + ∞ . Theorem 4 thusimplies that there is no function F ( x ) analytic in a right-half plane on which (1.11) holds,and such that a n = F ( n ) for all large n .Perelli-Zannier’s result and Theorem 4 are similar to P´olya’s celebrated theorem: if anentire function F ( x ) is such that lim sup R → + ∞ R max | x | = R | F ( x ) | < log(2) and F ( N ) ⊂ Z ,then F ( x ) is a polynomial. See [18] for a recent survey on P´olya type results, where aconnection with Rusza’s Conjecture 1 is also mentioned.Theorems 1, 2, 3 and 4 are proved in § § § § §
6, we presenta method to construct a non-polynomial primary pseudo-polynomial starting from everyprimary pseudo-polynomial generated by a G -function (Theorem 6). ( i ) Let ( a n ) n ≥ be a sequence of integers and consider its binomial transform ( b n ) n ≥ .Assume that for every non-negative integer n , P n divides b n . Let p be a fixed primenumber. Hence, for every integer n ≥ p , p divides b n . For every non-negative integer n , ityields a n + p = n + p X k =0 (cid:18) n + pk (cid:19) b k ≡ p − X k =0 (cid:18) n + pk (cid:19) b k mod p ≡ p − X k =0 (cid:18) nk (cid:19) b k mod p ≡ a n mod p, where we used Lucas’ congruence for binomial coefficients: for every u, v in { , . . . , p − } and every non-negative integers m and ℓ , we have (cid:18) u + mpv + ℓp (cid:19) ≡ (cid:18) uv (cid:19)(cid:18) mℓ (cid:19) mod p. It follows that ( a n ) n ≥ is a primary pseudo-polynomial.Conversely, assume that ( a n ) n ≥ is a primary pseudo-polynomial. Let p be a primenumber and n ≥ p be an integer. It suffices to show that p divides b n .7rite n = v + mp with v in { , . . . , p − } and m ≥
1. We obtain that b v + mp = v + mp X k =0 ( − v + mp − k (cid:18) v + mpk (cid:19) a k = p − X u =0 m − X ℓ =0 ( − v − u ( − ( m − ℓ ) p (cid:18) v + mpu + ℓp (cid:19) a u + ℓp + v X u =0 ( − v − u (cid:18) v + mpu + mp (cid:19) a u + mp ≡ p − X u =0 m − X ℓ =0 ( − v − u ( − ( m − ℓ ) p (cid:18) vu (cid:19)(cid:18) mℓ (cid:19) a u + v X u =0 ( − v − u (cid:18) vu (cid:19) a u mod p ≡ v X u =0 ( − v − u (cid:18) vu (cid:19) a u m X ℓ =0 ( − ( m − ℓ ) p (cid:18) mℓ (cid:19) mod p ( (cid:18) vu (cid:19) = 0 for u = v + 1 , . . . , p − ≡ v X u =0 ( − v − u (cid:18) vu (cid:19) a u (1 + ( − p ) m mod p ≡ p ( m ≥ . Hence p divides b n as expected. It follows that, for every non-negative integer n , P n divides b n . The first equivalence in Theorem 1 is proved.( ii ) The binomial transform ( b n ) n ≥ is eventually 0 if, and only if there exists a poly-nomial Q ( X ) in Q [ X ] such that a n = Q ( n ) for every large enough non-negative integer n .Hence, if the latter is false, then ( b n ) n ≥ is not eventually 0 and, since P n divides b n , itfollows that lim inf n → + ∞ | b n | /n ≥ e because P /nn → e as n → + ∞ . ( iii ) If the primary pseudo-polynomial a n is not eventually a polynomial, then by( ii ) above, lim inf n → + ∞ | b n | /n ≥ e . But since b n = P nk =0 ( − n − k (cid:0) nk (cid:1) a k , the assumptionlim sup n → + ∞ | a n | /n < e − n → + ∞ | b n | /n ≤ lim sup n → + ∞ (cid:18) n X k =0 (cid:18) nk (cid:19) | a k | (cid:19) /n < e. This proves that a n is eventually a polynomial.( iv ) The following argument generalizes Hall’s (sketchy) construction of a non-polynomialpseudo-polynomial with growth ≤ e n + o ( n ) in [5, p. 76]. By ( i ), we know that any sequence ofintegers ( B n ) n ≥ such that P n | B n defines a primary pseudo-polynomial A n := P nk =0 (cid:0) nk (cid:1) B k ,which is not eventually a polynomial if (and only if) B n = 0 for infinitely many n . Since A n − depends only on B , B , . . . , B n − , we will recursively construct B n = 0 and thus A n .We have A = B : choosing B = 1, we have ϕ (0) = A = ϕ (0) + 2 P . Let n ≥ B , B , . . . , B n − all non-zero and such that P k | B k and ϕ ( k ) ≤ A k ≤ ϕ ( k ) + 2 P k for all k ∈ { , . . . , n − } .We want to construct an integer B n = 0 such that P n | B n and ϕ ( n ) ≤ A n ≤ ϕ ( n ) + 2 P n .To do this, we first set C n := P n − k =0 (cid:0) nk (cid:1) B k , so that we will have A n = B n + C n . We nowperform the euclidean division of C n by P n : we have C n = u n P n + v n with u n ∈ Z , v n ∈ N and 0 ≤ v n < P n . We set B n := w n P n = 0 where the non-zero integer w n is defined asfollows: if ⌈ ϕ ( n ) − v n P n ⌉ 6 = u n , we take w n = (cid:24) ϕ ( n ) − v n P n (cid:25) − u n , while if ⌈ ϕ ( n ) − v n P n ⌉ = u n , we take w n = 1. Since A n = ( u n + w n ) P n + v n , we see that ϕ ( n ) ≤ A n ≤ ϕ ( n ) + P n in the former case, while ϕ ( n ) + P n ≤ A n ≤ ϕ ( n ) + 2 P n in thelatter case. This finishes the recursive construction of a non-polynomial primary pseudo-polynomial ( A n ) n ≥ such that ϕ ( n ) ≤ A n ≤ ϕ ( n ) + 2 P n for all integer n ≥ From the proof of Theorem 1( i ), we see that for any given prime number p , the assertions“for all n ≥ a n + p ≡ a n mod p ” and “for all n ≥ p , p | b n ” are equivalent. It follows thatthe assertions “for all p ∈ P and all n ≥ a n + p ≡ a n mod p ” and “for all p ∈ P andall n ≥ p , p divides b n ” are equivalent, where P is a same set of prime numbers, and thisgeneralizes Theorem 1( i ). We shall in fact prove Theorem 2 under the weaker assumptionthat there exists an infinite set P of prime numbers such that for all p ∈ P and all n ≥ a n + p ≡ a n mod p .Given u ∈ Z and a prime number p , we set u | p := u mod p . Given a power series F ( x ) := P ∞ n =0 u n x n ∈ Z [[ x ]], we set F | p ( x ) := P ∞ n =0 u n | p x n ∈ F p [[ x ]].We shall first prove ( ii ) for the series f b ( x ) := P ∞ n =0 b n x n . Let P denote an infinite setof prime numbers such that for all n ≥ p ∈ P , we have a n + p ≡ a n mod p . Asalready said, this is equivalent to the fact that for all p ∈ P and all n ≥ p , p divides b n . Itfollows in particular that for any p ∈ P , f b | p ( x ) is a polynomial in F p [ x ] of degree at most p −
1. For simplicity, we denote by Q p ( x ) this polynomial, and by q p its degree.Let us now assume that f b ( x ) is algebraic over Q ( x ). If f b ( x ) is a constant, there isnothing else to prove. We now assume that f b ( x ) is not a constant so that it has degree d ≥
1. There exist an integer δ ∈ { , . . . , d − } , some integers 0 ≤ j < j < . . . < j δ ≤ d −
1, and some polynomials A d ( x ), A j ( x ) , . . . , A j δ ( x ) ∈ Z [ x ] all not identically zero suchthat A d f db = δ X ℓ =1 A j ℓ f j ℓ b (3.1)in Z [[ x ]]. 9e fix p ∈ P such that p > H where H is the maximum of the modulus of thecoefficients of A d ( x ), A j ( x ) , . . . , A j δ ( x ). It follows thatdeg( A d | p ) = deg( A d ) , deg( A j | p ) = deg( A j ) , . . . , deg( A j δ | p ) = deg( A j δ ) . (3.2)We deduce from the reduction of (3.1) mod p that A d | p Q dp = δ X ℓ =1 A j ℓ | p Q j ℓ p (3.3)in F p [[ x ]], and in fact in F p [ x ] because Q p ( x ) ∈ F p [ x ].Case 1). If Q p is identically zero, this means that p divides the coefficients b n for all n ≥ Q p is not identically zero, we deduce from (3.2) and (3.3) thatdeg( A d ) + dq p ≤ max (cid:0) deg( A j ) + j q p , deg( A j ) + j q p , . . . , deg( A j δ ) + j δ q p (cid:1) ≤ max (cid:0) deg( A j ) , deg( A j ) , . . . , deg( A j δ ) (cid:1) + ( d − q p . Hence q p ≤ max (cid:0) deg( A j ) , deg( A j ) , . . . , deg( A j δ ) (cid:1) − deg( A d ) =: N. It follows that for any n > N , p divides b n , where N is independent of p .Since p ∈ P was simply assumed larger than a quantity H depending only on f b , theconclusion of Case 1 and Case 2 is that for any p ∈ P such that p > H and any n > N , p divides b n . Since P is infinite, b n is divisible by infinitely many primes when n > N .Hence b n = 0 for all n > N and f b ( x ) = P Nn =0 b n x n ∈ Z [ x ], as expected.Let us now prove ( i ). If f a ( x ) is algebraic over Q ( x ), then f b ( x ) as well by (1.8). Hence f b ( x ) ∈ Z [ x ] by ( ii ) just proven, i.e. there exists an integer M such that b n = 0 if n > M .Since, for all n ≥ a n = n X k =0 (cid:18) nk (cid:19) b k = min( n,M ) X k =0 (cid:18) nk (cid:19) b k , it follows that for all n ≥ M , we have a n = Q ( n ) with Q ( X ) = P Mk =0 (cid:0) Xk (cid:1) b k ∈ Q [ X ].The first part of Theorem 2 can be generalized has follows. Theorem 5.
Let ( a n ) n ≥ ∈ Z N be a primary pseudo-polynomial. Assume there exists m ≥ such that f ( m ) a ( x ) is algebraic over Q ( x ) . Then a n is eventually in Q [ n ] , and thus f a ( x ) ∈ Q ( x ) . The proof uses the following simple lemma.
Lemma 1.
Let R ( X ) ∈ Q ( X ) be such that R ( n ) ∈ Z for infinitely many integers. Then, R ( X ) ∈ Q [ X ] . roof. We write R = AB with A, B ∈ Q [ X ]. We assume that deg( B ) ≥ U, V ∈ Q [ X ] such that A = U B + V and deg( V ) < deg( B ).Let w ∈ Z \ { } be such that wU, wV ∈ Z [ X ]. Let N be the infinite set of integers n such that R ( n ) ∈ Z ; without loss of generality, we can assume that N contains infinitelymany positive integers. For every n ∈ N , we have wV ( n ) B ( n ) = wR ( n ) − wU ( n ) ∈ Z . Butlim x → + ∞ wV ( x ) B ( x ) = 0. Hence there exists M such that n ∈ N and n ≥ M imply that wV ( n ) B ( n ) = 0. Therefore, wV has infinitely many roots: it must be the null polynomial, sothat R = U ∈ Q [ X ]. Proof of Theorem 5.
We have f ( m ) a ( x ) = ∞ X n =0 ( n + m )( n + m − · · · ( n + 1) a n + m x n . Since ( a n ) n ≥ is a primary pseudo-polynomial, this is also the case of (( n + m )( n + m − · · · ( n + 1) a n + m ) n ≥ because it is a product of two primary pseudo-polynomials. Since f ( m ) a ( x ) is algebraic over Q ( x ), ( n + m )( n + m − · · · ( n + 1) a n + m is eventually in Q [ n ] byTheorem 2. Hence a n + m is eventually in Q ( n ), so that by Lemma 1, a n + m is eventually in Q [ n ]. This completes the proof. The proof of the Perelli-Zannier Theorem is based on the following lemma proved in [13].We shall also use it.
Lemma 2.
For k = ( k j ) j ≥ ∈ ( R + ) N , an integer N ≥ , set A ( N, k ) := { ( x , . . . , x N ) ∈ Z N : | x j | ≤ k j and ∀ p, ∀ n ≤ N − p, x n + p ≡ x n mod p } . Then A ( N, k ) ≤ N Y j =1 (cid:18) k j P j − (cid:19) . Let R ≥ , H ≥ , D ≥ q , . . . , q R ∈ Z [ X ] with max j H ( q j ) ≤ H andmax j deg( q j ) ≤ D −
1. Considering the coefficients of the q j ’s as indeterminates, there are(2 H + 1) RD functions F of the form F ( n ) := R − X j =0 q j ( n ) a n + j . (4.1)Such a function satisfies | F ( n ) | ≤ cRDH ( n + 1) D δ n + R for all n ≥ F ( n + p ) ≡ F ( n )mod p for all prime number p and all n ≥
0. Hence for all N ≥
1, ( F (0) , . . . , F ( N − ∈ ( N, K ) where K j := cRDHj D δ j + R − . Therefore, given N , if N Y j =1 (cid:18) K j P j − (cid:19) < (2 H + 1) RD (4.2)there exists two different functions F and F of the form (4.1) such that F ( n ) = F ( n )for all n ∈ { , , . . . , N − } . Hence the function G N := F − F is of the form (4.1) G N ( n ) = P R − j =0 q j ( n ) a n + j with q ( X ) , . . . , q R − ( X ) ∈ Z [ X ] not all identically zero, with | G N ( n ) | ≤ cRDH ( n + 1) D δ n + R for all n ≥ G N ( n ) = 0 for every integer n in { , , . . . , N − } . Note that G N depends on N which is fixed but can be as large as desiredin this construction.Eq. (4.2) holds if we assume the stronger condition ∞ Y j =1 (cid:18) K j P j − (cid:19) ≤ H RD , (4.3)because the assumption δ < e implies the convergence of the productΦ( D, x ) := ∞ Y j =1 (cid:18) x j D δ j P j − (cid:19) for all x ≥ D ≥
0, and obviously 1 + x j D δ j P j − ≥
1. We shall provide an upper boundfor Φ(
D, x ) in § H, R, D such that (4.3) holds. Itis important to observe here that (4.3) does not depend on N . G N ( n ) = 0 for all n ≥ Following the Perelli-Zannier method, we now want to prove that, provided N is largeenough, G N ( n ) vanishes for all n ≥
0. Assume this is not the case. Then for every N , let M N ≥ N denote the largest integer such that G N (0) = G N (1) = . . . = G N ( M N ) = 0 but G N ( M N + 1) = 0. We fix α ∈ (cid:0) , δ ) − (cid:1) .We shall first prove that G N ( m ) = 0 for m in I := [2 M N , (2 + α ) M N ]. Let m ∈ I .First assume that p is a prime and p < M N : since G N (0) = · · · = G N ( p ) = 0 and G N ( n + p ) ≡ G N ( n ) mod p , p divides G N ( m ). Assume now that M N ≤ p ≤ m/
2, so that0 ≤ m − p ≤ m − M N ≤ αM N ≤ M N , hence G N ( m ) ≡ G N ( m − p ) = 0 mod p . Assume to finish that m − M N < p ≤ m (suchprimes have not yet been considered) so that 0 ≤ m − p < M N and G N ( m ) ≡ G N ( m − p ) ≡ p . It follows that G N ( m ) si divisible by P m/ P m /P m − M N .Therefore, if G N ( m ) = 0 for some m ∈ I , then | G N ( m ) | ≥ e m/ M N + o ( M N ) ≥ e M N + o ( M N ) , o ( M N ) denotes a term such that o ( M N ) /M N becomes arbitrarily small when N istaken arbitrarily large. But on the other hand, we know that, for any m ∈ I , | G N ( m ) | ≤ cRDH ( m + 1) D δ m + R ≤ cRDH (2 M N + αM N + 1) D δ R δ (2+ α ) M N . We recall that we assume that
H, R, D are such that (4.3) holds, which is independentof N . Hence we can let N → + ∞ , hence a fortiori M N → + ∞ so that the above lowerand upper bounds for G N ( m ) = 0 imply that e ≤ δ α , i.e. that α ≥ δ ) −
2, which iscontrary to the assumption on α . Hence, provided N is large enough, we have G N ( m ) = 0for all m ∈ I .We thus have G N ( m ) = 0 for all integers m in [0 , . . . , M N ] or [2 M N , (2 + α ) M N ]. Itfollows that for any p ≤ (1 + α ) M N − p divides G N ( M N + 1). Indeed, if p ≤ M N , wewrite M N + 1 = n + p for some n ≤ M N − G N ( M N + 1) ≡ G ( n ) = 0 mod p ,while if M N < p ≤ (1 + α ) M N −
1, we have 0 = G N ( M N + 1 + p ) ≡ G ( M N + 1) mod p because M N + 1 + p ∈ [2 M N , (2 + α ) M N ]. Hence, because G N ( M N + 1) = 0, we have | G N ( M N + 1) | ≥ P (1+ α ) M N − ≥ e (1+ α ) M N + o ( M N ) . On the other hand, | G N ( M N + 1) | ≤ c RDH ( M N + 2) D δ M N +1+ R . As above, we take N large enough so that these two bounds imply that α ≤ log( δ ) − δ ) − < α was chosen positive.Therefore, there is no such M N such that G ( M N + 1) = 0, so that G ( n ) = 0 for allinteger n ≥ Φ( D, x ) In this section, x > D ≥ ε > δ < e − ε , and we let ω = δ/ ( e − ε ) <
1. By the Prime Number Theorem, P j − ≥ ( e − ε ) j for all j > J = J ( ε ) so thatΦ( D, x ) := ∞ Y j =1 (cid:18) xj D δ j P j − (cid:19) ≤ J Y j =1 (cid:18) xj D δ j P j − (cid:19) ∞ Y j = J +1 (cid:0) xj D ω j (cid:1) . We have J Y j =1 (cid:18) xj D δ j P j − (cid:19) ≤ J Y j =1 (cid:0) xj D δ j (cid:1) ≤ J Y j =1 (cid:0) x ) j D δ j (cid:1) ≤ J Y j =1 (cid:0) x ) j D δ j (cid:1) , since (1 + x ) j D δ j ≥ j in { , . . . , J } . Hence, J Y j =1 (cid:0) x ) j D δ j (cid:1) = 2 J (1 + x ) J J ! D δ J ( J +1) / ≤ J (1 + x ) J J ! D δ J .
13n addition, we have ∞ Y j = J +1 (cid:0) xj D ω j (cid:1) ≤ ∞ Y j =1 (cid:0) xj D ω j (cid:1) , which yields Φ( D, x ) ≤ J (1 + x ) J J ! D δ J ∞ Y j =1 (cid:0) xj D ω j (cid:1) . We now bound the infinite product Ψ(
D, x ) := Q ∞ j =1 (cid:0) xj D ω j (cid:1) . The maximum ofthe function t t D ω t/ is m ( D ) := (2 D/ ( e log(1 /ω ))) D , attained at j := 2 D/ log(1 /ω ).Hence, for all j ≥ j , we have j D ≤ m ( D ) ω − j/ . Moreover, t t D ω t/ is increasing on[0 , j ] and m ( D ) ≤ j D . Hence,Ψ( D, x ) ≤ Y ≤ j 12 log( y ) − Li ( − /y ) − ζ (2) , y > − ( − y ) ≤ log( y ) + 4 ζ (2) , y ≥ , because Li ( − /y ) + ζ (2) ≤ ζ (2) when y ≥ 1. We obtain that for y ≥ ∞ Y j =1 (cid:0) yω j/ (cid:1) ≤ c e log( y ) / log(1 /ω ) , c := exp(4 ζ (2) / log(1 /ω )) ≥ ≤ Φ( D, x ) ≤ J (1 + x ) J J ! D δ J (2(1 + x ) j D ) ⌊ j ⌋ c e log( xj D ) / log(1 /ω ) , (4.4)where we recall that ω = δ/ ( e − ε ) where ε > δ < e − ε . For ease of reading, we set d, r, h for D, R, H . We want to find conditions on d, r and h suchthat Φ( d, x ) ≤ h rd when x = 2 crdhδ r − (which corresponds to (4.3)). It will be enough tofind conditions on d, r, h and ε such that the right-hand side of (4.4) is ≤ h rd .From now on, we set ℓ := log( δ ) < 1. We assume that d and ρ depend on ℓ but areindependent of h , and we let r := ⌊ ρ log( h ) ⌋ + 1. Since J and j are also fixed, when h → + ∞ , we havelog (cid:16) c J (1 + x ) J J ! d δ J (2(1 + x ) j d ) ⌊ j ⌋ e log( xj d ) / log(1 /ω ) (cid:17) ∼ (1 + ρℓ ) log(1 /ω ) log( h ) , while log( h rd ) ∼ dρ log( h ) . Hence for our goal, it suffices to choose d, ρ and ε such that(1 + ρℓ ) log(1 /ω ) < dρ. (4.5)Recall that ω = δ/ ( e − ε ) so that log(1 /ω ) → − ℓ as ε → 0. So, by choosing d and ρ such that (1 + ρℓ ) − ℓ ≤ dρ, (4.6)we can choose ε > ℓ ρ + (2 ℓ − d (1 − ℓ )) ρ + 1 ≤ , which defines a polynomial in ρ whose discriminant is ∆ := d (1 − ℓ )( d (1 − ℓ ) − ℓ ). Taking d := max(1 , ⌈ ℓ − ℓ ⌉ ) ensures that (4.6) holds true for any choice of ρ > " d (1 − ℓ ) − ℓ − √ ∆2 ℓ , d (1 − ℓ ) − ℓ + √ ∆2 ℓ . For simplicity, we also restrict ρ to be ≤ /ℓ which is possible because the product of thoseroots is 1 /ℓ and, since d (1 − ℓ ) ≥ ℓ , we have d (1 − ℓ ) − ℓ + √ ∆2 ℓ ≥ ℓ and d (1 − ℓ ) − ℓ − √ ∆2 ℓ ≤ ℓ . ε , d and ρ , we now define H ( δ ) as the smallest integer h ≥ c J (1 + x ) J J ! d δ J (2(1 + x ) j d ) ⌊ j ⌋ e log( xj d ) / log(1 /ω ) ≤ h rd , (4.7)where x = 2 crdhδ r − . We then obtain (1.10) with max deg( p j ) ≤ d − , ⌈ δ ) − − log( δ ) ⌉ )and S ≤ r − ≤ ⌊ ρ log( h ) ⌋ ≤ log( H ( δ )) /ℓ . Notice that H ( δ ) also depends on the choiceof ρ and we now explain how to bound it.The left-hand side of (4.7) is an increasing function of h ≥ 1, which appears in theexpressions of r := ⌊ ρ log( h ) ⌋ + 1 and x := 2 crdhδ r − . Hence, H ( δ ) d (1+ ⌊ ρ log( H ( δ )) ⌋ ) is largerthan the value A (which is ≥ 1) of the left-hand side of (4.7) at h = 1, in which case r = 1and x = 2 cd . It follows that log( A ) ≤ d log( H ( δ )) + dρ log( H ( δ )) , so that H ( δ ) ≥ exp (cid:16) p d + 4 dρ log( A ) − ddρ (cid:17) . (4.8)Since A → + ∞ when δ → e and ε → / log( ω )), H ( δ ) can be verylarge.We now explain how to bound Y := H ( δ ) from above. We assume that H ( δ ) ≥ ≥ 1, so that wecan take the logarithms of both sides. After some transformations, we obtain a function S ( h ) ≥ h ≥ Y is the smallest integer h ≥ S ( h ) ≤ (cid:16) ρd − (1 + ρℓ ) log(1 /ω ) (cid:17) log( h ) . (4.9)Recall that γ := ρd − (1+ ρℓ ) log(1 /ω ) > 0. Moreover, there exist α, β that depend on ρ, d, δ, ε (andcould be explicited as well) such that S ( h ) ≤ α log( h ) + β for all h ≥ 1. Since Y ≥ γ log( Y − < s ( Y − ≤ α log( Y − 1) + β. Hence, log( Y − 1) is smaller than the largest solution of the quadratic equation γX − αX − β = 0, so that finally H ( δ ) ≤ (cid:16) α + p α + 4 βγ γ (cid:17) . (4.10) Let F ( z ) be as in the theorem such that F ( n ) = a n for all n > u . Notice that e a n := a n + ⌈ u ⌉ +2 , n ≥ 0, is a primary peudo-polynomial; it is eventually in Q [ n ] if and only if a n iseventually in Q [ n ]. The function e F ( z ) := F ( z + ⌈ u ⌉ + 2) is analytic in ℜ ( z ) > − 2, satisfies | e F ( z ) | ≤ ˜ c · exp( ρ ℜ ( z )) in ℜ ( z ) > − c > 0, and e a n = e F ( n ) for all n ≥ 0. Moreover, F ( z ) is in Q [ z ] if and only if e F ( z ) is in Q [ z ]. Therefore, without loss of16enerality, we can and will assume that F ( z ) is analytic in ℜ ( z ) > − F ( n ) = a n for all integers n ≥ C n denote the circle of center n and radius n oriented in the direct sense. Thefunction F ( z − 1) being analytic in ℜ ( z ) > − 1, the residue theorem yields( n − iπ Z C n F ( z − z − · · · ( z − n ) dz = n − X k =0 ( − n − k − (cid:18) n − k (cid:19) a k := b n − ∈ Z . We parametrize the circle C n as n + ne ix = 2 n cos( x ) e ix for x ∈ [ − π/ , π/ (cid:12)(cid:12)(cid:12)(cid:12) ( n − z − · · · ( z − n ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c ( n ) e − n cos( x ) ψ ( x ) , z = 2 n cos( x ) e ix ∈ C n , where c ( n ) > n and ψ ( x ) := cos( x ) log(2 cos( x )) + x sin( x ) . (See also the proof given in [15].) The minimum on [ − π/ , π/ 2] of ψ ( x ) is log(2) at x = 0.We have | F ( z − | ≤ c · e ρ ℜ ( z − = ce − ρ · e nρ cos( x ) , z = 2 n cos( x ) e ix ∈ C n . Hence | b n − | ≤ c ( n ) max x ∈ [ − π/ ,π/ e n cos( x )( ρ cos( x ) − ψ ( x )) where c ( n ) > n . Notice that 0 ≤ cos( x ) ≤ − π/ , π/ ρ > 0. Hence if x is such that ρ cos( x ) − ψ ( x ) ≥ 0, thencos( x )( ρ cos( x ) − ψ ( x )) ≤ ρ − ψ ( x ) ≤ ρ − log(2), whereas if x is such that ρ cos( x ) − ψ ( x ) ≤ x )( ρ cos( x ) − ψ ( x )) ≤ 0. Therefore, for all x ∈ [ − π/ , π/ e x )( ρ cos( x ) − ψ ( x )) ≤ max(1 , e ρ − log(2)) )and | b n − | ≤ c ( n ) max(1 , e ρ − log(2)) ) n . Since 2( ρ − log(2)) < 1, it follows thatlim sup n → + ∞ | b n | /n < e. But because ( a n ) n ≥ is a primary pseudo-polynomial, we know by Theorem 1( ii ) that if b n is not eventually equal to 0 then lim inf n → + ∞ | b n | /n ≥ e. This implies that b n is indeed eventually equal to 0, thus that there exist P ( X ) ∈ Q [ X ]and an integer N ≥ a n = P ( n ) for all n ≥ N .Consider now the function g ( z ) := F ( z + N ) − P ( z + N ) which is analytic in ℜ ( z ) > − g ( n ) = 0 for every integer n ≥ 1. Moreover, since ρ > 0, thereexists a constant d > | g ( z ) | ≤ d · exp( ρ | z | ) for any z such that ℜ ( z ) > 0. Since ρ < + log(2) < π , we can then apply a classical result of Carlson (see Hardy [6, p. 328])and deduce that g ( z ) = 0 identically. Hence F ( z ) reduces to a polynomial function in Q [ z ].This completes the proof of Theorem 4. 17 Construction of non-polynomial primary pseudo-polynomials We conclude this paper by presenting a method to construct a non-polynomial primarypseudo-polynomial starting from a given primary pseudo-polynomial ( a n ) n ≥ such that a = 1. The justification of the method uses a non-trivial property satisfied by E -functions.Let as usual ( b n ) n ≥ be the binomial transform (1.1) of ( a n ) n ≥ . Let F b ( x ) := P ∞ n =0 b n n ! x n .We assume that a = 1, so that b = 1 as well. We define the sequence ( c n ) n ≥ formally by F c ( x ) := ∞ X n =0 c n n ! x n = 1 F b ( x ) . Let us now define ( u n ) n ≥ as the inverse binomial transform (1.2) of ( c n ) n ≥ , i.e. u n := n X k =0 (cid:18) nk (cid:19) c k . Then we have the following. Theorem 6. Let ( a n ) n ≥ ∈ Z N be a primary pseudo polynomial such that a = 1 . Then, ( u n ) n ≥ ∈ Z N is a primary pseudo-polynomial. Moreover, assuming also that f a ( x ) is a G -function, if lim sup n | u n | /n < e , then a n = u n = 1 for all n ≥ . Consequently, if f a ( x ) is a G -function not equal to − x , then ( u n ) n ≥ ∈ Z N is a primarypseudo-polynomial such that lim sup n | u n | /n ≥ e (hence not a polynomial). We explainafter the proof of the theorem why u n grows like n ! in this case.So far, the assumption that f a ( x ) is a G -function is known to be satisfied only whenlim sup n | a n | /n < e (when the Perelli-Zannier Theorem can be applied), which in turnimplies that a n should eventually be in Q [ n ] by Conjecture 2. Hence, in practice thesecond assertion of Theorem 6 is useful only when a n is already known to be eventually in Q [ n ] in which case F b ( x ) ∈ Q [ x ]. For instance, if a n = n + 1, then ∞ X n =0 c n n ! x n = 11 + x and u n = n X k =0 ( − k (cid:18) nk (cid:19) k ! . (6.1) Proof of Theorem 6. We first prove that for all n ≥ c n ∈ Z and for every prime p ≤ n , p divides c n . By definition of the c n ’s, we have1 = ∞ X n =0 b n n ! x n ! ∞ X n =0 c n n ! x n ! = ∞ X n =0 n ! n X k =0 (cid:18) nk (cid:19) b k c n − k ! x n , so that c = 1 and, for every integer n ≥ 1, we have n X k =0 (cid:18) nk (cid:19) b k c n − k = 0 . b = 1) c n = − n X k =1 (cid:18) nk (cid:19) b k c n − k , ( n ≥ . It follows that, for all n ≥ c n is an integer.Let n be a positive integer such that, for every positive integer m < n and every prime p ≤ m , p divides c m . Consider a prime number p ≤ n and an integer k in { , . . . , n } .If p ≤ k then p divides b k . If p ≤ n − k , then p divides c n − k . If p > max( k, n − k ),then p divides (cid:0) nk (cid:1) because p divides n ! but neither k ! nor ( n − k )!. In all cases, p divides (cid:0) nk (cid:1) b k c n − k , so that p divides c n . By strong induction on n , it follows that, for every integer n ≥ p ≤ n , p divides c n (this property holds trivially if n = 0 or 1). ByTheorem 1( i ), the sequence ( u n ) n ≥ ∈ Z N is a primary pseudo-polynomial. Notice that sofar we had no need to assume that f a is a G -function.We now complete the proof of Theorem 6 when f a is also a G -function. We first observethat F b ( z ) is an E -function: indeed, f b ( x ) is a G -function because f b ( x ) := ∞ X n =0 b n x n = 11 + x f a (cid:18) x x (cid:19) and f a ( x ) is a G -function. ( ) Let us assume that lim sup n | u n | /n < e . By the Perelli-Zannier Theorem quoted in the Introduction, f u ( x ) := P ∞ n =0 u n x n is a G -function. Fromthe equation c n := P nk =0 ( − n − k (cid:0) nk (cid:1) u k ( ∀ n ≥ f c ( x ) := ∞ X n =0 c n x n = 11 + x f u (cid:18) x x (cid:19) is also a G -function. Hence F c ( x ) := 1 /F b ( z ) is also an E -function. Therefore, F b ( x ) is aunit of the ring of E -functions, i.e. it is of the form βe αz with α ∈ Q and β ∈ Q ∗ by [1, p.717] (see also Footnote 4). Therefore, b n = βα n for all n ≥ 0, with the usual conventionthat α = 1 if α = 0. Since all primes p ≤ n divide b n , we deduce that α = 0 is the onlypossibility. Hence 1 = b = β and b n = 0 for all n ≥ 0, so that a n = 1 for all n ≥ 0. Itnow remains to observe the following facts: “ a n = 1 for all n ≥ 0” is equivalent to “ b = 1and b n = 0 for all n ≥ 1” which is equivalent to “ c = 1 and c n = 0 for all n ≥ u n = 1 for all n ≥ u n as n → + ∞ .Since F u ( x ) := P ∞ n =0 u n n ! x n = e x F c ( x ) = e x /F b ( x ), the asymptotic behavior of u n /n ! isdetermined by the zeroes of smallest modulus of F b ( x ), when it has at least one. Noticethat an E -function F ( x ) with no zero in C must be of the form βe αx with α ∈ Q , β ∈ Q ∗ .Indeed, an E -function is an entire function satisfying | F ( x ) | ≪ e ρ | x | for some ρ > 0, so Given a G -function f ( x ) and α ( x ) an algebraic function over Q ( x ) regular at x = 0, f ( xα ( x )) is a G -function. F ( x ) = βx m e αx Q j ≥ (cid:0) (1 − xx j ) e x/x j (cid:1) wherethe x j ’s are the zeroes of F ( x ), m ∈ N and α, β ∈ C . ( ) Therefore, a “no zero” assumptionimplies that F ( x ) = βe αx with β = 0, and since F ( x ) ∈ Q [[ x ]], α, β must be in Q . Comingback to F b ( x ), we have seen during the proof of Theorem 6 that this case implies that α = 0 , β = 1, hence that f a ( x ) = − x . Therefore, assuming that f a ( x ) is a G -function different of − x , the E -function F b ( x )has at least one zero. Let x , . . . , x m denote the zeroes of F b ( x ) of the same moduluswhich is the smallest amongst all modulus of the zeroes. Classical transfer theorems in [3,Chapter VI] enable to deduce the asymptotic behavior of u n . For instance, if the x j ’s aresimple zeroes of F b ( x ), then u n = n ! m X j =1 e x j F ′ b ( x j ) 1 + o (1) x nj . This is coherent with (6.1) above (where F b ( x ) = 1 + x ) because we can rewrite it as u n = ( − n n ! n X k =0 ( − n − k ( n − k )! ∼ ( − n e n ! , n → + ∞ . Because of the different arguments of the x j ’s, oscillations can occur. In presence of zeroesof F b ( x ) of higher multiplicities, similar but more complicated expressions can be given.Finally, even though F b ( x ) /e x is an E -function, we don’t expect e x /F b ( x ) to be D -finite ingeneral, ( ) but this is obviously the case if F b ( x ) is a polynomial. Acknowledgements. Both authors have partially been funded by the ANR project DeRerum Natura (ANR-19-CE40-0018). 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