Abstract 1 Introduction 2 Background 3 Large Zeros and a Universal Skolem Set of Density One 4 Bad Primes and Good Primes 5 The Cramér Argument References

On Large Zeros of Linear Recurrence Sequences

Florian Luca ORCID Mathematics Division, Stellenbosch University, South Africa
Max Planck Institute for Software Systems, Saarbrücken, Germany
Joël Ouaknine ORCID Max Planck Institute for Software Systems, Saarbrücken, Germany James Worrell ORCID Department of Computer Science, Oxford University, UK
Abstract

The Skolem Problem asks to determine whether a given integer linear recurrence sequence (LRS) has a zero term. This problem, whose decidability has been open for many decades, arises across a wide range of topics in computer science, including loop termination, formal languages, automata theory, and probabilistic model checking, amongst many others.

In the present paper, we introduce a notion of “large” zeros of (non-degenerate) linear recurrence sequences, i.e., zeros occurring at an index larger than a sixth-fold exponential of the size of the data defining the given LRS. We establish two main results. First, we show that large zeros are very sparse: the set of positive integers that can possibly arise as large zeros of some LRS has null density. This in turn immediately yields a Universal Skolem Set of density one, answering a question left open in the literature. Second, we define an infinite set of prime numbers, termed “good”, having density one amongst all prime numbers, with the following property: for any large zero of a given LRS, there is an interval around the large zero together with an upper bound on the number of good primes possibly present in that interval. The bound in question is much lower than one would expect if good primes were distributed similarly as ordinary prime numbers, as per the Cramér model in number theory. We therefore conjecture that large zeros do not exist, which would entail decidability of the Skolem Problem.

Keywords and phrases:
Skolem Problem, linear recurrence sequences, decidability, Cramér conjecture
Funding:
Florian Luca: Supported by ERC grant DynAMiCs (101167561).
Joël Ouaknine: Also affiliated with Keble College, Oxford as emmy.network Fellow, and supported by ERC grant DynAMiCs (101167561) and DFG grant 389792660 as part of TRR 248.
James Worrell: Supported by UKRI Frontier Research Grant EP/X033813/1.
Copyright and License:
[Uncaptioned image] © Florian Luca, Joël Ouaknine, and James Worrell; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Mathematics of computing Discrete mathematics
Editors:
Paweł Gawrychowski, Filip Mazowiecki, and Michał Skrzypczak

1 Introduction

An (integer) linear recurrence sequence (LRS) unn=0 is a sequence of integers satisfying a recurrence of the form

un+k=a1un+k1++akun (1)

where the coefficients a1,,ak are integers. The celebrated theorem of Skolem, Mahler, and Lech [28, 20, 15] describes the set of zero terms of such a recurrence:

Theorem 1.

Given an integer linear recurrence sequence unn=0, the set {n:un=0} is a union of finitely many arithmetic progressions together with a finite set.

The statement of Thm. 1 can be refined by considering the notion of non-degeneracy of an LRS. An LRS is non-degenerate if in its minimal recurrence no quotient of two distinct roots of the characteristic polynomial is a root of unity.111For basic definitions, facts, and properties concerning linear recurrence sequences, we refer the reader to standard texts such as [9, Chaps. 1 and 2], [14, Chap. 4], or [29, Chap. 4]. A given LRS can be effectively decomposed as the interleaving of finitely many non-degenerate sequences, some of which may be identically zero. The core of the Skolem-Mahler-Lech theorem is the fact that a non-zero non-degenerate linear recurrence sequence has finitely many zero terms. Unfortunately, all known proofs of this last result are ineffective: it is not known how to compute the finite set of zeros of a given non-degenerate linear recurrence sequence. It is readily seen that existence of a procedure to do so is equivalent to the existence of a procedure to determine whether an arbitrary given LRS has a zero term; the latter is known as the Skolem Problem. We refer to [4, Chap. 6] and [30, Chap. X] for expository accounts of the Skolem-Mahler-Lech theorem and discussion of the ineffectiveness of known proofs.

In computer science, the Skolem Problem lies at the heart of key decision problems in formal power series [26, 3], stochastic model checking [25], control theory [6, 10], and loop termination [24]. The problem is also closely related to membership problems on commutative matrix groups and semigroups, as considered in [7, 13]. We note that in several of the above-mentioned citations, the Skolem Problem is used as a reference benchmark to establish hardness of other open decision problems.

Decidability of the Skolem Problem is known only for certain special cases, based on the relative order of the absolute values of the characteristic roots. Say that a characteristic root λ is dominant if its absolute value is maximal among all the characteristic roots. Decidability is known in case there are at most 3 dominant characteristic roots, and also for recurrences of order at most 4 [21, 31]. However for LRS of order 5 it is not currently known how to decide the Skolem Problem. For a (highly restricted) subclass of LRS, the paper [1] obtains nearly matching complexity lower and upper bounds for the problem.

Some recent lines of research have succeeded in establishing conditional decidability of the Skolem Problem for simple LRS (i.e., LRS none of whose characteristic roots are repeated), assuming certain classical number-theoretic conjectures [16, 5]. Nevertheless, to the best of our knowledge, no putative algorithm has to date been proposed to solve the Skolem Problem in full generality.

A different approach was initiated in [18, 19, 17] via the notion of Universal Skolem Sets. An infinite, recursive set 𝒮 is a Universal Skolem Set if there is some algorithm which, given any LRS, determines whether or not the LRS has a zero in 𝒮. Decidability of the Skolem Problem is then of course equivalent to the assertion that is itself a Universal Skolem Set. The authors of [18] succeded in exhibiting a sparse Universal Skolem Set, i.e., a set having null density, and left open the question of whether Universal Skolem Sets of strictly positive density, or even density one, could be constructed (the interest in high-density Universal Skolem Sets being that they approximate more closely). The question was partially answered in [19], which presented a positive-density Universal Skolem Set, albeit restricted to simple LRS, and in [17], which exhibited a Universal Skolem Set of strictly positive density, and even established density 1 subject to the Bateman-Horn conjecture in number theory.

In this paper we propose an explicit bound for the largest zero of a non-degenerate LRS in terms of the data describing the LRS. We call zeros that exceed this bound large zeros of the LRS. Evidently, decidability of the Skolem Problem would follow from a proof that large zeros do not exist. Using known upper bounds on the cardinality of the set of zeros of non-degenerate LRS, it is relatively straightforward to show that the set of integers arising as large zeros of some non-degenerate LRS has null density, which in turn yields a Universal Skolem Set of unconditional density one; this is the first of our two main contributions.

While a proof that large zeros do not exist currently seems well out of reach, we give a heuristic argument as to why this should nevertheless be expected. This argument is based on an analogue of the well-known Cramér conjecture on gaps between consecutive primes. This conjecture, originally formulated by Cramér in 1936 [8] and subsequently refined by various number theorists into its present form, asserts that, for some constant κ>1, for every prime p the distance to the next largest prime is at most κ(logp)2. The conjecture is based on the heuristic that the sequence of prime numbers behaves similarly to a Poisson-like random process in which the probability of a number x being prime is 1/logx. The largest observed prime gaps are of the order of 0.5(logp)2 [23], however the best known upper bound on prime gaps is O(p0.525), due to Baker, Harman, and Pintz [2], which is far from Cramér’s conjectured bound. Cramér himself proved that, under the Riemann hypothesis, prime gaps are bounded above by O(p0.5logp) [8]. On the other hand, the best known lower bound is Ω(logploglogp), which is some way from the conjectured upper bound. We refer to [12] for a discussion of Cramér’s conjecture and its refinements.

Here we define a subset of so-called good primes based on divisibility properties of LRS. We show that the set of good primes has density one in the set of all primes, or in other words that, asymptotically speaking, almost all primes are good primes. We further show that if the Cramér conjecture applies also to gaps between consecutive good primes, then large zeros of LRS cannot exist. The proof of the latter result (our second main contribution) proceeds by establishing an upper bound on the number of good primes in the neighbourhood of a large zero that violates the conjectured upper bound on gaps between good primes. In other words, if good primes are distributed according to Cramér’s heuristic then large zeros cannot exist and the Skolem Problem is decidable.

2 Background

We will need some basic notions concerning algebraic numbers. All material can be found in [11]. Recall that a number field 𝕂 is a subfield of that is finite dimensional as a vector space over . We assume that 𝕂 is a Galois extension of , that is, it arises as the splitting field of a polynomial with integer coefficients. All elements of 𝕂 are algebraic over , that is, they arise as roots of polynomials with integer coefficients. Those elements that arise more specifically as roots of monic polynomials with integer coefficients are called algebraic integers. The algebraic integers in 𝕂 form a subring, denoted 𝒪𝕂.

For a number field 𝕂, we denote by Gal(𝕂/) the group of field automorphisms of 𝕂. Given α𝕂, the norm of α is defined by

𝒩𝕂/(α)=σGal(𝕂/)σ(α).

The norm 𝒩𝕂/(α) is rational for all α𝕂; moreover 𝒩𝕂/(α)=0 iff α=0, and 𝒩𝕂/(α) is an integer if α𝒪𝕂. Clearly we have |𝒩𝕂/(α)|Md𝕂, where d𝕂 is the degree of 𝕂 and

M:=maxσGal(𝕂/)|σ(α)|

is the house of α.

We recall that every ideal in 𝒪𝕂 can be written uniquely up to the order of its factors as the product of prime ideals. Given a rational prime P, we say that a prime ideal 𝔭 lies above P if 𝔭 is a factor of P𝒪𝕂. In this case we have that P𝒩𝕂/(α) for all α𝔭.

Let 𝔭 be a prime ideal of 𝒪𝕂 lying above P. Recall that the Frobenius automorphism σGal(𝕂/) corresponding to 𝔭 is such that σ(α)αPmod𝔭 for all α𝒪𝕂; in fact it is the unique Galois automorphism with this property.

3 Large Zeros and a Universal Skolem Set of Density One

For an LRS 𝐮=unn=0 as in (1), define its size222Note that we consider here the magnitude of the numbers defining a given LRS (rather than their bit size as is more common in complexity theory). An alternative definition in terms of bit size would of course be possible, only requiring altering (2) into a seventh-fold exponential. to be

C𝐮:=max{k,|a1|,,|ak|,|u0|,,|uk1|,12}.

Given a (partial) function f: and a positive integer , let f(x)=fff(x), where the iteration is -fold (thus f1=f). We say that n is a zero of 𝐮 if un=0, and we say that it is a large zero if the inequality

n<2exp6(C𝐮) (2)

fails.

As we argue later on, there are good reasons to expect that (2) holds for all zeros of all non-degenerate LRS, which in turn would establish decidability of the Skolem Problem.333The sixth-fold exponential in (2) has of course been chosen in order for our mathematical argument to go through. In actual fact, it is plausible to expect that a single exponential would suffice: as far as we are aware, there is currently no known construction of a family of non-degenerate LRS having zeros at indices of magnitude even a single exponential in terms of the size of the LRS, as defined above. Unfortunately, we are unable to prove this assertion. Nevertheless, we now show that large zeros are very sparse, i.e., have null density amongst the positive integers. To this end, let

:={n:there exists a non-degenerate LRSusuch thatun=0and(2)fails}.

Thus is the set of large zeros of some non-degenerate LRS.

Theorem 2.

The set has null density. In fact, writing (X)=[0,X], the inequality

#(X)=O(X(logX)B)

holds with any constant B>0.

Proof.

Let X>2exp7(1) be an integer, and put C:=log6(X/2). We say that an LRS 𝐮 is small at level X if it has size C𝐮log6(X/2). We now wish to count the number of large zeros in the interval [0,X]. By definition, any such zero originates from an LRS 𝐮 which is small at level X, i.e., having size C𝐮C. Let us count the number of such LRS. Its coefficients a1,,ak and initial values u0,,uk1 are all in [C,C], an interval containing at most 2C+1<3C integers. Altogether for fixed k there are at most (3C)2k(3C)2C 2k-tuples, and summing up over k we derive an upper bound of C(3C)2C<C3C distinct possible LRS of size at most C.

On the other hand, Schmidt [27] proved that any LRS of order k or less has at most exp3(3klogk)<exp4(C) zeros, using the fact here that k is taken to be at most C. Hence the total number of zeros emanating from such LRS in the interval [0,X] is at most exp4(C)C3C, whence the inequality

#(X)=O(X(logX)B)

easily follows for any B>0.

Corollary 3.

The set 𝒮:= is a Universal Skolem Set of density one.

Proof.

It is clear that the set is recursive, and hence that 𝒮 is recursive as well.

Density one follows from Thm. 2, and universality follows from the fact that 𝒮, by definition, doesn’t contain any large zeros. Thus given any non-degenerate LRS 𝐮 of size C𝐮, its only possible zeros in 𝒮 can only lie in the interval [0,2exp6(C𝐮)], which can readily be checked.

4 Bad Primes and Good Primes

We first define what it means for a prime number to be bad. As before, let X>2exp7(1) be an integer, and recall than an LRS 𝐮 is small at level X if it has size C𝐮log6(X/2). Let us write C:=C𝐮 (that is, we omit the dependence on 𝐮). We can express the general term ut of 𝐮 as

ut=i=1sQi(t)αit, (3)

where sC and α1,,αs are the roots of the characteristic polynomial

xka1xk1ak

of 𝐮 and Q1,,Qs are univariate polynomials. Note that all characteristic roots are algebraic integers since the characteristic polynomial is monic and comprises exclusively coefficients in . Recall that if αi has multiplicity μi as a characteristic root then Qi has degree at most μi1. Let 𝕂:=(α1,,αs). The coefficients of each Qi are in 𝕂 and can straightforwardly be computed from the initial values u0,,uk1 of the sequence by solving a system of k linear equations, thanks to (3). By Cramer’s determinant rule,444This rule is named after the 18th-century Genevan mathematician Gabriel Cramer, who is presumably unrelated to the 20th-century Swedish mathematician Harald Cramér, whose work plays an important role in motivating the present article. each of the coefficients of Qi is the quotient of an algebraic integer by the determinant Δ:=det(M) of the matrix555The matrix M has s blocks, one for each characteristic root. For {1,,s} the -th block has dimension k×μ and has (i,j)-th element (i1)(j1)α(i1) for i{1,,k} and j{1,,μ}.

M:=[10101α1α1α2αs1αsα1k1(k1)μ11α1k1α2k1(k1)μs1αs1k1αsk1].

By the Cauchy root bound we have |αi|1+C for i{1,,s}. It follows that the squared Euclidean norm of each column vector above is at most

k(k1)2(k1)(1+C)2k<k2k(1+C)2k.

Thus, by the Hadamard inequality,

|Δ|2<(k2k(1+C)2k)k=(k(1+C))2k2.

The determinant Δ is in general, of course, a complex number. Note however that any Galois automorphism σGal(𝕂/) will permute the characteristic roots, and thus when applied to M will have the effect of permuting its columns. As a result, for any such σ, σ(Δ)=±Δ, and therefore the quantity Δ2 is stable under Galois automorphisms. We conclude that Δ2 must be a rational number, and since it is also by construction an algebraic integer,666Note that every entry of M is an algebraic integer. we must have Δ2.

Let us now consider the LRS 𝐯:=Δ2𝐮, noting that 𝐮 and 𝐯 share the same zeros. Writing

vt=i=1sPi(t)αit,

we observe that all the coefficients of each of the Pi are algebraic integers. We therefore have, for each 1is,

Pi(t)=Δ2Qi(t)=j=0μi1ci,jtj.

We wish to estimate the size of each ci,j𝒪𝕂. From our earlier calculation via Cramer’s determinant rule, noting that |u0|,,|uk1| are all bounded above by C1+C, and invoking the Hadamard inequality once more, we conclude that the house of each ci,j is bounded above by

|Δ|(kk(1+C)k)k<(k(1+C))2k2<(1+C)4C2<CC3.

Let σΣs be any permutation of the first s integers and let

βi:=ασ(i)fori=1,,s.

For some nonnegative integer m consider the algebraic integer

vm,σ=i=1sPi(m)βiαim. (4)
Definition 4.

We say that P[X,2X] is bad, if there exists an LRS 𝐮 which is small at level X, a permutation σΣs, and an integer m[0,X1/4], such that

  • The algebraic integer vm,σ defined in (4) above is non-zero, and

  • P is a prime factor of 𝒩𝕂/(vm,σ).

Let 𝒫bad(X) be the set of bad primes in [X,2X].

Proposition 5.

We have

#𝒫bad(X)<X2/3

for all X>X0, where X0 is some effective absolute constant.

Proof.

In order to estimate the size of 𝒫bad(X), we first need to find out:

  1. 1.

    How many such expressions (4) are there?

  2. 2.

    How large are they?

For (1), recall from the proof of Thm. 2 that there are at most C3C distinct possible LRS of size at most C. This in turn is an upper bound on the number of s-tuples ((Qi,αi))i=1s. We must then multiply this quantity with the number of possible permutations of the characteristic roots, which is at most C!<CC. There are therefore at most C4C linear recurrence sequences 𝐰=wmm=0 whose m-th term is given by

wm=i=1sPi(m)βiαimfor allm0.

This answers (1). As for (2), recall that the coefficients of Pi are of size at most CC3. There are at most C terms, the largest monomial involved in Pi(m) is at most mC<XC and the largest root has magnitude at most 1+C<2C. Thus each individual term is of absolute value at most

CC3+1(2C)XC(2C)X1/4 =exp((C3+1)logC+log(2C)+ClogX+X1/4log(2C))
<exp(X0.26)

for X>X0, since C is tiny in comparison to X. Hence the norm of the number shown in (4) is of size at most

exp(C!X0.26)<exp(X0.27)forX>X0,

since the degree of 𝕂 is at most C! (as 𝕂 is the splitting field of a polynomial of degree at most C). Moreover, as noted earlier, there are at most C4C such expressions. Thus a bad prime P divides an integer which is a product of such numbers and is of size at most

exp(C4CX0.27)<exp(X0.28)forX>X0.

Therefore the number of possible choices for P is at most X0.28. Since the number of choices for m is at most X0.25, we conclude that, for X>X0, the cardinality of 𝒫bad(X) is at most

X0.25+0.28<X2/3,

as required.

Finally, let us write 𝒫={p1,p2,} to denote the set of prime numbers, enumerated in increasing order, and let 𝒫good:=𝒫𝒫bad={g1,g2,} denote the set of good primes, again enumerated in increasing order. Note that, by Prop. 5 along with the prime number theorem, the set of bad primes has null density amongst the prime numbers. This in turn entails that good primes have density one amongst all prime numbers.

5 The Cramér Argument

In this section we present a heuristic argument supporting the assertion that large zeros do not exist, or in other words that the set is empty. The strategy is as follows. Assuming that good primes are distributed roughly similarly as ordinary primes, according to the Cramér model in number theory, one would expect that Cramér’s conjecture on gaps between primes applies also to good primes. More precisely, this conjecture postulates the existence of precise upper bounds on the largest possible gap between consecutive primes, and is predicated on the heuristic that the primes behave as a set of randomly distributed integers with asymptotic density conforming to the prime number theorem. However we show that around any large zero of an LRS there is an interval and an upper bound on the number of good primes in the interval that together contradict the above Cramér-type conjecture on gaps between good primes. We therefore surmise that large zeros do not exist.

Recall that 𝒫={p1,p2,} denotes the set of prime numbers, enumerated in increasing order, and likewise let us write 𝒫good:={g1,g2,} to denote an enumeration of the set of good primes in increasing order.

Conjecture 6 (Cramér-Granville).

For some κ>1,

lim supjpj+1pjlog2pj=κ.

Cramér initially suggested that the constant κ in Conjecture 6 might be 1 [8], but several decades later, building on substantial developments in the field, Granville produced evidence that κ2eγ1.1229, where γ is the Euler–Mascheroni constant [12]. There is in any event considerable computational evidence in support of the Cramér-Granville conjecture [23, 22].

As noted earlier, thanks to Prop. 5 and the prime number theorem, good primes have density one amongst all prime numbers:

limX#(𝒫good[0,X])#(𝒫[0,X])=1.

In other words, asymptotically speaking, almost all primes are good primes. Accordingly, it seems reasonable to suppose that good primes should behave similarly to ordinary primes, or at least should exhibit similar “statistical” properties. We therefore formulate:

Conjecture 7.

For some η>1,

lim supjgj+1gjlog2gj=η.

We now have the following result.

Theorem 8.

Conjecture 7 implies that large zeros of LRS do not exist; or more precisely, that is a finite set.

Proof.

Conjecture 7 can be reformulated as follows: there exist η>1 and n0 such that, for all nn0, the interval

[nη(logn)2,n]

always contains some good prime. In turn, this implies that the interval [nη(logn)3,n] must contain at least logn distinct good primes for n sufficiently large (say nn1max{n0,2exp7(1)}).

Thus let nn1, put C:=log6(n/2), and suppose that there is some LRS 𝐮 with C𝐮C such that un=0  – in other words, n is a large zero of 𝐮. Write n=P+m, where P[nη(logn)3,n] is a good prime and 0m<η(logn)3<n1/4. As in the previous section, let α1,,αs be the characteristic roots of 𝐮, put 𝕂:=(α1,,αs), and let Δ2 be the smallest positive integer such that, writing 𝐯:=Δ2𝐮, every term of vt of 𝐯 has a representation as an exponential polynomial

vt=i=1sPi(t)αit

in which all polynomials Pi have algebraic-integer coefficients.

Since un=vn=0, we get

0=i=1sPi(P+m)αiP+m.

We now reduce the above equation modulo 𝔭, where 𝔭 is some prime ideal of 𝒪𝕂 dividing P, from which we deduce that P divides

𝒩𝕂/(i=1sPi(m)βiαim), (5)

where each βi=σ(αi) is obtained from applying the Frobenius automorphism induced by 𝔭 in 𝕂 to αi. If the above expression (5) were non-zero, we would have to conclude that P𝒫bad(n/2), contradicting our choice of P. Thus the expression (5) is zero.

Let us count how many expressions of the form (5) can vanish. More precisely, consider the (complex-valued) LRS 𝐰=wjj=0 whose j-th term is given by

wj=i=1sPi(j)βiαijfor allj0,

and whose order is at most C. Schmidt [27] proves that the number of distinct positive integers m such that wm=0 is at most

exp3(3ClogC)<exp4(C).

Of course, given 𝐮, the s-tuple (β1,,βs) can be chosen in at most s!<CC ways. Thus the total number of possible zeros for expression (5) is at most CCexp4(C)<exp5(C). Since distinct choices of P give rise to distinct such zeros,777Recall that n=P+m, and thus distinct choices of P entail distinct values of m. and (as noted earlier) there are at least logn possible choices for P, we conclude that

logn<exp5(C),

or equivalently n<exp6(C)=n/2, a contradiction. It thus follows, as claimed, that Conjecture 7 prohibits the existence of large zeros of LRS that are greater than the absolute constant n1.

Thanks to Thm. 8, Conjecture 7 implies the existence of an algorithm to solve the Skolem Problem, as follows. Given an LRS 𝐮, first decompose 𝐮 into finitely many non-degenerate LRS, and check that none of these is identically zero. Next, for each sub-LRS 𝐯 of size C𝐯, simply search for a zero up to index 2exp6(C𝐯).888Technically speaking, the algorithm should examine all terms up to index max{2exp6(C𝐯),n1}, where n1 is the absolute constant appearing in the proof of Thm. 8. The existence of n1 is implied by Conjecture 7, but its effectivity would depend on the effectivity of Conjecture 7. If at the end of this process no zero has been found for any of the LRS, return that 𝐮 has no zeros.

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