TIGHT INSTANCES OF THE LONELY RUNNER

Luis Goddyn^{1}

Department of Mathematics, Simon Fraser University, Burnaby, BC, V5A 1S6, Canada

                            goddyn@math.sfu.ca

Erick B. Wong

Department of Mathematics, Simon Fraser University, Burnaby, BC, V5A 1S6, Canada

                               erick@sfu.ca

Received: , Accepted: , Published:

Abstract

The Lonely Runner Conjecture of J. Wills asserts the following.

For any vector v ∈ (ℝ − {0})^{n−1} there exists t ∈ ℝ such that every component of tv has distance at least 1/n to the nearest integer.

We study those vectors v for which the bound of this conjecture is attained. In particular, we construct an infinite family of such tight vectors. This family, plus three sporadic examples, constitute all known tight vectors. We completely characterize a subfamily: those tight vectors obtained from v = ⟨1, 2, …, n − 1⟩ by scaling one entry by a positive integer. The characterization motivates the problem of finding the least positive integer which has a common factor with every integer in the interval [a, b]. We solve this problem when b ≥ 2a.

1. Introduction

The Lonely Runner Conjecture of Wills [12] asserts the following.

For any vector v = ⟨v_{1}, …, v_{n−1}⟩ ∈ (ℝ − {0})^{n−1} there exists t ∈ ℝ such that
every component of tv has its fractional part in [\frac{1}{n}, \frac{n−1}{n}].

This conjecture has relevance to Diophantine approximation [1, 4, 11, 12, 13, 14, 15], viewobstruction theory [5, 6, 7, 8], and flows in graphs and matroids [2].

1The first author would like to thank Laboratoire Leibniz – IMAG and NSERC Canada for support during the course of this research.

A physical interpretation has n − 1 runners on a circular track of unit length, with v being the vector of angular speeds. The fractional part of v_{r}t is the position of runner r at time t. The distance from r to the origin at time t is ∥v_{r}t∥ := min{v_{r}t − ⌊v_{r}t⌋, ⌈v_{r}t⌉ − v_{r}t}. We write

LR(v) := sup_{t∈ℝ} min_{r} ∥v_{r}t∥

and

LR(n) := inf{LR(v) : v ∈ (ℝ − {0})^{n−1}}.

Thus Wills’s conjecture is that LR(n) ≥ 1/n for n ≥ 2. An easy argument [5] shows LR(⟨1, 2, …, n − 1⟩) = 1/n so the conjectured bound is best possible. It is known that LR(n) = 1/n for 2 ≤ n ≤ 6 [8, 2, 3, 9], and that LR(n) > \frac{1}{2(n−1)} for n ≥ 3 [12]. The original formulation of Wills’s conjecture is actually for the special case that

v is an (n − 1)-vector whose entries are distinct positive integers in

increasing order and having no common divisor.

(1)

Bohman et al. [3] have shown that this formulation is sufficient to imply the general conjecture, and that in fact LR(v) ≥ LR(n − 1) unless v is a scalar multiple of a vector of rationals. We henceforth limit our discussion to integral speed vectors.

In an attempt to understand the extremal instances of the conjecture, we strive to classify those (n − 1)-vectors v for which LR(v) = 1/n. Such vectors v are said to be tight. We use the notation [k] := ⟨1, 2, …, k⟩. As mentioned above, [n − 1] is tight for n ≥ 2. It is known [8] that, for 2 ≤ n ≤ 4,

v = [n − 1] is the unique tight vector satisfying (1).  (2)

Statement (2) is false for n ∈ {5, 6, 8}, but is probably true for n = 7. A computer experiment reveals the following tight vectors which are different from [n − 1].

All non-trivial tight speed vectors with n <= 20 and maximum speed <= 40
T1: n= 5:  1  3  4  7
T2: n= 6:  1  3  4  5  9
T3: n= 8:  1  4  5  6  7 11 13
T4: n= 8:  1  2  3  4  5  7 12
T5: n=14:  1  2  3  4  5  6  7  8  9 10 11 13 24
T6: n=20:  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 19 36

The first three of these tight speed vectors appear to be sporadic. However, there is a pattern to the last three; each is obtained from [n − 1] by doubling the speed of the second-fastest runner. Further tests reveal other tight vectors, but all are of a similar form: each is obtained from [n − 1] by multiplying some of the fastest speeds by small positive integers.

In Section 2 we determine exactly when accelerating a single runner in this fashion yields a tight speed vector. The characterization involves a number-theoretic condition that we have not seen before. The condition typically requires large values of n, whose minimum value is determined by an instance of the optimization problem described in the abstract. This problem is solved (for the case b ≥ 2a) in Section 4.

We treat the case of accelerating more than one runner in Section 3, where we present only a sufficiency condition for a vector of the form ⟨m_{1}, 2m_{2}, …, (n − 1)m_{n−1}⟩ (where each m_{i} is a positive integer) to be a tight vector.

2. Accelerating a single runner

We denote by [n − 1]_{r↦r′} the speed vector obtained from [n − 1] by replacing the speed r by r′. In this section, we characterize those integer triples (n, r, m) for which the speed vector [n − 1]_{r↦mr} is tight.

The substitution r ↦ mr is called an acceleration (of r). A runner with speed mr where r ∈ [n − 1] is called an accelerated runner, and mr is properly accelerated if m ≥ 2. If X = {x_{1}, …, x_{n−1}} is a set of real numbers, then LR(X) is well defined to equal LR(⟨x_{1}, …, x_{n−1}⟩). Accordingly, [n − 1] denotes either the vector ⟨1, 2, …, n − 1⟩ or the set {1, 2, …, n − 1}, depending on the context. We denote by ℕ the set of positive integers. We use [a, b] and [a, b) to denote closed and semi-closed real intervals.

Lemma 2.1 If r ∈ ℕ, then any set of ⌊r/2⌋ + 1 consecutive integers contains a number relatively prime to r.

Proof. The lemma is easily verified for r < 6, and we assume r ≥ 6. If r = 2p for some prime p, then any run of 4 consecutive integers contains an odd number not divisible by p, so we may assume r/2 is not prime.

Now, consider the moduli 1, …, r − 1 with respect to r. By Bertrand’s postulate, there exists a prime p ∈ [ ⌊r/3⌋ + 1 , 2⌊r/3⌋ + 1 ]. Since r ≠ 2p and p < r < 3p, the integers p and r − p are both relatively prime to r. The set {0, 1, …, r − 1} − {1, p, r − p, r − 1} contains no interval of ⌊r/2⌋ consecutive moduli, proving the lemma.

Lemma 2.2 If X is a proper subset of [n − 1], then LR(X) > 1/n.

Proof. We may assume without loss of generality that X = [n − 1] − {x} for some x ∈ [n−1]. If 2x ≥ n, then no speed in X is divisible by x. By considering the time t = 1/x, we find LR(X) ≥ 1/x > 1/n. We may assume that 2x < n.

We claim that at time t_{0} := \frac{1}{x} + \frac{1}{2xn}, exactly one runner in X has distance 1/n, and all other runners have distance > 1/n. It then follows that at some time sufficiently close to t_{0}, all runners have distance > 1/n. Let r be a speed in X. If r = kx for some k ∈ ℕ, then 2 ≤ k < n, so

∥rt_{0}∥ = ∥\frac{k}{2n}∥ ≥ \frac{1}{n},

with equality achieved if and only if k = 2. Now suppose r is not divisible by x. Then at time 1/x, the runner r has distance ≥ 1/x, so

∥rt_{0}∥ ≥ ∥\frac{r}{x}∥ − |r (t_{0} − \frac{1}{x})| ≥ \frac{1}{x} − \frac{r}{2xn} > \frac{1}{2x} > \frac{1}{n}.

Thus, the runner of speed 2x has distance exactly 1/n, and all other runners have distance > 1/n at time t, as claimed.

Theorem 2.3 For integers n ≥ 2, r ∈ [n − 1] and m ≥ 1, we have

LR([n − 1]_{r↦mr}) ≥ 1/n

with equality holding if and only if either n=2, or (n, r, m) = (3, 1, 4), or GCD(r, b) > 1 for every b ∈ {n − r, n − r + 1, …, m(n − r) − 1}.

Proof. Let X := [n − 1]_{r↦mr}. We first treat some special cases. The theorem is trivially true when m = 1. The cases n ∈ {2, 3, 4} follow from (2).

We assume m ≥ 2 and n ≥ 5. For the case r = 1, we claim that X = {m, 2, 3, …, n − 1} is never a tight speed set. At all times in the open interval T := (\frac{1}{2n}, \frac{1}{n}), every runner in {2, 3, …, n − 1} is at distance > 1/n. Supposing that X is tight, we have m ≥ n (by Lemma 2.2) and ∥mt∥ ≤ 1/n for all t ∈ T. Since T has length \frac{1}{2n}, the integer m is bounded above by \frac{2}{n}/\frac{1}{2n} = 4. Therefore n ≤ 4, a contradiction proving the claim. We henceforth assume r ≥ 2.

Under these assumptions, we first demonstrate the necessity of the GCD conditions. Define

s := n − r.

Suppose that GCD(r, b) = 1 for some b ∈ {s, s+1, …, ms−1}. Without loss of generality let b be the smallest such value. As GCD(r, b) = 1, there exists a ∈ ℕ such that ab ≡ 1 (mod r). Let

t_{ϵ} := \frac{a}{r} − \frac{ϵ}{nr}.

We aim to show that for some ϵ ∈ {1/m} ∪ [1/2, 2/3], all runners in X at time t_{ϵ} have distance ≥ 1/n, and at most one runner has distance exactly 1/n. At some time sufficiently close to t_{ϵ}, all runners will have distance > 1/n, thus proving the necessity of GCD(r, b) > 1.

We partition X into three sets according to the residues modulo r.

• X_{0} = {x ∈ X | x ≡ 0 (mod r)}

• X_{1} = {x ∈ X | x ≡ b (mod r)}

• X_{2} = {x ∈ X | x ≢ 0, b (mod r)}

If x ∈ X_{2}, then x < n. At time t_{0} = a/r, the runner x is at some position of the form c/r with 2 ≤ c ≤ r − 1. Since x(t_{0} − t_{1}) = x/nr < 1/r, this runner lies in the interval ((c − 1)/r, c/r] throughout the time interval [t_{1}, t_{0}], and therefore has distance at least 1/r > 1/n.

If x ∈ X_{1}, then runner x is at position 1/r at time t_{0}. Now x ≡ b (mod r) and x < s + r ≤ b + r together imply x ≤ b. So, provided ϵ < s/b, we have

∥xt_{ϵ}∥ ≥ ∥\frac{ab}{r} − \frac{bϵ}{nr}∥ = ∥\frac{1}{r} − \frac{bϵ}{nr}∥ > ∥\frac{1}{r} − \frac{s}{nr}∥ = \frac{1}{n} .

We have shown the following.

For any ϵ ∈ [ 0, \frac{s}{b}), every runner in X_{1} ∪ X_{2} has distance > \frac{1}{n} at time t_{ϵ}.  (3)

We consider those runners in X_{0} under two cases.

Case 1: s ≤ r

Here we have n ≤ 2r, so X_{0} contains only the properly accelerated runner mr. At time t_{1/m} this runner has distance 1/n. Since b < ms, we have 1/m < s/b, so ϵ = 1/m satisfies the condition in (3). Therefore every runner in X − {mr} has distance > 1/n at time t_{1/m}.

Case 2: s > r

By the choice of b and Lemma 2.1 we have b ≤ s + ⌊r/2⌋ < 3s/2, so s/b > 2/3. We may assume m ≥ 3, for otherwise mr ∈ [n − 1] and X is not tight by Lemma 2.2. We have

X_{0} = \{mr, 2r, 3r, …, ⌊\frac{n − 1}{r}⌋ r\} .

At the time t_{ϵ}, a runner of speed kr is at distance ∥kϵ/n∥. For the non-properly accelerated runners in X_{0} we have 2 ≤ k < n/2, so for any ϵ ∈ [1/2, 2/3] we have

\frac{1}{n} ≤ \frac{kϵ}{n} < 1/3.

This implies ∥kϵ/n∥ ≥ 1/n with equality only when ϵ = 1/2 and k = 2. In view of ϵ ≤ 2/3 < s/b and (3), we have shown that, throughout the time interval [t_{2/3}, t_{1/2}], all runners in X − {mr} have distance ≥ 1/n with equality holding only for runner 2r at time t_{1/2}.

It remains to show that for some ϵ ∈ [1/2, 2/3], runner mr has distance ≥ 1/n at time t_{ϵ}, with strict inequality holding if ϵ = 1/2. At time t_{ϵ}, runner mr has distance ∥ϵm/n∥. If m < 2n − 2, then ϵ = 1/2 suffices since

\frac{1}{n} < \frac{m}{2n} < \frac{n − 1}{n} .

We therefore assume m ≥ 2n − 2. If m ≥ 12, then runner mr traverses an interval of length m/6n ≥ 2/n during the time interval [t_{2/3}, t_{1/2}]. Thus, for some ϵ ∈ (1/2, 2/3], runner mr will have distance ≥ 1/n at time t_{ϵ}, as required.

Suppose instead that m < 12, whence 5 ≤ n ≤ ⌊m/2⌋+1 ≤ 6. In particular, either n = 5 and 8 ≤ m ≤ 11, or else n = 6 and 10 ≤ m ≤ 11. Since r < s = n − r we have r < 3, hence r = 2. It follows that either n = 5 and X ⊂ X_{5} := {1, 3, 4} ∪ {16, 18, 20, 22}, or n = 6 and X ⊂ X_{6} := {1, 3, 4, 5} ∪ {20, 22}. We claim that LR(X_{5}) > 1/5 and LR(X_{6}) > 1/6, hence X cannot be a tight speed set. To see this, observe that X_{5} reduces modulo 19 to the set \{±1, ±3, 4\} and that all runners in X_{5} have distance at least \frac{5}{19} > \frac{1}{4} at time t = \frac{8}{19}; similarly, X_{6} reduces modulo 25 to \{1, ±3, 4, ±5\} and all runners in X_{6} have distance at least \frac{1}{5} at time t = \frac{11}{25} (these bounds are actually tight). This completes Case 2 and proves the GCD condition to be necessary.

For the proof of sufficiency, we refer the reader to Theorem 3.1 where a more general result is proved.

In the above proof, we identified explicit intervals in which the runner of speed r is the sole runner at distance ≤ 1/n (such intervals must exist by Lemma 2.2). As we saw in Case 2, an accelerated runner with speed at least 12r must have distance ≥ 1/n at some point in these intervals. This observation holds even if the speed is not an integer multiple of r; we can use it to prove the following:

Theorem 2.4 For any fixed speed r there are only finitely many pairs (n, r′) such that [n − 1]_{r↦r′} is tight.

Proof. Let r be fixed. Suppose [n − 1]_{r↦r′} is tight with n ≥ 12r. Then we have s > r which is Case 2 in the proof of Theorem 2.3. By the argument in Case 2 we have r′ < 12r, so [n − 1]_{r↦r′} is a proper subset of [n − 1], contradicting Lemma 2.2. Thus there are only finitely many n such that [n − 1]_{r↦r′} is tight. For any n < 12r, there exists by Lemma 2.2 an interval of positive length in which all runners in [n − 1] − {r} have distance > 1/n, so again the speed set [n − 1]_{r↦r′} cannot be tight for sufficiently large r′ depending on n.

We believe Theorem 2.4 partially explains why there are no further sporadic speed sets resembling T1 and T2 where the speed 2 is replaced by an odd integer.

3. Accelerating several runners

Theorem 2.3 gives a necessary and sufficient “GCD condition” for a single-runner-accelerated speed set [n − 1]_{r↦mr} to be tight. We deferred the proof of “sufficiency” to this section, because we prove something more general here. We show that, if each of several runners are properly accelerated r ↦ m_{r}r and each individual acceleration [n − 1]_{r↦m_{r}r} satisfies the GCD condition, then accelerating all of these runners simultaneously also results in a tight speed set. For example, each of [73]_{70↦140} and [73]_{72↦144} is a tight speed set, therefore accelerating both runners also results in a tight speed set.

Later we demonstrate that tight speed sets exist where arbitrarily many runners have been properly accelerated in this way. We also discuss the missing converse to Theorem 3.1 (when there is more than one properly accelerated runner).

Let m = ⟨m_{1}, m_{2}, …, m_{n−1}⟩ be a vector of “speed multipliers”. Then [n − 1]m denotes the vector (or set) of accelerated speeds ⟨m_{1}, 2m_{2}, …, (n − 1)m_{n−1}⟩.

Theorem 3.1 Let m_{1}, m_{2}, …, m_{n−1} ∈ ℕ. Then LR([n − 1]m) = 1/n if the following condition holds:

For each r ∈ [n − 1] and b ∈ {n − r, n − r + 1, …, m_{r}(n − r) − 1}, we have GCD(r, b) > 1.

Note that the condition of Theorem 3.1 is vacuously true for those non-properly accelerated r with m_{r} = 1.

We shall use the following lemma. Althought it holds true for any n > 1, we shall only use the case n = |X| + 1.

Lemma 3.2 Let X be a speed set satisfying (1) such that LR(X) ≤ 1/n. Let r ∈ X and let X′ be obtained from X by replacing r with mr, where m is an integer greater than 1. Then LR(X′) ≤ 1/n provided the following hold.

1. For each prime divisor p of r, we have \frac{r}{p} ∈ X′.

2. For each integer q relatively prime to r, there exists b ∈ X′ such that m(n − r) ≤ b ≤ r + n and b ≡ q (mod r).

Proof. For each a ∈ ℤ we consider the following union of two time intervals, centered about a/r.

T_{a} := \{\frac{a}{r} + Δt : \frac{1}{nmr} < |Δt| ≤ \frac{1}{nr}\} .  (4)

Thus ⋃{T_{a} : a ∈ ℤ} is the set of times at which the accelerated runner having speed mr has distance more than 1/n, but would have had distance at most 1/n were he to have speed r instead. As LR(X) ≤ 1/n, we have only to prove the following statement for each a ∈ ℤ.

For every t ∈ T_{a}, there exists x ∈ X′ such that ∥xt∥ ≤ \frac{1}{n}.  (5)

Given a ∈ ℤ and t ∈ T_{a}, we write t = a/r + Δt where \frac{1}{nmr} < |Δt| ≤ \frac{1}{nr}. Let d = GCD(a, r). We first suppose d > 1. Let p be a prime divisor of d, and set x = r/p. By the hypothesis, x ∈ X′. Now x satisfies (5) according to the following calculation.

∥xt∥ = ∥\frac{r}{p} · \frac{a}{r} + \frac{r}{p}Δt∥ = ∥\frac{r}{p}Δt∥ ≤ \frac{r}{p} · \frac{1}{nr} < \frac{1}{n}.

We now assume d = GCD(a, r) = GCD(−a, r) = 1. There exist by hypothesis b, b′ ∈ X such that mn − mr ≤ b, b′ ≤ n + r and ab ≡ −ab′ ≡ 1 (mod r).

We claim that setting x to equal to either b or b′, according to the sign of Δt, satisfies (5). If Δt < 0, then for some q ∈ ℕ we have the following.

bt ∈ [\frac{ab}{r} − \frac{b}{nr}, \frac{ab}{r} − \frac{b}{nmr}] = [q + \frac{1}{r} − \frac{b}{nr}, q + \frac{1}{r} − \frac{b}{nmr}] .

Since \frac{1}{r} − \frac{b}{nr} ≥ −\frac{1}{n} and

\frac{1}{r} − \frac{b}{nmr} = \frac{nm − b}{nmr} ≤ \frac{mr}{nmr} = \frac{1}{n},

we have ∥bt∥ ≤ \frac{1}{n}. If Δt > 0, then a nearly identical calculation shows that x = b′ satisfies (5), as claimed. This completes the proof of Lemma 3.2.

Proof of Theorem 3.1. Suppose that m = ⟨m_{1}, m_{2}, …, m_{n−1}⟩ satisfies the hypothesis. Let r ∈ [n − 1] be such that m_{r} ≥ 2. We have n − r ≥ 2 since no integer has a common factor with 1. Also r is even, since there is always a power of 2 in [ n − r , 2(n − r) − 1 ]. Since r is not relatively prime to any integer in [ n − r , m_{r}(n − r) ), we have by Lemma 2.1 that r/2 ≥ (m_{r} − 1)(n − r) ≥ n − r, so 2n/3 ≤ r < n. We have established the following.

If r, r′ ∈ [n − 1] are distinct and m_{r}, m_{r′} ≥ 2, then 1 < GCD(r, r′) < r′.  (6)

Let us define the sequence of speed vectors X_{0}, X_{1}, …, X_{n−1}, where X_{0} := [n − 1], and each X_{i} is equal to X_{i−1} with its ith entry replaced by m_{i}i. We will show inductively that LR(X_{r}) ≤ 1/n, for r = 0, 1, …, n − 1. We have that LR(X_{0}) = 1/n. Assume that LR(X_{r−1}) ≤ 1/n for some r ∈ [n − 1]. If m_{r} = 1, then X_{r} = X_{r−1} and the induction step is trivial. We assume that m_{r} ≥ 2.

We shall apply Lemma 3.2 with X = X_{r−1} and X′ = X_{r}. Let p be a prime divisor of r. Putting r′ = \frac{r}{p} in (6) shows that m\frac{r}{p} = 1, so \frac{r}{p} ∈ X_{r}. This establishes the first condition of Lemma 3.2. Now suppose q is relatively prime to r. Then q is congruent modulo r to a unique representative b in {n − r, n − r + 1, …, n − 1} ⊂ X_{0}. Since GCD(r, b) = 1, we have by (6) that m_{b} = 1, so b ∈ X_{r}. Thus both conditions of Lemma 3.2 are satisfied, completing the induction step. We have proved LR([n − 1]m) = LR(X_{n−1}) ≤ 1/n.

To prove that LR([n − 1]m) ≥ 1/n, we show that, at time 1/n, every runner m_{r}r ∈ [n − 1]m has distance at least 1/n. It suffices to show that m_{r}r is not divisible by n, since this implies ∥m_{r}r\frac{1}{n} ∥ ∈ \{\frac{1}{n}, \frac{2}{n}, …, \frac{n−1}{n}\}. This is clearly true if m_{r} = 1, since 1 ≤ m_{r}r < n.

Suppose that m_{r} ≥ 2. Again by Lemma 2.1 we have (m_{r} − 1)(n − r) ≤ r/2. In particular 1 ≤ m_{r}(n − r) ≤ r/2 + (n − r) < n, so n does not divide m_{r}(n − r). Therefore n does not divide m_{r}r. This completes the proof of Theorem 3.1.

One naturally wonders whether the converse of Theorem 3.1 is true (when more than one runner is properly accelerated). In the most general sense, this is certainly not the case: given an arbitrary rational speed vector v ∈ (ℚ − {0})^{n−1}, it is easy to choose m so that [n − 1]m is a scalar multiple of v. Thus, in order to obtain a necessary condition one might need to restrict m such that the entries of [n − 1]m are relatively prime as in (1), or perhaps bound the cardinality of {r ∈ [n − 1] : m_{r} ≥ 2} by a suitable function of n. The status of either of these formulations is currently open.

It is not obvious that the GCD conditions of Theorem 3.1 can be simultaneously satisfied by several properly accelerated runners. We conclude this section by showing that tight speed sets can indeed be constructed in which the set of properly accelerated speeds is arbitrarily large.

Corollary 3.3 Let (a_{1}, a_{2}, …, a_{k}) be a sequence of positive integers. Then there are infinitely many tight speed vectors of the form [n−1]m where (a_{k}, a_{k−1}, …, a_{1}) is a subsequence of the sequence (m_{1}, m_{2}, …, m_{n−1}) of entries in m.

Proof. Let r(m, s) be any function such that r(m, s) is a positive integer having a common factor with every element of {s, …, ms − 1}. There is considerable freedom in the choice of r(m, s) — in Section 4 we determine the smallest possible value of r — but here we may simply take r(m, s) to be any multiple of (ms − 1)!. Let s_{1} be any integer greater than one, and define s_{2}, …, s_{k+1} as follows.

s_{i+1} = s_{i} + LCM( r(a_{1}, s_{1}), …, r(a_{i}, s_{i}) ) (i = 1, 2, …, k).

Now take n = s_{k+1}, and define m = ⟨m_{1}, …, m_{n−1}⟩ by

m_{i} = \{ \begin{matrix}a_{j} & if i = n − s_{j} for some j ∈ \{1, 2, …, k\} \\ 1 & otherwise. \end{matrix}

By construction, we have for each j ∈ {1, …, k} that n = s_{k+1} ≡ s_{j} (mod r(a_{j}, s_{j})), so n − s_{j} is divisible by r(a_{j}, s_{j}) and thereby satisfies the condition of Theorem 3.1. Therefore [n − 1]m is a tight speed vector. By varying the choice of s_{1} and the choice of the function r(m, s), we obtain infinitely many different sequences {s_{i}} and thus infinitely many distinct tight speed vectors.

In this construction, n grows very quickly with k. For example, using s_{1} = 2 and values of r(m, s) provided in Section 4 we find: if [n − 1]m is tight and m contains the subsequence (2, 3), then the construction requires that

n ≥ 2 + 2 · 3 · 5 + 2 · 3 · 5 · 7 · 37 · 41 · 43 · 47 · 53 · 59 · 61 ≈ 1.2 · 10^{14}.

Applying the above construction with the prescribed subsequence (2, 2) requires that n ≥ 2 + 2 · 3 + 2 · 3 · 11 · 13 = 866. At the start of this section, we described the tight speed vector [n − 1]m where n = 74 and m = ⟨1, 1, …, 1, 1, 2, 1, 2, 1⟩. Thus the values of n provided by this construction are likely far from optimal.

4. Optimization

The arithmetic condition in the statements of Theorems 2.3 and 3.1 gives rise to an optimization problem which is of independent interest. If a speed set of the form [n − 1]_{r↦mr} is tight, then r (and thus n) must be quite large compared to integers m and s := n − r. We solve here the problem of finding, for a given pair m, s, the least possible value of r.

We say a triple (r, m, s) is tight if [r + s − 1]_{r↦mr} satisfies the condition of Theorem 2.3. That is, (r, m, s) is tight if r has a common factor with each integer in {s, s + 1, …, ms − 1}. For example, (30, 3, 2) is tight since 30 has a common factor with each of 2, 3, 4, 5; and hence ⟨1, 2, …, 29, 90, 31⟩ is a tight speed vector. It is clear that if (r, m, s) is tight then so is (dr, m, s) for any d ∈ ℕ. We say (r, m, s) is minimally tight if it is tight, but (r/d, m, s) is not tight for any prime divisor d of r. Table 1 gives all minimal tight triples (r, m, s) for some small values of m and s. Also listed are those integers s, s + 1, …, ms − 1, required to have a common factor with r. Notice that for some pairs (m, s), such as (2, 10) and (2, 11), there is more than one value of r for which (r, m, s) is minimally tight. Such values of r are easy to construct for a given (m, s) by considering the possible prime factorizations of r.

For integers m, s ≥ 2, we let r(m, s) denote the least positive integer r such that (r, m, s) is tight. Since any integer in {s, s + 1, …, ms − 1} which is not prime has a prime factor less than \sqrt{ms}, we have that r(m, s) is at most the product of all the primes in the two half-open intervals [2, \sqrt{ms}) ∪ [s, ms). Table 1 shows that, for small values of s and m, r(m, s) is exactly equal to the product of these primes.

In fact r(m, s) is exactly equal to this upper bound, with just two exceptions: r(2, 26) and r(2, 27) are smaller than this bound by a factor of five. Indeed we can prove a more general result. If S is a set of real numbers then we denote by by PS the set of primes in S. Thus ∏ P[α, β) is the product of all primes p with α ≤ p < β.

m  s | all r st (r,m,s) is minimally tight|  s s+1 s+2    ...    ms-1
----
2  2 |         6 = 2*3                    |  2  3
2  3 |        30 = 2*3*5                  |  3  4  5
2  4 |        70 = 2  *5*7                |  4  5  6  7
2  5 |       210 = 2*3*5*7                |  5  6  7  8  9
2  6 |       462 = 2*3  *7*11             |  6  7  8  9 10 11
2  7 |      6006 = 2*3  *7*11*13          |  7  8  9 10 11 12 13
2  8 |       858 = 2*3    *11*13          |  8  9 10 11 12 13 14 15
2  9 |     14595 = 2*3    *11*13*17       |  9 10 11 12 13 14 15 16 17
2 10 | a  277134 = 2*3    *11*13*17*19    | 10 11 12 13 14 15 16 17 18 19
     | b  461890 = 2  *5  *11*13*17*19    |
2 11 | a  277134 = 2*3    *11*13*17*19    | 11 12 13 14 15 16 17 18 19 20 21
     | b 3233230 = 2  *5*7*11*13*17*19    |
----
3  2 |        30 = 2*3*5                  |  2  3  4  5
3  3 |       210 = 2*3*5*7                |  3  4  5  6  7  8
3  4 |      2310 = 2*3*5*7*11             |  4  5  6  7  8  9 10 11
3  5 |     30030 = 2*3*5*7*11*13          |  5  6  7  8  9 10 11 12 13 14
3  6 |    102102 = 2*3  *7*11*13*17       |  6  7  8  9 10 11 12 13 14 15 16 17
3  7 |   1939938 = 2*3  *7*11*13*17*19    |  7  8  9 10 11 12 13 14 15 16 17 18 19 20
3  8 |   6374082 = 2*3    *11*13*17*19*23 |  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Table 1: Factorizations of r such that (r, m, s) is minimally tight

Theorem 4.1 Let s, t be positive integers with s ≥ 2 and t/s ≥ 2. If (s, t) ∉ {(26, 52), (26, 53), (26, 54), (26, 55), (27, 54), (27, 55)} then the smallest integer having a prime factor in common with each integer in [s, t) equals ∏ P( [2, \sqrt{t}) ∪ [s, t) ).

Proof. Let s, t be as in the hypothesis. Let r be the smallest integer having a common prime factor with each integer in [s, t). Let R be the set of primes dividing r so by minimality r = ∏ R. As in the above discussion, each integer in [s, t) has a prime divisor in P( [2, \sqrt{t}) ∪ [s, t) ). We aim to show that, aside from the listed exceptions, we have P( [2, \sqrt{t})∪[s, t) ) ⊆ R. Each prime p ∈ P[2, t/s] must divide r since some power of p lies in [s, t). Also, each prime in [s, t) or whose square belongs to [s, t) must divide r. We have established that P( [2, t/s] ∪ [\sqrt{s}, \sqrt{t}) ∪ [s, t) ) ⊆ R. We define

X = P(t/s, \sqrt{s}) − R.

Our plan is to show that X = ∅ if \sqrt{t} ≥ 90, and to use a computer search for the case \sqrt{t} < 90.

Suppose that X ≠ ∅. We define the non-empty finite union of intervals

I = ⋃_{x∈X} I_{x} where I_{x} = [s/x, t/x).

From X ⊆ (t/s, \sqrt{s}) we have that I ⊆ (\sqrt{s}, s). Let x ∈ X. By considering those multiples of x lying in [s, t), we see that each prime in I_{x} must divide r. This implies PI ⊆ R. Let I′ Figure 1: A log-scale diagram for the proof of Theorem 4.1. The relative positions of \sqrt{t} and s/β may be reversed.

be the maximal interval contained in I and satisfying sup I′ = sup I. Let

X′ = {x ∈ X : I_{x} ⊆ I′}

and let α = min X′(= min X) and β = max X′. Figure 1 may help the reader with these definitions. Let

R′ = R ∪ X′ − P( I′ ∩ [\sqrt{t}, s) )

and let r′ = ∏ R′. Our goal is to establish two claims: first, that r′ has a common factor with every integer in [s, t); second, if \sqrt{t} ≥ 90, then r′ < r, contradicting the choice of r.

Suppose, for contradiction, that some integer w ∈ [s, t) has no prime factor in R′. Then w has a prime factor p ∈ P( I′ ∩ [\sqrt{t}, s) ). As \sqrt{w} < p < w, the integer w has another prime factor, say q < \sqrt{w}. Because p ∈ I′, we have that p ∈ I_{x}, for some x ∈ X′. In particular s/x ≤ p, so

q ≤ \frac{w}{p} < \frac{t}{p} ≤ \frac{tx}{s} ≤ \frac{tβ}{s} .  (7)

Since q ∉ R′, we have that q ∈ X − X′, so I_{q} ⊆ I − I′. By the definition of I′, we have sup I_{q} < min I′, so t/q < s/β. This contradicts (7) and establishes the first claim.

For the second claim, we aim to show that ln ∏ X′ < ln ∏ P( I′ ∩ [\sqrt{t}, s) ) for \sqrt{t} ≥ 90. Using the facts X′ ⊆ P[α, β] and I′ ∩ [\sqrt{t}, s) ⊇ [ max\{s/β, \sqrt{t}\}, t/α), it is sufficient to establish that

ln ∏ P[α, β] < ln ∏ P[ max\{s/β, \sqrt{t}\} , t/α).  (8)

We will make use of the following estimates for the Prime Number Theorem which are readily verified using the effective bounds given by Rosser and Schoenfeld [10].

For any real ℓ < 1 there exists z_{0} such that, for all real z ≥ z_{0}, we have

ℓz < ln ∏ P[2, z] < 1.017z. (9)

Some pairs (ℓ, z_{0}) which are valid for (9) are given in the following table.

\begin{matrix}ℓ & .485 & .595 & .662 & .703 & .722 & .761 & .792 & .807 & .816 & .828 & .843 & .849 \\ z_{0} & 5 & 11 & 17 & 29 & 37 & 41 & 59 & 67 & 71 & 97 & 101 & 127 \end{matrix}  (10)

The following estimates use the hypothesized bound t/s ≥ 2 and assume \sqrt{t} ≥ 90. We have that t/α > t/\sqrt{s} ≥ \sqrt{2t} > 127, so ln ∏ P[2, t/α) ≥ .849t/α by (10). Thus the right hand side of (8) is bounded below by

.849 \frac{t}{α} − 1.017 max\{\sqrt{t} , \frac{s}{β} \} .  (11)

If \sqrt{t} ≤ s/β, then using the fact α ≤ β < \sqrt{s} we may bound (11) from below.

.849 \frac{t}{α} − 1.017\frac{s}{β} ≥ .849 \frac{2s}{β} − 1.017\frac{s}{β} = .681\frac{s}{β} ≥ .681\sqrt{t}
≥ 1.362 \frac{s}{\sqrt{t}} ≥ 1.362 β > ln ∏ P[2, β].

We have proved (8) in this case.

We henceforth assume \sqrt{t} > s/β. The left hand side of (8) is bounded above by

\{ \begin{matrix}1.017β − .761α & if α ≥ 41 \\ 1.017β & if α < 41. \end{matrix}  (12)

Since (12) is increasing with β, and (11) is independent of β, we may assume that β = \sqrt{s}.

Suppose first that α ≥ 41. Then the difference of (11) and (12) is bounded below by

.849 \frac{t}{α} − 1.017\sqrt{t} − 1.017\sqrt{s} + .761α ≥ .849 \frac{t}{α} + .761α − 1.017 (1 + \frac{1}{\sqrt{2}})\sqrt{t}
≥ (.849\sqrt{2} + \frac{.761}{\sqrt{2}} − 1.737) \sqrt{t} > .001\sqrt{t} > 0.

(We have used the fact .849 t/α + .761α is decreasing in α for α ≤ \sqrt{t}, and that α ≤ \sqrt{t/2} .) Therefore (8) holds for α ≥ 41.

Now suppose α < 41 and \sqrt{t} ≥ 90. Then (11) is bounded below as follows.

.849 \frac{t}{α} − 1.017\sqrt{t} ≥ .849 \frac{90}{41}\sqrt{t} − 1.017\sqrt{t} ≥ .846\sqrt{t}
≥ 1.196\sqrt{s} > ln ∏ P[2, \sqrt{s}] ≥ ln ∏ P[α, β].

We have established the theorem for all but a finite number of pairs (s, t). We now outline an efficient computational method to verify the remaining cases in which 2s ≤ t < 8100 and α, β ∈ P(t/s, \sqrt{s}) ⊆ P[3, 61]. We select pairs α ≤ β from P[3, 61] and verify (8) for all integers t ∈ [ 2β^{2} + 2 , 8100 ] and where s = ⌊t/2⌋. It suffices to test this single value of s since the right hand side of (8) increases as s decreases. The search is further accelerated by observing that a short interval of primes contained as terms in the right hand side of (8) can serve to simultaneously check several values of t. The resulting computation reveals the six exceptions in the statement of the theorem, where α = β = 5 and 52 ≤ t ≤ 55.

It is easy to see that Theorem 4.1 remains true, with a finite number of exceptions, when the bound t/s ≥ 2 in the statement is replaced by t/s ≥ 1 + ϵ, for any ϵ > 0. Finding these exceptions becomes computationally expensive for small values of ϵ, since the bound \sqrt{t} ≥ 90 in the proof would have to be substantially increased.

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