Moments of Gaps Between k-free Numbers

S.W. Graham*

Department of Mathematics, Michigan Technological University Houghton, MI 49931

Let k be a fixed positive integer, and let s_{n} denote the nth k-free number. If k ≥ 3 and 0 ≤ γ < 2k − 2 + 4/(k + 1), then

∑_{s_{n}≤x}(s_{n+1} − s_{n})^{γ} ≪ x.

1. Introduction

Let k be a fixed integer with k ≥ 2. We say that an integer n is k-free if n is divisible by no kth power of a prime. Let s_{n} denote the nth k-free number. A well–known and elementary result states that the number of k-free integers up to x is asymptotic to x/ζ(k). Thus the average gap between k-free numbers is bounded. One expects that long gaps are rare, and a strong quantitative formulation of this expectation is that the conjecture that

∑_{s_{n}≤x}(s_{n+1} − s_{n})^{γ} ≪ x  (1)

for all γ ≥ 0.

The case k = 2 (i.e. squarefree numbers) was first considered by Erdős[1], who proved that (1) is true for 0 ≤ γ ≤ 2. Hooley [5] extended this to 0 ≤ γ ≤ 3, and Filaseta [2] made a further improvement to 0 ≤ γ < 29/9.

In the case of aribitrary k, Hooley[5] remarked that his method generalized immediately to give the range 0 ≤ γ ≤ k + 1. He also remarked that in the case k ≥ 3, it was possible to extend the analysis further. The purpose of this paper is to present such an extension.

Theorem. If k ≥ 3 and 0 ≤ γ < 2k − 2 + 4/(k + 1) then

∑_{s_{n}≤x}(s_{n+1} − s_{n})^{γ} ≪ x.  (2)

The implied constant depends only on γ and k.

By combining the Theorem with work of Mirsky [6], one can show that

∑_{s_{n}≤x}(s_{n+1} − s_{n})^{γ} ∼ B(γ, k)x

* Supported in part by a grant from the National Security Agency. for 0 ≤ γ < 2k − 2 + 4/(k + 1). We do not give the details here, but we remark that the approach in the last section of Hooley [5] generalizes immediately to the k-free case.

In the following, the letter p denotes a positive prime integer, the letters h, m, n and r denote positive integers, x denotes a real variable tending to infinity, and ϵ denotes a small positive number. The meanings of the other variables will be explicitly stated or will be clear from context. Constants implied by ≪- and O-symbols may depend on γ, k, r and ϵ, but are independent of other variables.

2. Reduction to Moments

Instead of dealing with (2) directly, we will prove that

∑_{\frac{1}{2} x<s_{n}≤x}(s_{n+1} − s_{n})^{γ} ≪ x.  (3)

for appropriate values of γ. The truth of (2) follows easily from the above and a splitting– up argument.

Let L(h) be the number of n for which h ≤ s_{n+1} − s_{n} ≤ 2h and \frac{1}{2}x < s_{n} ≤ x. Information about L(h) can be used to estimate the left-hand side of (3), since

∑_{\frac{1}{2} x<s_{n}≤x}(s_{n+1} − s_{n})^{γ} ≪ ∑_{j=0}^{∞} L(2^{j})2^{γj}.  (4)

Henceforth, we shall assume that h is larger than some large fixed constant and that

h ≪ x^{1/(2k+1)} log x.  (5)

The upper bound of (5) is motivated by a result of Trifonov [8], who proved that if s_{n} ≤ x then

s_{n+1} − s_{n} ≪ x^{1/(2k+1)} log x.  (6)

With regard to the lower bound, we note that the trivial bound L(h) ≪ x suffices for our purposes when h is bounded.

In analogy with other work on gaps between elements of sequences (for example, [4], [7]), it would be natural to derive information about L(h) from estimates for

M_{r}^{∗}(x, h) = ∑_{\frac{1}{2} x<n≤x} ( ∑_{n<m≤n+h} χ(m) − \frac{1}{ζ(k)}h)^{r} ,  (7)

where χ(m) denotes the characteristic function of the k-free numbers. However, a better way to get information about L(h) is to consider

M_{r}(x, h) = ∑_{\frac{1}{2} x<n≤x}( ∑_{n<m≤n+h} ∑_{{p^{k}|m \atop p>H}} 1)^{r},

where

H = \frac{1}{12}h log h.  (8)

Our first two lemmas show how upper bounds for M_{r}(x, h) can be translated into upper bounds for L(2h).

Lemma 1. Let H be as in (8). If h is sufficiently large and if there are no k-free numbers in the interval (n, n + h] then

∑_{n<m≤n+h} ∑_{{p^{k}|m \atop p>H}} 1 ≥ \frac{1}{6}h.  (9)

Proof. As before, we use χ to denote the characteristic function of the k-free numbers. By the hypothesis,

\begin{aligned}0 & = ∑_{n<m≤n+h} χ(m) ≥ ∑_{n<m≤n+h}(1 − ∑_{p^{k}|m} 1) = h − ∑_{p} ∑_{{n<m≤n+h \atop p^{k}|m}} 1 \\ & ≥ h − ∑_{p≤H} (\frac{h}{p^{k}} + 1) − K ≥ h (1 − ∑_{p} \frac{1}{p^{k}} ) − π(H) − K, \end{aligned}

where K is the left–hand side of (9). Now we note that

1 − ∑_{p} \frac{1}{p^{k}} ≥ 1 − ∑_{r=2}^{∞} \frac{1}{r^{2}} = 2 − \frac{π^{2}}{6} > 2 − \frac{5}{3} = \frac{1}{3}.

Moreover, we have the Chebyshev estimate π(H) < 2H/ log H < \frac{1}{6}h provided h is sufficiently large.

Lemma 2. If r is any positive integer then

L(2h) ≪ \frac{M_{r}(x, h)}{h^{r+1}}.

Proof. If s_{j+1} − s_{j} ≥ 2h, then there are no k-free numbers in the interval (n, n + h] for n = s_{j}, s_{j} + 1, …, s_{j} + h − 1. Such a gap therefore contributes ≫ h^{r+1} to M_{r}(x, h).

In the next section, we shall give a heuristic argument to support

Conjecture M. If r is a positive integer then

M_{r}(x, h) ≪ \frac{hx}{H^{k−1} log H} ≪ \frac{x}{h^{k−2} log^{k} h}.

Conjecture M is immediate in the case r = 1, for

\begin{aligned}M_{1}(x, h) & = ∑_{\frac{1}{2} x<n≤x} ∑_{n<m≤n+h} ∑_{{p^{k}|m \atop p>H}} 1 = ∑_{p>H} ∑_{{\frac{1}{2} x<m≤x+h \atop p^{k}|m}} ∑_{{m−h≤n<m \atop \frac{1}{2} x<n≤x}} 1 \\ & ≪ ∑_{p>H} \frac{hx}{p^{k}} ≪ \frac{hx}{H^{k−1} log H} ≪ \frac{x}{h^{k−2} log^{k} h}. \end{aligned}  (10)

Suppose Conjecture M is true for a given r. From Lemma 2, we see that

L(2h) ≪ \frac{x}{h^{r+k−1} log^{k} h},

and so (4) gives

∑_{\frac{1}{2} x<s_{k}≤x}(s_{k+1} − s_{k})^{γ} ≪ ∑_{j=1}^{∞} L(2^{j})2^{γj} ≪ x.  (11)

for 0 ≤ γ ≤ r + k − 1. In particular, statement (10) gives (11) for 0 ≤ γ ≤ k.

With these observations in mind, we can now see the advantage of using M_{r} in place of the more natural moment M_{r}^{∗} defined in (7). We can obtain an analog of Lemma 2 for M_{r}^{∗}, but only for even positive integers r. Moreover, we conjecture that

M_{r}^{∗}(x, h) ≪ xh^{r/2k}  (12)

whenever r is even. This conjecture was proved by Hall [3] in the case r = 2, k = 2, and his argument is easily generalized to r = 2 and arbitrary k. But the truth of (12) would give (11) only for 0 ≤ γ < r + k − 1 − r/2k. Thus M_{r}^{∗} has two deficiencies when compared to M_{r}: the former is useful only for even r, and it gives a narrower range of γ.

3. Preliminary transformations of M_{r}

We start by writing

M_{r}(x, h) = ∑_{\frac{1}{2} x<n≤x} F^{r}(n),

where F (n) = F (n, h) is defined as

F (n) = ∑_{n<m≤n+h} ∑_{{p^{k}|m \atop p>H}} 1.

It is convenient to use a splitting–up argument on F (n), so we write

F (n) = ∑_{X∈X} G(n, X),

where X = {2^{j}H : j = 0, 1, 2, …; 2^{j}H < x^{1/k}}, and

G(n, X) = ∑_{n<m≤n+h} ∑_{{p^{k}|m \atop X<p≤2X}} 1 = ∑_{X<p≤2X} ∑_{\frac{n}{p^{k}} <l≤ \frac{(n+h)}{p^{k}}} 1.

Now h/p^{k} < h/X^{k} ≤ h/H^{k} < 1, so the inner sum is at most 1. Consequently, G(n, X) is the cardinality of the set

G(n, X) = {p : X < p ≤ 2X and there exists l such that n < p^{k}l ≤ n + h}.

Since \frac{1}{2}x < n ≤ x and h ≤ x, the summation conditions imply that L < l ≤ L_{1}, where

L = \frac{x}{2^{k+1}X^{k}} and L_{1} = 2^{k+2}L.

Now let ϵ > 0 be a real number to be chosen in due course. By Hölder’s inequality,

F^{r}(n) ≤ ( ∑_{X∈X} X^{− \frac{ϵ}{r−1}} )^{r−1} ∑_{X∈X} X^{ϵ}G^{r}(n, X) ≪ H^{−ϵ} ∑_{X∈X} X^{ϵ}G^{r}(n, X).

If G(n, X) < r, then G^{r}(n, X) < r^{r−1}G(n, X). If G(n, X) ≥ r, then

G^{r}(n, X) ≪ \binom{G(n, X)}{r}.

In any case,

G^{r}(n, X) ≪ G(n, X) + \binom{G(n, X)}{r}.

Using the definition of G(n, X) as the cardinality of G(n, X), we find that

\binom{G(n, X)}{r} = |\{(p_{1}, …, p_{r}) : p_{1} < … < p_{r} and p_{i} ∈ G(n, X), for i = 1, …, r\}|.

Let Θ denote the conditions

X < p_{1} < … < p_{r} ≤ 2X , L < l_{1}, …, l_{r} ≤ L_{1}.  (13)

Then

∑_{\frac{1}{2} x<n≤x} \binom{G(n, X)}{r} ≤ ∑_{Θ} ∑_{{\frac{1}{2} x<n≤x \atop p_{i}^{k}l_{i}−h≤n<p_{i}^{k}l_{i};i=1,...,r}} 1.

The inner sum is at most h, and it is 0 unless |p_{1}^{k}l_{1} − p_{i}^{k}l_{i}| < h for i = 2, …, r. Therefore

∑_{\frac{1}{2} x<n≤x} \binom{G(n, X)}{r} ≤ hS_{r}(X),  (14)

where S_{r}(X) = S_{r}(X, h) is the number of 2r-tuples (p_{1}, …, p_{r}, l_{1}, …, l_{r}) satisfying (13) and

|p_{1}^{k}l_{1} − p_{i}^{k}l_{i}| < h for i = 2, …, r.  (15)

Combining the above, we see that

∑_{\frac{1}{2} x<n≤x} G^{r}(n, X) ≪ h(S_{1}(X) + S_{r}(X)),

and therefore

M_{r}(x, h) ≪ \frac{h}{H^{ϵ}} ∑_{X∈X} X^{ϵ}(S_{1}(X) + S_{r}(X)).

In the conditions defining S_{r}, there are

≪ (\frac{XL}{log X} )^{r} ≪ (\frac{x}{X^{k−1} log X} )^{r}

2r-tuples that satisfy (13). The probability that (15) is satisfied is ≪ h/x for each i, i = 2, …, r. These heuristics suggest Conjecture S. If r is a positive integer and X ≥ H, then

S_{r}(X) ≪ (\frac{XL}{log X} )^{r} (\frac{h}{x})^{r−1} ≪ \frac{h^{r−1}x}{X^{(k−1)r} log^{r} X} .

Note that Conjecture S is trivially true for r = 1. Moreover, if Conjecture S is true for a given value of r, then Conjecture M follows for the same r, and (2) is true for γ ≤ r + k − 1. With a small variant of Lemma 2, one can show that Conjecture S implies (2) for 0 ≤ γ ≤ (k − 1)r + 1. We shall elaborate on this point further in Section 5.

4. Bounds for S_{r}

The following lemma uses congruence arguments to bound S_{r}. Note that it yields Conjecture S when X ≤ x^{1/kr}. Filaseta [2] used this result in the special cases k = 2 and r = 2, 3.

Lemma 3. If X ≥ H and r ≥ 2 then

S_{r}(X) ≪ \frac{h^{r−1}x}{X^{(k−1)r} log^{r} X} + \frac{h^{r−1}X^{r}}{log^{r} X} .

Proof. Suppose X < p_{1} < … < p_{r} ≤ 2X. Let V (p_{1}, …, p_{r}) be the number of r-tuples l_{1}, …, l_{r} with L < l_{1}, …, l_{r} ≤ L_{1}, and

|p_{1}^{k}l_{1} − p_{i}^{k}l_{i}| < h for i = 2, …, r.  (16)

From equation (16), we see that l_{1} lies in one of < 2h + 1 residue classes modp_{i}^{k}. By the Chinese Remainder Theorem, l_{1} is in one of ≪ h^{r−1} residue classes modp_{2}^{k} … p_{r}^{k}. The number of integers l_{1} in a given residue class is

≪ \frac{L}{X^{k(r−1)}} + 1 ≪ \frac{x}{X^{kr}} + 1.

We further note the inequalities

\frac{p_{1}^{k}l_{1} − h}{p_{i}^{k}} < l_{i} < \frac{p_{1}^{k}l_{1} + h}{p_{i}^{k}}.

Since 2h/p_{i}^{k} < 1, we see that l_{i} is uniquely determined by p_{1}, p_{i}, and l_{1}. Therefore

V (p_{1}, …, p_{r}) ≪ (\frac{x}{X^{kr}} + 1) h^{r−1},

and

S_{r}(X) ≪ (\frac{x}{X^{kr}} + 1) h^{r−1} \frac{X^{r}}{log^{r} X} ≪ \frac{h^{r−1}x}{X^{(k−1)r} log^{r} X} + \frac{h^{r−1}X^{r}}{log^{r} X} .

This completes the proof.

In the proof of the Theorem, we will use only the case r = 2 of Lemma 3. However, we have chosen to present the theorem in full generality because the proof is simple and because of the connection with Conjecture S.

Lemma 4. If X ≥ H, then

S_{2}(X) ≪ \frac{x}{X^{k−1} log X} .

Although this is weaker than Conjecture S, it is still sufficient to prove Conjecture M in the case r = 2 and (2) for γ ≤ k + 1. In the squarefree case, Lemma 4 was proved by Hooley [5], and our proof is a straight-forward modification of his argument.

Proof. By our definition, S_{2} is the number of four-tuples p_{1}, p_{2}, l_{1}, l_{2} satisfying the relations

|p_{2}^{k}l_{2} − p_{1}^{k}l_{1}| < h , X < p_{1} < p_{2} ≤ 2X , L < l_{1}, l_{2} ≤ L_{1}.

We write d = (l_{1}, l_{2}). Then l_{1} = l′_{1}d, l_{2} = l′_{2}d, and (l′_{1}, l′_{2}) = 1. Substituting for l_{1}, l_{2} and suppressing superscripts, we find that

S_{2}(X) = ∑_{d} V (d),

where V (d) is the number of solutions of

|p_{2}^{k}l_{2} − p_{1}^{k}l_{1}| < h/d , X < p_{1} < p_{2} ≤ 2X , L/d < l_{1}, l_{2} ≤ L_{1}/d , (l_{1}, l_{2}) = 1.

We have two separate methods of estimating V (d). First, we write

V (d) = ∑_{X<p_{1}<p_{2}≤2X} V_{1}(d; p_{1}, p_{2}),

where V_{1}(d; p_{1}, p_{2}) is the number of solutions in l_{1}, l_{2} of

|p_{2}^{k}l_{2} − p_{1}^{k}l_{1}| < h/d, \frac{L}{d} < l_{1}, l_{2} ≤ \frac{L_{1}}{d} , (l_{1}, l_{2}) = 1.

From the first of these conditions, we see that

|\frac{l_{2}}{l_{1}} − \frac{p_{1}^{k}}{p_{2}^{k}} | < \frac{h}{dl_{1}p_{2}^{k}} ≪ \frac{h}{x}.

Now any two fractions l_{1}/l_{2} are separated by at least d^{2}/L_{1}^{2}, so

V_{1}(d; p_{1}, p_{2}) ≪ \frac{hL^{2}}{d^{2}x} + 1.

Summing over p_{1} and p_{2}, we get

V (d) ≪ (\frac{hL^{2}}{d^{2}x} + 1) \frac{X^{2}}{log^{2} X} ≪ \frac{hx}{d^{2}X^{2k−2} log^{2} X} + \frac{X^{2}}{log^{2} X} .  (17)

A second approach to V (d) is to first count solutions for fixed values of l_{1} and l_{2}. Accordingly, we write

V (d) = ∑_{{L/d<l_{1},l_{2}≤L_{1}/d \atop (l_{1},l_{2})=1}} V_{2}(d; l_{1}, l_{2}),

where V_{2}(d; l_{1}, l_{2}) is the number of solutions of

|p_{2}^{k}l_{2} − p_{1}^{k}l_{1}| < h/d , X < p_{1} < p_{2} ≤ 2X.

From these conditions, we see that

|\frac{p_{2}^{k}}{p_{1}^{k}} − \frac{l_{1}}{l_{2}} | ≪ \frac{h}{X^{k}L} ≪ \frac{h}{x}.

Since p_{2}/p_{1} ≫ 1, we also have

|\frac{p_{2}}{p_{1}} − \frac{l_{1}^{1/k}}{l_{2}^{1/k}} | ≪ \frac{h}{x}.

Now any two values \frac{p_{2}}{p_{1}} are separated by at least 1/(4X^{2}), so

V_{2}(d; l_{1}, l_{2}) ≪ \frac{hX^{2}}{x} + 1.

Summing over l_{1} and l_{2}, we derive

V (d) ≪ \frac{L^{2}}{d^{2}} (\frac{hX^{2}}{x} + 1) ≪ \frac{hx}{d^{2}X^{2k−2}} + \frac{x^{2}}{d^{2}X^{2k}} .  (18)

From our two estimates, (17) and (18), for V (d), we see that

V (d) ≪ \frac{hx}{d^{2}X^{2k−2}} + min (\frac{x^{2}}{d^{2}X^{2k}} , \frac{X^{2}}{log^{2} X} ) .

Summing over d, we obtain

S_{2}(X) ≪ \frac{xh}{X^{2k−2}} + ∑_{d≤xX^{−k−1} log X} \frac{X^{2}}{log^{2} X} + ∑_{d≥xX^{−k−1} log X} \frac{x^{2}}{d^{2}X^{2k}}

≪ \frac{xh}{X^{2k−2}} + \frac{x}{X^{k−1} log X} ≪ \frac{x}{X^{k−1} log X} .

This completes the proof of Lemma 4.

5. A Modified Bound for L(2h) To take full advantage of the results of the last section, we need the following modifi-

cation of Lemma 2. Lemma 5. Let H be as in (8) and let S_{2} be as defined in Section 3. If ϵ > 0, then

L(2h) ≪ \frac{x}{h^{2k−1} log^{k} h} + \frac{1}{h^{2}H^{ϵ}} ∑_{X∈X} X^{ϵ}S_{2}(X).

Proof. Let F (n) and G(n, X) be as defined in Section 3. Define

F_{0}(n) = ∑_{{X∈X \atop G(n,X)≤1}} G(n, X) and F_{1}(n) = ∑_{{X∈X \atop G(n,X)>1}} G(n, X).

Then F (n) = (F_{0} + F_{1})(n). From Lemma 1, we see that if there are no k-free integers in the interval (n, n + 2h], then F (m) ≥ \frac{1}{6}h for m = n, n + 1, …, n + h − 1. At least one of F_{0}(m) or F_{1}(m) exceeds \frac{1}{12}h for at least \frac{1}{2}h values of m. Thus either

∑_{n<m≤n+h} F_{0}^{k}(m) ≫ h^{k+1} or ∑_{n<m≤n+h} F_{1}^{2}(m) ≫ h^{3}.

We therefore obtain

L(2h) ≪ \frac{1}{h^{k+1}} ∑_{\frac{1}{2} x<n≤x} F_{0}^{k}(n) + \frac{1}{h^{3}} ∑_{\frac{1}{2} x<n≤x} F_{1}^{2}(n) = \frac{1}{h^{k+1}} U_{0} + \frac{1}{h^{3}} U_{1},

say.

To bound U_{0}, we use Hölder’s Inequality to get

F_{0}^{k}(n) ≪ ( ∑_{{X∈X \atop G(n,X)≤1}} X^{−1/(k−1)})^{k−1} ∑_{{X∈X \atop G(n,X)≤1}} XG^{k}(n, X) ≪ H^{−1} ∑_{X∈X} XG(n, X).

In the last step, we use the simple but crucial fact that G(n, X) ≤ 1 implies that G(n, X)^{k} = G(n, X). Now we sum over n, and we use (14) and the trivial estimate S_{1}(X) ≪ xX^{−k+1} log^{−1} X to deduce that

U_{0} ≪ hH^{−1} ∑_{X∈X} XS_{1}(X) ≪ \frac{x}{h^{k−2} log^{k} h}.

To bound U_{1}, we again use Hölder’s Inequality, and we get

F_{1}^{2}(n) ≪ ( ∑_{X∈X} X^{−ϵ}) ∑_{{X∈X \atop G(n,X)>1}} X^{ϵ}G^{2}(n, X) ≪ H^{−ϵ} ∑_{X∈X} X^{ϵ}\binom{G(n, X)}{2}.

We now sum over n and use (14) to get

U_{1} ≪ \frac{h}{H^{ϵ}} ∑_{X∈X} X^{ϵ}S_{2}(X).

Combining the above estimates for U_{0} and U_{1} completes the proof.

The proof of Lemma 5 can be easily modified to give

L(2h) ≪ \frac{x}{h^{(k−1)r+1} log^{k} h} + \frac{1}{h^{r}H^{ϵ}} ∑_{X∈X} X^{ϵ}S_{r}(X).

This together with Conjecture S would imply that

L(2h) ≪ \frac{x}{h^{(k−1)r+1} log^{k} h}.

Consequently, as we stated at the end of Section 3, Conjecture S implies (2) for 0 ≤ γ ≤ (k − 1)r + 1.

6. Proof of the Theorem

From Lemma 3 with r = 2 and Lemma 4, we conclude that

S_{2}(X) ≪ \frac{hx}{X^{2k−2} log^{2} X} + min (\frac{x}{X^{k−1} log X} , \frac{hX^{2}}{log^{2} X} )

≪ \frac{hx}{X^{2k−2} log^{2} X} + (\frac{x}{X^{k−1} log X} )^{2/(k+1)} (\frac{hX^{2}}{log^{2} X} )^{(k−1)/(k+1)}

≪ \frac{hx}{X^{2k−2} log^{2} X} + \frac{h^{(k−1)/(k+1)}x^{2/(k+1)}}{log^{2k/(k+1)} X} .

In the next-to-last line, we have used the simple fact that if A, B are positive and if 0 ≤ a ≤ 1, then

min(A, B) ≤ A^{a}B^{1−a}.

Now we apply Lemma 5 to deduce that

L(2h) ≪ \frac{x}{h^{2k−1}} + \frac{x^{2/(k+1)+ϵ}}{h^{(k+3)/(k+1)}} .  (19)

We are ignoring log factors since they are irrelevant to our estimates. Also, the crude estimate (x^{1/2k}/H)^{ϵ} < x^{ϵ} suffices for our purposes. As we noted in (5), we may assume that h ≪ x^{1/(2k+1)} log x. Consequently, L(2h) ≪ xh^{−γ−ϵ} whenever γ < 2k − 1 − ϵ and

γ < 2k − 2 + \frac{4}{k + 1} − ϵ′,  (20)

where ϵ′ is some function of ϵ and k that tends to 0 as ϵ tends to 0. (In fact, the choice ϵ′ = 10kϵ suffices.) The proof is completed by taking ϵ small enough to satisfy (20) and using (4).

As we stated in the introduction, the constants implied by ≪- and O-symbols depend at most on γ, k, r and ϵ. Note ϵ is chosen in terms of γ and k, and that we use Lemma 3 only for r = 2. This justifies our statement that the implied constant in the Theorem depends only on γ and k.

In conclusion, we make the following remarks. It is only at the last stage that our argument requires the hypothesis k ≥ 3. For (19) is valid for k ≥ 2, but it actually yields 0 ≤ γ < min(2k − 1, 2k − 2 + 4/(k + 1)). Thus our argument gives only 0 ≤ γ < 3 in the case k = 2, while Filaseta [2] has obtained 0 ≤ γ < 29/9. The crucial extra ingredient in Filaseta’s argument is an upper bound for G(n, X) that is derived from the method of divided differences. Such a bound can also be proved in the case of k ≥ 3, but it is not helpful for the problem considered here. That bound might become useful if a better exponent becomes available in the gap result (6).

Acknowledgments This paper was written while I was a visitor in the Mathematics Department of the

University of Illinois at Urbana–Champaign, and I thank them for their hospitality. I also thank Michael Filaseta for many helpful comments and for supplying me with a preprint of [2].

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