Grimm's conjecture and smooth numbers

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SHANTA LAISHRAM AND M. RAM MURTY

ABSTRACT. Letg(n)be the largest positive integerk such that there are distinct primesp_i for1 ≤ i ≤kso thatp_i|n+i. This function is related to a celebrated conjecture of C.A. Grimm. We establish upper and lower bounds forg(n)by relating its study to the distribution of smooth numbers. Standard conjectures concerning smooth numbers in short intervals implyg(n) = O(n) for any > 0. We also prove unconditionally thatg(n) = O(n^α)with0.45< α <0.46. The study ofg(n)and cognate functions has some interesting implications for gaps between consecutive primes.

1. INTRODUCTION

In 1969, C.A. Grimm [8] proposed a seemingly innocent conjecture regarding prime factors of consecutive composite numbers. We begin by stat- ing this conjecture.

Let n ≥ 1 and k ≥ 1 be integers. Suppose n+ 1,· · · , n +k are all composite numbers. Then there are distinct primesP_i such thatP_i|(n+i) for 1 ≤ i ≤ k. That this is a difficult conjecture having several interesting consequences was first pointed out by Erd¨os and Selfridge [5]. For example, the conjecture implies there is a prime between two consecutive square numbers, something which is out of bounds for even the Riemann hypothesis. In this paper, we will pursue this theme. We will relate several results and conjectures regarding smooth numbers (defined below) to Grimm’s conjecture.

To begin, we say that Grimm’s conjecture holds forn andk if there are distinct primesPisuch thatPi|(n+i)for1≤i≤kwhenevern+1, . . . , n+

k are all composites. For positive integersn > 1andk, we say that(n, k) has a prime representation if there are distinct primes P₁, P₂, . . . , P_k with P_j|(n+j),1≤j ≤k. We defineg(n)to be the maximum positive integer k such that (n, k) has a prime representation. It is an interesting problem to find the best possible upper bounds and lower bounds for g(n). If n⁰

2010Mathematics Subject Classification. 11N36 (primary), 11N25 (secondary).

Key words and phrases. Grimm’s conjecture, smooth numbers, Selberg’s sieve.

Research of the second author partially supported by an NSERC Discovery grant.

1

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is the smallest prime greater thann, Grimm’s conjecture would imply that g(n)> n⁰−n. On the other hand, it is clear thatg(2^m)<2^m form >3.

The question of obtaining lower bounds for g(n) was attacked using methods from transcendental number theory by Ramachandra, Shorey and Tijdeman [15] who derived

g(n)≥c

logn log logn

3

for n > 3 and an absolute constant c > 0. In other words, for any sufficiently large natural numbern, (n, k)has a prime representation ifk (logn/log logn)³.

We prove:

Theorem 1.

(i) There exists anα < ¹₂ such thatg(n)< n^α for sufficiently largen.

(ii) For >0, we have|{n≤X :g(n)≥n}| Xexp(−(logX)¹³⁻) where the implied constant depends only on.

We show in Section 3 that0.45 < α < 0.46is permissible in Theorem 1(i).

For realx, y, letΨ(x, y)denote the number of positive integers≤ xall of whose prime factors do not exceedy. These arey-smooth numbersand have been well-studied. In 1930, Dickman [3] proved that for anyα ≤1,

x→∞lim

Ψ(x, x^α) x

exists and equalsρ(1/α)whereρ(t)is defined fort ≥ 0as the continuous solution of the equationsρ(t) = 1for0 ≤ t ≤ 1and−tρ⁰(t) = ρ(t−1) for t ≥ 1. Later authors derived refined results. We refer to [11] for an excellent survey on smooth numbers. An important conjecture on smooth numbers in short intervals is the following.

Conjecture 1.1. Let >0. For sufficiently largex, we have Ψ(x+x, x)−Ψ(x, x)x.

This is still open. Assuming Conjecture 1.1, we have the following.

Theorem 2. Let > 0. Theng(n) < n for largen assuming Conjecture 1.1.

Letp_i denote theith prime. As a consequence of Theorem 2, we obtain Corollary 1.2. Assume Grimm’s conjecture and Conjecture 1.1. Then for any >0,

p_i+1−p_i < p_i (1)

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for sufficiently largei.

If we assume Grimm’s conjecture alone, then Erd˝os and Selfridge[5]

have shown that

p_i+1−p_i (p_i/logp_i)^1/2,

which is something well beyond what the Riemann hypothesis would imply about gaps between consecutive primes. Indeed, the Riemann hypothesis implies an upper bound of O(p^1/2_i (logp_i)). It was conjectured by Cram´er [1] in 1936 that

p_i+1−p_i (logp_i)²

If Cram´er’s conjecture is true, then the result of Ramachandra, Shorey and Tijdeman [15] would imply Grimm’s conjecture, at least for sufficiently large numbers. In [13], Laishram and Shorey verified Grimm’s conjecture for alln <1.9×10¹⁰. They also checked thatp_i+1−p_i <1 + (logp_i)²for i≤8.5×10⁸.

It is worth mentioning that there are several weaker versions of Grimm’s conjecture that have also been attacked using methods of transcendental number theory. For an integer ν > 1, we denote by ω(ν) the number of distinct prime divisors ofνand letω(1) = 0. A weaker version of Grimm’s conjecture states that if n + 1, n + 2, . . . , n+ k are all composite numbers, thenω(Qk

i=1(n+i))≥ k. This conjecture is also open though much progress has been made towards it by Ramachandra, Shorey and Tijdeman [16].

We defineg₁(n)to be the maximum positive integerk such that ω(

l

Y

i=1

(n+i))≥l

for all1≤l≤k. Observe thatg₁(n)≥g(n). We prove Theorem 3. There exists aγwith0< γ < ¹₂ such that

g(n)≤g₁(n)< n^γ (2)

for large values ofn.

We show in Section 5 thatγ = ¹₂−₃₉₀¹ is permissible. This result will be proved as a consequence of the following theorem which is of independent interest.

Theorem 4. Suppose there exists0< α < ¹₂ andδ >0such that X

j≤m^α

π(m+m^α

j )−π(m j )

≥δm^α (3)

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holds for largem. Theng1(n)< n^γwith γ = max(α,1−δ(1−α)

2−δ )< 1 2. for largen.

A conjecture coming from primes in short intervals states that(see for example Maier [12]):

π(x+x^α)−π(x)∼ x^α

logx asx→ ∞.

Assuming this conjecture, we obtain form → ∞, X

j≤m^α

π(m+m^α

j )−π(m j )

∼ X

j≤m^α m^α

j

log ^m_j = m^α logm

X

j≤m^α

1 j(1− _log^log_m^j )

∼ m^α logm

Z m^α 1

dt t(1− _log^log_m^t). Takingu= _log^log_m^t, we get

X

j≤m^α

π(m+m^α

j )−π(m j )

∼m^α Z α

0

du

1−u =m^α[−log(1−u)]^α₀ =−m^αlog(1−α)

asm → ∞. Continuing as in the proof of Theorem 4, we obtaing₁(n) <

n^α¹ with

α₁ = max(α,1 + (1−α) log(1−α) 2 + log(1−α) ).

Sincelog(1−α)≈ −αfor0< α <1, we see that 1 + (1−α) log(1−α)

2 + log(1−α) ≈ 1−α(1−α) 2−α = 1

2(1−α+α²)(1− α 2)⁻¹

≈ 1

2(1−α+α²)(1 + α 2) = 1

4(2−α+α²+α³)

and the function ¹₄(2−α+α²+α³)attains its maximum atα = ¹₃ where the value ofα₁ ≈0.4567. Hence, it is unlikely that we can get a result with g₁(n)< n^γwithγ < .4567, by these methods. As such, this valueg₁(n) = O(n^α) seems to agree with the permissible value of 0.45 < α < 0.46in g(n) = O(n^α).

It was noted by Erd¨os and Selfridge in [5] that “the assertionγ < ¹₂ seems to follow from a recent result of Ramachandra[14]but we do not give the details here.” In [6], Erd¨os and Pomerance noted again that “Indeed from

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the proof in[14], it follows that there is an α > 0such that for all large n a positive proportion of the integers in (n, n+ n^α] are divisible by a prime which exceeds n¹⁵²⁶. Using this result with the method in [5] gives g(n) < n¹²^−c for some fixed c > 0 and all largen.” However there is no proof anywhere in the literature about this fact. We give a complete proof in this paper by generalizing the result of Ramachandra [14] in Lemma 2.5.

2. PRELIMINARIES ANDLEMMAS

We introduce some notation. We shall always writepfor a prime number.

LetΛ(n)be the von Mangoldt function which is defined as Λ(n) = logp if n = p^r for some positive integer r and 0 otherwise. We writeθ(x) = P

p≤xlogp. For realx, y, letΨ(x, y)denote the number of positive integers

≤ xall of whose prime factors do not exceed y. We also writelog₂xfor log logx. We begin with some results from prime number theory.

Lemma 2.1. Letk, t∈Zandx∈R. We have (i) π(x)< _log^x_x(1 + ^1.2762_log_x )forx >1.

(ii) p_t> t(logt+ log₂t−c₁)for somec₁ >0and for larget.

(iii) θ(x)≤1.00008xforx >0.

(iv) θ(pt)> t(logt+ log₂t−c2)for somec2 >0and for larget.

(v) k!>√

2πk e^−kk^ke^12k+1¹ fork > 1.

The estimate (ii) is due to Rosser and Schoenfeld [18]. Inequalities (i),(iii) and(iv)are due to Dusart [4]. The estimate(v) is Stirling’s formula, see [17].

The following results are due to Friedlander and Lagarias [7].

Lemma 2.2. Let0 < < 1be fixed. Then there are positive constantsc₀ andc₁depending only onsuch that there are at mostc₁Xexp(−(logX)¹³⁻) manynwith1≤n ≤Xwhich do not satisfy

Ψ(n+n, n)−Ψ(n, n)≥c₀n. (4)

Lemma 2.3. There exist positive absolute constantsαandc₁ with³₈ < α <

1

2 such that

Ψ(n+n^α, n^α)−Ψ(n, n^α)> c₁n^α. (5)

for sufficiently largen.

Lemma 2.2 is obtained by taking α = β = in [7, Theorem 5] and Lemma 2.3 is obtained by taking x = n, y = z = n^α with α = ¹₂ − ^η₂ in [7, Theorem 2.4]. From [10, Theorem 2] and the remarks after that, a permissible value ofαin Lemma 2.3 is given by anαwith0.45< α <0.46.

The following is the key lemma which follows from the definition ofg(n) and relates the study ofg(n)to smooth numbers.

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Lemma 2.4. Letx, y, z ∈ Rbe such that Ψ(x+z, y)−Ψ(x, y) > π(y).

Theng(bxc)< z.

Proof. Letx ≤ n₁ < n₂ < · · · < n_t ≤ x+z be ally-smooth numbers witht > π(y). Then,(n₁, n_t−n₁)does not have a prime representation. In particular,(bxc,bzc)has no prime representation. Thusg(bxc)< z.

The next result is a generalization of a result of Ramachandra [14].

Lemma 2.5. Let ₃₃¹ < λ < ₂₉¹. Forα= ^1−λ₂ and for sufficiently largex, we have

X

n≤x^α

π(x+x^α

n )−π(x n)

≥(1 4 +λ

2 −⁰)x^α (6)

where⁰ >0is arbitrary small.

We postpone the proof of Lemma 2.5 to Section 4.

3. PROOF OFTHEOREMS 1AND2

Proof of Theorem 1: (i)Letαbe given by Lemma 2.3. We apply Lemma 2.4 by takingx = n, z =y =n^α. Sinceπ(y) = π(n^α) <2_αⁿ_log^α_n < c₁n^α for sufficiently largen, the assertion follows from Lemma 2.4 and Lemma 2.3. As remarked after Lemma 2.3, a permissible value of α is given by 0.45< α <0.46.

(ii)Let > 0be given. By(i), we may assume that < ¹₂. Sinceπ(n)<

2_logⁿ_n < c₀n for sufficiently largenwherec₀ is given by Lemma 2.2, the assertion now follows from Lemma 2.4 by takingx = n, z = y = n and

Lemma 2.2.

Proof of Theorem 2: Let >0be given. We apply Lemma 2.4 by taking x = n, z = y = n. Sinceπ(y) = π(n) < 2_logⁿ_n n for sufficiently largen, the assertion follows from Lemma 2.4 and Conjecture 1.1.

4. PROOF OFLEMMA 2.5

We follow the proof of Ramachandra in [14] and fill in the details as we go along. Letα < ¹₂ and0 < β < ¹₂. By taking = x^α−1 in [14, Lemma 1], we obtain

X

n≤x^1−α

π(x+x^α

n )−π(x n)

log x

n = (1−α)x^αlogx+O(x^α).

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We divide the interval[β,1−α]as0< β=β0 < β1 < . . . < βm = 1−α for somem. For0< r < s <1, let

S(r, s) = X

x^r≤n≤x^s

π(x+x^α

n )−π(x n)

logx

n. (8)

We would like to get an upper bound forS(β,1−α) = Pm−1

i=0 S(β_i, β_i+1).

We first prove the following lemma which is minor refinement of [14, Lemma 3].

Lemma 4.1. Letx≥1and1≤R ≤S≤x^1−α. For an integerd≥1, let

R_d= X

R≤n≤S

x+x^α nd

− x nd

(9) .

Then

X

R≤n≤S

π(x+x^α

n )−π(x n)

≤ (2−)x^α logz log(S

R + 2)

1 +O(1

R + 1

logz)

+O(zmax

d≤z |R_d|) (10)

wherez ≥3is an arbitrary real number and >0is arbitrary small.

Proof. Let

T = ∪

R≤n≤S

(x

n,x+x^α n ]∩Z

.

FromT, we remove those which are divisible by primes≤ √

z and let T₁ be the remaining set. We note that for eachd, the number of integers inT divisible bydis

x^α d

X

R≤n≤S

1 n +R_d

Using Selberg’s sieve as in [14], we obtain the assertion of lemma.

Letφ(u) =u−[u]−¹₂. Then we can write x+x^α

nd

− x nd

= x^α

nd−φ(x+x^α

nd ) +φ( x nd).

The following result is a restatement of [14, Lemma 2] which follows from a result of van der Corput (see [14]).

Lemma 4.2. Let u ≥ 1, V, V₁ be real numbers satisfying 3 ≤ V < V₁ ≤ 2V, V₁ ≥V + 1andu≤η ≤2u. Then

X

V≤n≤V1

φ(η

n) =O(V ¹² logV +V ³²u⁻¹² +u¹³).

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To get an upper bound for S(βi, βi+1), we take R = x^βⁱ, S = x^βⁱ⁺¹ in Lemma 4.1. Recall that β_i+1 ≤ 1−α. We subdivide(R, S]into intervals of type(V,2V]and at most one interval of type(V, V₁]withV₁ ≤2V. We apply Lemma 4.2 twice by takingη= ^x_d andη= ^x+x_d^α to get

R_d =O

x¹²^βⁱ⁺¹+x¹²^(3βⁱ⁺¹⁻¹⁾d¹² + (x d)¹³

(logx)²

=O

x¹⁻³²^αd¹² +x¹³

(logx)² ,

sinceβ_i+1 ≤1−αLet3α− ⁴₃ < δ < ^5α−2₃ and takez =x^δ. Then zmax

d≤z |R_d|=O(x¹⁻³²^α+³²^δ(logx)²) and1− ³₂α+ ³₂δ < α. From (10), we obtain

X

x^βi≤n≤x^βi+1

π(x+x^α

n )−π(x n)

≤ 2x^α

δ (β_i+1−β_i).

Therefore an upper bound for X

x^β≤n≤x^1−α

π(x+x^α

n )−π(x n)

logx

n

is

2x^αlogx

δ ×

{(β₁−β₀)(1−β₀) + (β₂−β₁)(1−β₁) +· · ·+ (β_m−βm−1)(1−βm−1)}.

We takeβ_i’s to be equally spaced and takemsufficiently large. Since 2x^αlogx

δ

Z 1−α β

(1−t)dt = x^αlogx

δ (1−α²−β(2−β)), we obtain with (7) that

X

n≤x^β

π(x+x^α

n )−π(x n)

≥(1−α−⁰ −1−α²−β(2−β)

δ )x^α.

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where1− ³₂α+ ³₂δ < αand⁰ >0is arbitrary small.

Let ₃₃¹ < λ < ₂₉¹ and we putα =β = ^1−λ₂ andδ = 4λ. Then1− ³₂α+

3

2δ < αand hence we obtain (6) from (12).

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5. PROOF OFTHEOREM4AND THEOREM 3 We begin with the proof of Theorem 4.

Proof. Recall thatg₁(n)is the largest integerk such that ω(

l

Y

i=1

(n+i))≥l

for1≤ l ≤ k. Suppose thatg₁(n) > n^γ. Theng₁(n) > n^α. Letk = [n^α].

Then

ω(P)≥k, P =

k

Y

i=1

(n+i).

By (3),

X

j≤k

π

n+k j

−π n

j

≥δk.

Now the intervals [n, n+ k], [n/2,(n +k)/2], ... are disjoint intervals.

In fact, if we write I_j = [n/j,(n+k)/j] = [a_j, b_j](say), then it is easily seen b₁ > a₁ > b₂ > a₂ > b₃ > a₃· · · by virtue of the condition that k < n^α with α < 1/2. A prime q_i (say) lying in the interval I_j satisfies n < jq_i < n+kand consequently is a prime dividingP. Since these primes q_i are all distinct, and all of these primes are greater thann/k ≥ n^1−α, we deduce that there are at leastδkdistinct primes greater thann^1−α dividing P. Letδ⁰ ≥δ be such thatδ⁰k =dδke. Since ω(P)≥ k, there are at least (1−δ⁰)k other primes dividingP and(1−δ⁰)k ∈ Z. Alsok!|P sinceP is a product ofk consecutive numbers. All the prime factors ofk!are less than or equal tok < n^α < n^1−αsinceα <1/2. Hence we get

P ≥k!





Y

k<p<p_(1−δ0)k

p



(n^1−α)^δ⁰^k.

Now we apply the bounds provided by Lemma 2.1. By Lemma 2.1 (iii) and(iv), we obtain

log



 Y

k<p≤p_(1−δ0)k

p



=θ(p(1−δ⁰)k)−θ(k)

≥(1−δ⁰)klog(1−δ⁰)k+ (1−δ⁰)k{log₂(1−δ⁰)k−c₂}

−1.00008k

>(1−δ⁰)klog(1−δ⁰)k+k(c₃log₂c₃k−c₄)

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wherec₃, c₄are positive constants. This together withk!>(^k_e)^kby Lemma 2.1(v)andP <(2n)^kimply

2n > k

e(1−δ⁰)^1−δ⁰k^1−δ⁰c₅(logc₃k)^c³n^δ⁰^(1−α)

= 1

e(1−δ⁰)^1−δ⁰c₅(logc₃k)^c³( k

n^γ)^2−δ⁰n^γ(2−δ⁰⁾n^δ⁰^(1−α)

>2n^γ(2−δ⁰^)+δ⁰^(1−α) ≥2nγ(2−δ)+δ(1−α)

2n

for largen sinceδ⁰ > δ and1−α > ¹₂ > γ. This is a contradiction. Thus

g₁(n)< k≤n^α ≤n^γ.

Proof of Theorem 3: From Lemma 2.5, we obtain (3) withα = ^1−α₂ and δ= ¹₄+^λ₂−es⁰for some ₃₃¹ < λ < ₂₉¹ . Now the assertion follows from (4).

Takingλ= ₃₀¹ + 2⁰for instance, we getγ ≤ ¹₂ − ₃₉₀¹ . Remark: It is possible to improve the result we have obtained. However the improvement is not substantial. Indeed the result of van der Corput has been improved and using methods of Harman and Baker [2], it is possible to obtain a small refinement. The details are rather technical and will be discussed in a future paper by the junior author.

ACKNOWLEDGMENTS

We thank the referee for helpful remarks and careful reading of our paper. We would also like to thank Sanoli Gun and Purusottam Rath for their careful reading and corrections on an earlier version of this paper.

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STAT-MATHUNIT, INDIANSTATISTICALINSTITUTE, NEWDELHI, INDIA

E-mail address:[email protected]

DEPARTMENT OFMATHEMATICS, QUEEN’SUNIVERSITY, KINGSTON, ON, CANADA

E-mail address:[email protected]