aa r X i v : . [ m a t h . N T ] F e b PRIME-REPRESENTING FUNCTIONS ANDHAUSDORFF DIMENSION
KOTA SAITO
Abstract.
In 2010, Matom¨aki investigated the set of
A > A c k is a prime number for every k ∈ N , where c ≥ Introduction
Let us N denote the set of all positive integers. We say that a function f : N → N is prime-representing if f ( k ) is a prime number for each k ∈ N . The existence andconstruction of prime-representing functions have been studied since long ago. Forexample, Mills showed that there exists a constant A > ⌊ A k ⌋ is prime-representing [Mil47], where let us ⌊ x ⌋ denote the integer part of x ∈ R . Kuıpers showedthat the exponent 3 k can be replaced with c k for all integers c ≥
3, that is, for all integers c ≥
3, there exists
A > ⌊ A c k ⌋ is prime-representing [Kuı50]. Further, Nivenextended this result for real c > / . · · · [Niv51]. Wright investigated differenttypes of prime-representing functions. He showed that there exists α > ⌊ α ⌋ , ⌊ α ⌋ , ⌊ α ⌋ , · · · are prime numbers [Wri51]. Those examples of prime-representing functions can bereformed as the iterated composition of certain functions. Indeed, defining λ M ( x ) = x c ,we observe that ⌊ A c n ⌋ = ⌊ λ M ◦ · · · ◦ λ M ( A ) ⌋ . Additionally, defining λ W ( x ) = 2 x , eachterm of the sequence (1.1) can be reformed as ⌊ λ W ◦ · · · ◦ λ W ( α ) ⌋ . Wright studied aclass of such prime-representing functions, and he generalized and arranged the theoryof representing functions. By his work [Wri54, Section 6], we immediately obtain: Theorem 1.1.
Let ǫ be an arbitrarily small positive real number, let ( c k ) k ∈ N be a realsequence satisfying c k ≥ /
348 + ǫ for all k ∈ N , and let C k = c · · · c k . The set of Mathematics Subject Classification.
Primary:11K55, Secondary:11A41.
Key words and phrases. prime-representing function, Hausdorff dimension.
A > satisfying that ⌊ A C k ⌋ is prime-representing is uncountable, nowhere dense, andhas Lebesgue measure . Note that 925 / ≈ . c k ) k ∈ N , we define C k = c · · · c k and let(1.2) W ( c k ) = { A > ⌊ A C k ⌋ is prime-representing } . In view of the work of Wright, we can discern the geometric structure of W ( c k ).Matom¨aki extended Theorem 1.1 for all real sequence ( c k ) k ∈ N with c k ≥ k ∈ N )[Mat10, Theorem 3]. She evaluated the number of short intervals [ n, n + n γ ] ⊆ [ x, x ]which contains at least Cn γ / (log n ) prime numbers, and arrived at the condition c k ≥ W ( c k ) in view of fractal geometry. More precisely, we evaluate the Hausdorffdimension of W ( c k ). For all sets F ⊆ R , let us dim H F denote the Hausdorff dimensionof F . We will give this definition in Section 2. We say that a set F ⊆ R has fullHausdorff dimension if dim H F = 1. Theorem 1.2.
Let ( c k ) k ∈ N be a bounded real sequence satisfying c k ≥ for all k ∈ N .Let C k = c · · · c k for all k ∈ N . Then we have dim H W ( c k ) = dim H { A > ⌊ A C k ⌋ is prime-representing } = 1 . We will prove Theorem 1.2 in Section 6. By combining Matom¨aki’s theorem [Mat10,Theorem 3] and Theorem 1.2, the set W ( c k ) is nowhere dense, has Lebesgue measure0, and has full Hausdorff dimension if ( c k ) k ∈ N satisfies the conditions in Theorem 1.2.Remark that if F ⊆ R has positive Hausdorff dimension, then F is uncountable. Bysubstituting c k = 2 for all k ∈ N , we have Corollary 1.3.
The set of
A > satisfying that ⌊ A k ⌋ is prime-representing is nowheredense, has Lebesgue measure 0, and has full Hausdorff dimension. Note that we impose boundedness on ( c k ) k ∈ N in Theorem 1.2. We can not removethis condition due to technical problems on calculating the Hausdorff dimension. Wepropose two questions which are related with this problem: Question 1.4.
If a real sequence ( c k ) k ∈ N is unbounded and satisfies c k ≥ for all k ∈ N , then what is the Hausdorff dimension of W ( c k ) ? Question 1.5.
What is the Hausdorff dimension of the set of α > such that all terms (1.1) are prime numbers? We can not get any answers of these question.The rest of the article is organized as follows. Firstly, in Section 2, we prepare with amain tool for calculating the Hausdorff dimension of a general Cantor set. In Section 3,
RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 3 we consider a simple case and show that dim H W (3) = 1. In Section 4, we considera general case and propose a lemma for constructing a general Cantor set which is asubset of W ( c k ). In Section 5, we present a proposition for calculating the Hausdorffdimension of W ( c k ). Finally, in Section 6, we provide a proof of Theorem 1.2. Notation 1.6.
Let N be the set of all positive integers, Z be the set of all integers, Q be the set of all rational numbers, and R be the set of all real numbers. For all sets X ,let us X denote the cardinality of X . We write o (1) ( k → ∞ ) for a quantity whichgoes to 0 as k → ∞ . As is customary, we often abbreviate o (1) g ( k ) to o ( g ( k )) for anon-negative function g ( k ). 2. Preparations
We firstly introduce the Hausdorff dimension. For every U ⊆ R , write the diameterof U by diam( U ) = sup x,y ∈ U | x − y | . Fix δ >
0. For all F ⊆ R and s ∈ [0 , H sδ ( F ) = inf ( ∞ X j =1 diam( U j ) s : F ⊆ ∞ [ j =1 U j , diam( U j ) ≤ δ for all j ∈ N ) , and H s ( F ) = lim δ → +0 H sδ ( F ) is called the s - dimensional Hausdorff measure of F . Fur-ther, dim H F = inf { s ∈ [0 ,
1] : H s ( F ) = 0 } is called the Hausdorff dimension of F . We refer Falconer’s book [Fal14] for the readerswho want to know more details on fractal dimensions. In Falconer’s book [Fal14, (4.3)],we can see a general construction of Cantor sets and a technique to evaluate the Haus-dorff dimension of them as follows: Let [0 ,
1] = E ⊇ E ⊇ E · · · be a decreasingsequence of sets, with each E k a union of a finite number of disjoint closed intervalscalled k -th level intervals , with each interval of E k containing at least two intervals of E k +1 , and the maximum length of k -th level intervals tending to 0 as k → ∞ . Then let(2.1) F = ∞ \ k =0 E k . Lemma 2.1 ( [Fal14, Example 4.6 (a)]) . Suppose in the general construction (2.1) each ( k − -st level interval contains at least m k ≥ k -th level intervals ( k = 1 , , . . . ) whichare separated by gaps of at least ǫ k , where < ǫ k +1 < ǫ k for each k . Then dim H F ≥ lim k →∞ log( m · · · m k − ) − log( m k ǫ k ) . K. SAITO
Remark 2.2.
It is an exercise that Lemma 2.1 is still true if we replace “closed inter-vals” with “half-open intervals” in the construction (2.1). A proof of this exercise canbe found in [MS20, Lemma 2.3].3.
On a simple case
In this section, we firstly discuss a simple case because a general case is much morecomplicated. We show that(3.1) dim H W (3) = dim H { A > ⌊ A k ⌋ is prime-representing } = 1 . Let P be the set of all prime numbers, and let π ( x ) = P ∩ [1 , x ]) for all x >
1. Westart with the result given by Baker, Harman, and Pintz [BHP01].
Lemma 3.1.
There exists a constant d > such that π ( x + x / ) − π ( x ) ≥ d x / log x for sufficiently large x > . By this lemma, we have
Lemma 3.2.
There exists d > such that x, x + x / ] ∩ P ) ≥ d x / log x for sufficiently large x > .Proof. Let us take a sufficiently large x >
0. For all 1 ≤ j ≤ x / − / , we have[ x + 2( j − x / , x + 2 jx / ) ⊆ [ x, x + x / ] . Since ( x +2 jx / ) − ( x +2( j − x / ) = 2 x / > ( x +2( j − x / ) / , Lemma 3.1implies that x, x + x / ] ∩ P ) ≥ X ≤ j ≤ x − x + 2( j − x / , x + 2 jx / ) ∩ P ) ≥ X ≤ j ≤ x − d ( x + 2( j − x / ) / log( x + 2( j − x / ) ≥ X ≤ j ≤ x − d x / log x ≥ d x / log x . (cid:3) RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 5
Choose any 0 < δ <
1. Let n > δ , andlet p be a prime number with p ≥ n . Define I = { } . Set a (1) = p . Let m (1) be thecardinality of the set(3.2) [ a (1) , a (1) + a (1) ] ∩ P . By Lemma 3.2 with x = a (1) , we obtain m (1) ≥ d a (1) log a (1) ≥ d a (1) − δ since n is sufficiently large. Further, let a (1 , < a (1 , < · · · < a (1 , m (1))be all elements of (3.2). Note that a (1) ≤ a (1 , < · · · < a (1 , m (1)) ≤ ( a (1) + 1) − . Define I = { (1 , j ) : 1 ≤ j ≤ m (1) } .We repeat this argument by replacing a (1) with a (1 , j ) for every 1 ≤ j ≤ m (1). Forexample, let us consider the case j = 1. Let m (1 ,
1) be the cardinality of the set(3.3) [ a (1 , , a (1 , + a (1 , ] ∩ P . By Lemma 3.2 with x = a (1 , , we obtain m (1 , ≥ d a (1 , / (log a (1 , ≥ a (1 , − δ since a (1 , ≥ a (1) ≥ n and n is sufficiently large. Further, let a (1 , , < a (1 , , < · · · < a (1 , , m (1 , a (1 , ≤ a (1 , , < · · · < a (1 , , m (1)) ≤ ( a (1 ,
1) + 1) − . In general, assume that I k − ⊆ N k − is given for some integer k ≥
2, and a primenumber a ( j ) ≥ n is also given for each j ∈ I k − . Choose any j ∈ I k − . Then let m ( j )be the cardinality of the set(3.4) [ a ( j ) , a ( j ) + a ( j ) ] ∩ P . By Lemma 3.2 with x = a ( j ) , we obtain m ( j ) ≥ d a ( j ) − δ since n is sufficiently large.Further, let a ( j , < a ( j , < · · · < a ( j , m ( j )) be all elements of (3.4). We also have(3.5) a ( j ) ≤ a ( j , < · · · < a ( j , m ( j )) ≤ ( a ( j ) + 1) − . Define I k = { ( j , j k ) : j ∈ I k − , ≤ j k ≤ m ( j ) } .By induction, we have Lemma 3.3.
For every < δ < , there exists n ∈ N such that for every prime number p ≥ n , we can construct I k ⊆ N k ( k = 1 , , . . . ) , a : S ∞ k =1 I k → P with a (1) = p , and m : S ∞ k =1 I k → N satisfying the following: (A1) I = { } , I k = { ( j , j ) : j ∈ I k − , ≤ j ≤ m ( j ) } for all k ≥ K. SAITO (A2) for all k ≥ and for all j ∈ I k − , a ( j ) ≤ a ( j , < · · · < a ( j , m ( j )) ≤ ( a ( j ) + 1) − for all k ≥ and for all distinct j , j ′ ∈ I k − , | a ( j ) − a ( j ′ ) | ≥ for all k ≥ and for all j ∈ I k − , m ( j ) ≥ d a ( j ) − δ .Proof of (3.1) . Let us take any 0 < δ <
1. By Lemma 3.3, for every prime number p ≥ n , we construct I k ⊆ N k ( k = 1 , , . . . ), a : S ∞ k =1 I k → P with a (1) = p , and m : S ∞ k =1 I k → N satisfying (A1) to (A4). We define W = ∞ \ k =1 [ j ∈I k L ( j ) , L ( j ) := [ a ( j ) / k , ( a ( j ) + 1) / k ) . If A ∈ W , then for all k ∈ N , there exists j ∈ I k such that a ( j ) / k ≤ A < ( a ( j ) + 1) / k , which yields that ⌊ A k ⌋ is a prime number for each k ∈ N . Hence we havedim H W ≤ dim H W (3)by the monotonicity of the Hausdorff dimension. Note that the union S j ∈I k L ( j ) is finiteand disjoint for each k ∈ N . Additionally, by (A2) for each integer k ≥ j ∈ I k − , and1 ≤ j k ≤ m ( j ), we have[ a ( j , j k ) / k , ( a ( j , j k ) + 1) / k ) ⊆ [ a ( j ) / k − , ( a ( j ) + 1) / k − ) . This yields that S j ∈I k +1 L ( j ) ⊆ S j ∈I k L ( j ) for every k ∈ N . Therefore the way ofthe construction of W is same as (2.1). Let us take any integer k ≥ j = ( j , . . . , j k − ) ∈ I k − . Each ( k − L ( j ) contains at least m ( j ) k -thlevel intervals. By (A2), we see that m ( j ) ≥ d a ( j ) − δ ≥ d a ( j , . . . , j k − ) − δ ) ≥ · · · ≥ d a (1) k − (2 − δ ) . Therefore each ( k − m k k -th level intervals, where m k := d p k − (2 − δ ) . Further, from (A2) and (A3), for each 1 ≤ j k < m ( j ), k -th levelintervals L ( j , j k ) and L ( j , j k + 1) are separated by gaps of a ( j , j k + 1) / k − ( a ( j , j k ) + 1) / k ≥ k a ( j , j k + 1) / k − ≥ k ( a ( j , . . . , j k − ) + 1) / k − − ≥ k ( a ( j , . . . , j k − ) + 1) / k − − ≥ · · · ≥ k ( a (1) + 1) / − k − . RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 7
This yields that k -th level intervals are separated by gaps of at least ǫ k , where ǫ k := k ( p + 1) / − k − . By applying Lemma 2.1 to W , we obtaindim H W ≥ lim k →∞ log( m m · · · m k − ) − log( ǫ k m k )= lim k →∞ log( d k − p (1+3+ ··· +3 k − )(2 − δ ) )log(3 k ( p + 1) k − − / d − p − k − (2 − δ ) )= (1 − δ/
2) log p (1 + δ ) log p + 3 log(1 + 1 /p ) ≥ − δ/
21 + δ + 3 / ( p log p ) . Therefore for every 0 < δ <
1, there exists n ∈ N such that for all p ≥ n , we havedim H W (3) ≥ − δ/
21 + δ + 3 / ( p log p ) . By taking p → ∞ , and δ →
0, we obtain dim H W (3) ≥
1. Hence we conclude (3.1)since dim H W (3) ≤ dim H R = 1. (cid:3) On a general case
In this section, we study a general case, and construct I k , m , a as we discussed inSection 3. More precisely, we will prove the following lemma at the end of this section: Lemma 4.1.
Let
R > , and let ( c k ) k ∈ N be a real sequence of real numbers satisfying ≤ c k ≤ R for every k ∈ N . Then there exists M = M ( R ) ∈ N such that for every M ≥ M , we can find a prime number p ∈ [ M, M ] , and we can construct I k ⊆ N k ( k = 1 , , . . . ) , a : S ∞ k =1 I k → P with a (1) = p , and m : S ∞ k =1 I k → N satisfying thefollowing: (B1) I = { } , I k = { ( j , j ) : j ∈ I k − , ≤ j ≤ m ( j ) } for all k ≥ for all k ≥ and for all j ∈ I k − , a ( j ) c k ≤ a ( j , < · · · < a ( j , m ( j )) ≤ ( a ( j ) + 1) c k − for all k ≥ and for all distinct j , j ′ ∈ I k − , | a ( j ) − a ( j ′ ) | ≥ there exists an absolute constant d > such that for all k ≥ and for all j ∈ I k − , m ( j ) ≥ d a ( j ) c k − ( c k log a ( j )) − . We will see that the properties (B1) to (B4) imply dim H W ( c k ) = 1 in Section 6. Inthe case c k ≥
2, Lemma 3.1 does not work. Alternatively, we use the following usefullemma given by Matom¨aki:
K. SAITO
Lemma 4.2.
There exist positive constants d < and D such that, for every sufficientlylarge x and every γ ∈ [1 / , , the interval [ x, x ] contains at most Dx / − γ disjointintervals [ n, n + n γ ] for which π ( n + n γ ) − π ( n ) ≤ dn γ log n . Proof.
See [Mat10, Lemma 9]. (cid:3)
Let p be a variable running over the set P . By this lemma, we obtain Lemma 4.3.
Let d be so as in Lemma 4.2. Let B > , β ≥ / , c ≥ , and X ≥ .Assume that at least BX β (log X ) − intervals [ p c , p c + p c − ] are completely contained in [ X c , X c ] . Then there exists M ′ = M ′ ( B ) > such that if X ≥ M ′ , then at least BX β (2 log X ) − intervals [ p c , p c + p c − ] ⊆ [ X c , X c ] satisfy p c , p c + p c − ] ∩ P ) > dp c − c log p . Proof.
We observe that ( c − /c = 1 − /c ∈ [1 / ,
1] by the condition c ≥
2. Thusby Lemma 4.2 with γ = ( c − /c and x = X c , the number intervals [ p c , p c + p c − ] ⊆ [ X c , X c ] such that p c , p c + p c − ] ∩ P ) > dp c − c log p is at least BX β log X − DX c (2 / − ( c − /c ) = BX β log X − DX − c/ ≥ BX β log X − DX / ≥ BX β log X (cid:18) − DB X / − / log X (cid:19) ≥ BX β X if X ≥ M ′ and M ′ is sufficiently large. (cid:3) Proof of Lemma 4.1.
Let ( c k ) k ∈ N be a real sequence satisfying 2 ≤ c k ≤ R for all k ∈ N .Let M be a sufficiently large parameter depending on R , and let us take M ≥ M .Since each p ∈ [ M, (3 / /c M ] satisfies M c ≤ p c ≤ p c + p c − = p c (1 + 1 /p ) ≤ (3 / M c (1 + 1 /M ) ≤ M c if M is sufficiently large. Hence by the prime number theorem, the number of intervals[ p c , p c + p c − ] ⊆ [ M c , M c ] is at least M, (3 / /c M ] ∩ P ) ≥ ((3 / /c − M M .
RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 9
Therefore, by Lemma 4.3 with B = ((3 / /c − / β = 1, c = c , and X = M , thereexists a prime number p ∈ [ M, M ] such that[ p c , p c + p c − ] ⊆ [ M c , M c ] , p c , p c + p c − ] ∩ P ) > dp c − c log p . Let us fix such a prime number p , and set a (1) = p . Define(4.1) I = { } . Assume that I k − ⊆ N k − is given for some integer k ≥
2, and a prime number a ( j ) ≥ M is also given for each j ∈ I k − . Additionally, suppose that a ( j ) c k , a ( j ) c k + a ( j ) c k − ] ∩ P ) > da ( j ) c k − c k log a ( j )for each j ∈ I k − . Let us choose any j ∈ I k − . Then let A ( j ) = [ a ( j ) c k , a ( j ) c k + a ( j ) c k − ] , P ( j ) = A ( j ) ∩ P . For all p ∈ P ( j ), intervals [ p c k +1 , p c k +1 + p c k +1 − ] are subsets of [ a ( j ) c k c k +1 , a ( j ) c k c k +1 ].To see why, it is clear that a ( j ) c k c k +1 ≤ p c k +1 , and we see that p c k +1 + p c k +1 − = p c k +1 (1 + p − ) ≤ ( a ( j ) c k + a ( j ) c k − ) c k +1 (1 + M − ) ≤ a ( j ) c k c k +1 (1 + M − ) R (1 + M − ) ≤ a ( j ) c k c k +1 if M is sufficiently large. Therefore, Lemma 4.3 with c = c k +1 , β = ( c k − /c k , B = d , X = a ( j ) c k implies that at least da ( j ) c k − (2 c k log a ( j )) − prime numbers p ∈ P ( j ) satisfy p c k +1 , p c k +1 + p c k +1 − ] ∩ P ) > dp c k +1 − c k +1 log p . Let m ( j ) be the cardinality of the set (cid:26) p ∈ P ( j ) : p c k +1 , p c k +1 + p c k +1 − ] ∩ P ) > dp c k +1 − c k log p (cid:27) , and let a ( j , < a ( j , < · · · < a ( j , m ( j )) be the all elements of the set. Then we have a ( j ) c k ≤ a ( j , < a ( j , < · · · < a ( j , m ( j )) ≤ ( a ( j ) + 1) c k − , (4.2) m ( j ) ≥ da ( j ) c k − (2 c k log a ( j )) − . (4.3)Further, we define(4.4) I k = { ( j , j k ) : j ∈ I k − , ≤ j k ≤ m ( j ) } . By induction, we construct I k ⊆ N k (for all k ∈ N ) , m : ∞ [ k =1 I k → N , a : ∞ [ k =1 I k → P , satisfying (B1) to (B4) from (4.1), (4.2), (4.3), and (4.4). (cid:3) Calculation of the Hausdorff dimension
Let
B ⊆ N and let ( c k ) k ∈ N be any real sequence. In this section, we will calculate theHausdorff dimension of the set of A > ⌊ A C k ⌋ ∈ B for all k ∈ N .Let R, θ, L, Q >
0. Assume that 1 + θ ≤ c k ≤ R for all k ∈ N . Let I k ⊆ N k ( k = 1 , , . . . ), m : S ∞ k =1 I k → N , and a : S ∞ k =1 I k → B . Let us consider the followingconditions corresponding to (A1) to (A4) and (B1) to (B4):(C1) I = { } , I k = { ( j , j ) : j ∈ I k − , ≤ j ≤ m ( j ) } for all k ≥ k ≥ j ∈ I k − , a ( j ) c k ≤ a ( j , < · · · < a ( j , m ( j )) ≤ ( a ( j ) + 1) c k − k ≥ j , j ′ ∈ I k − , | a ( j ) − a ( j ′ ) | ≥ k ≥ j ∈ I k − , m ( j ) ≥ Qa ( j ) c k − ( c k log a ( j )) − L . Proposition 5.1.
Assume that there exist I k ⊆ N k ( k = 1 , , . . . ) , m : S ∞ k =1 I k → N ,and a : S ∞ k =1 I k → B satisfying (C1) , (C2) , (C3) , and (C4) . There exists a large M = M ( R, θ, L, Q ) > such that if a (1) ≥ M , then we have dim H { A ∈ [ a (1) /c , ( a (1)+1) /c ) : ⌊ A C k ⌋ ∈ B for all k ∈ N } ≥ (cid:18) Ra (1) log a (1) (cid:19) − . We will prove Proposition 5.1 at the end of this section.Let us assume that there exist I k , m, a satisfying (C1) to (C4). Choose a largeparameter M = M ( R, θ, L, Q ) > k ∈ N and j ∈ I k , let L ( j ) = [ a ( j ) /C k , ( a ( j ) + 1) /C k ) . From (C1), (C2), it is clear that for anyfixed k ∈ N , intervals L ( j ) ( j ∈ I k ) are disjoint. Further, L ( j , j k +1 ) ⊆ L ( j ) for all k ∈ N , j ∈ I k , and 1 ≤ j k +1 ≤ m ( j ). Let W := ∞ \ k =1 [ j ∈I k L ( j ) . Then the following holds:
RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 11
Lemma 5.2.
The set W is a subset of { A ∈ [ a (1) /c , ( a (1) + 1) /c ) : ⌊ A C k ⌋ ∈ B for all k ∈ N } . Proof.
Let us take any A ∈ T ∞ k =1 S j ∈I k L ( j ). Then for every k ∈ N there exists j k ∈ I k such that A ∈ L ( j k ). Hence we have(5.1) a ( j k ) /C k ≤ A < ( a ( j k ) + 1) /C k for every k ∈ N . In particular, by substituting k = 1, A ∈ [ a (1) /c , ( a (1) + 1) /c )holds. Further, from (5.1), we see that ⌊ A C k ⌋ ∈ B for all k ∈ N . Therefore we obtainthe lemma. (cid:3) Hence by the monotonicity of the Hausdorff dimension, it suffices to evaluate lowerbounds of dim H W to prove Proposition 5.1. Note that the way of the construction of W is same as (2.1). Thus we can apply Lemma 2.1 to W . We now evaluate m k and ǫ k in Lemma 2.1. Lemma 5.3.
For every k ≥ , disjoint k -th level intervals of W are separated by gapsof at least (5.2) 1 C k ( a (1) + 1) (1 − C k ) /C . Proof.
Fix an arbitrary integer k ≥
2. Let us take any j = ( j , . . . , j k − ) ∈ I k − . Then,for all 1 ≤ j k < m ( j ), by the mean value theorem and (C3), k -th level intervals L ( j , j k )and L ( j , j k + 1) are separated by gaps of a ( j , j k + 1) /C k − ( a ( j , j k ) + 1) /C k ≥ ( a ( j , j k + 1) − a ( j , j k ) − · C k a ( j , j k + 1) /C k − ≥ C k a ( j , j k + 1) /C k − . By applying (C2) iteratively, we obtain1 C k ( a ( j , j k ) + 1) /C k − ≥ C k ( a ( j , . . . , j k − ) + 1) /C k − − c k ≥ C k ( a ( j , . . . , j k − ) + 1) /C k − − c k − c k ... ≥ C k ( a (1) + 1) /c − c c ··· c k = 1 C k ( a (1) + 1) (1 − C k ) /C . Therefore we conclude Lemma (5.3). (cid:3)
We will apply Lemma 2.1 with ǫ k = C − k ( a (1) + 1) (1 − C k ) /C . To evaluate m k , wepresent the following lemma: Lemma 5.4.
Let < s < t . Define f ( x ; t, s ) = x t − s ( t log x ) − L for all real numbers x ≥ . Then f ( x ; t, s ) is increasing in the range x ≥ e L/ ( t − s ) .Proof. Since ddx f ( x ; t, s ) = ( t − s ) x t − s − ( t log x ) L − x t − s L ( t log x ) L − · t · /x ( t log x ) L , we have ddx f ( x ; t, s ) ≥ x ≥ e L/ ( t − s ) . Therefore we conclude Lemma 5.4. (cid:3) Lemma 5.5.
Assume that a (1) ≥ M . For every integer k ≥ , each ( k − -st levelinterval of W completely contains at least Qa (1) ( C k − C k − ) /C ( C k log a (1)) L k -th level intervals if M is sufficiently large.Proof. Let us fix an arbitrary integer k ≥
2. Take any j = ( j , . . . , j k − ) ∈ I k − . Let f ( x ; t, s ) be so as in Lemma 5.4. By (C4), the interval L ( j ) completely contains at least m ( j ) ≥ Q a ( j ) c k − ( c k log a ( j )) L intervals L ( j ′ ) ( j ′ ∈ I k ). By the condition c k ≥ θ and Lemma 5.4, f ( x ; c k ,
1) isincreasing in the range x ≥ e L/θ . Note that from (C2), M ≤ a (1) ≤ a ( j ) for all j ∈ S ∞ k =1 I k . Hence by (C2), we have f ( a ( j ); c k , ≥ f ( a ( j , . . . , j k − ) c k − ; c k , f ( a ( j , . . . , j k − ); c k − c k , c k − )if M ≥ e L/θ . Note that c k − c k − c k − = c k − ( c k − ≥ (1 + θ ) θ ≥ θ . Thus if M ≥ e L/θ ,then we can repeat the above argument as follows: f ( a ( j ); c k , ≥ f ( a ( j , . . . , j k − ); c k − c k , c k − ) ≥ f ( a ( j , . . . , j k − ); c k − c k − c k , c k − c k − )... ≥ f ( a (1); c · · · c k , c · · · c k − )= f ( a (1); C k /C , C k − /C ) . Therefore we get Lemma 5.5. (cid:3)
RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 13
For all k ∈ N , let m k = Qa (1) ( C k − C k − ) /C ( C k log a (1)) L . Lemma 5.6.
Assume that a (1) ≥ M . Then m k ≥ holds for all k ≥ if M > issufficiently large.Proof. Since C k − C k − ≥ (1 + θ ) k − θ and C k ≤ R k , Lemma 5.4 implies m k = Q a (1) ( C k − C k − ) /C ( C k log a (1)) L ≥ Q e L (1+ θ ) k − /R ( R k L/θ ) L if M ≥ e L/θ . Therefore there exists k = k ( R, L, θ ) > k ≥ k , m k ≥ ≤ k ≤ k , we see that m k = Q a (1) ( C k − C k − ) /C ( C k log a (1)) L ≥ QM θ/R ( R k log M ) L . Hence if M is sufficiently large, m k ≥ ≤ k ≤ k . Therefore we conclude thelemma. (cid:3) Proof of Proposition 5.1.
Let us choose I k ⊆ N k ( k = 1 , , . . . ), m : S ∞ k =1 I k → N , and a : S ∞ k =1 I k → B satisfying (C1), (C2), (C3), and (C4). Let M = M ( R, θ, L, Q ) > a (1) ≥ M . By Lemma 5.3, Lemma 5.5,and Lemma 5.6, we apply Lemma 2.1 to W with ǫ k = 1 C k ( a (1) + 1) (1 − C k ) /C , m k = Qa (1) ( C k − C k − ) /C ( C k log a (1)) L . Then we havedim H W ≥ lim k →∞ log( m m · · · m k − ) − log m k ǫ k = lim k →∞ log( Q k − (log a (1)) − L ( k − a (1) ( C k − − C ) /C ( C · C · · · C k − ) − L )log( C k ( a (1) + 1) ( C k − /C Q − a (1) ( C k − − C k ) /C C Lk (log a (1)) L ) . Since C k − ≥ (1 + θ ) k − for any k ≥
2, the numerator of this fractional islog( Q k − (log a (1)) − L ( k − a (1) ( C k − − C ) /C ( C · C · · · C k − ) − L )= ( k −
2) log Q − L ( k −
2) log log a (1) + ( C k − /C −
1) log a (1) − L k − X j =2 log C j = ( C k − /C ) log a (1) + o ( C k − ) (as k → ∞ ) , where the last inequality follows from k − X j =2 log C j ≤ k − X j =2 log R j ≤ k R = o ( C k − ) (as k → ∞ ) . Further, since C k − ≤ R k − for all k ∈ N , the denominator islog( C k ( a (1) + 1) ( C k − /C Q − a (1) ( C k − − C k ) /C C Lk (log a (1)) L )= ( L + 1) log C k + ( C k /C − /C ) log(1 + 1 /a (1)) − log Q + ( C k − /C ) log a (1) + L log log a (1)= ( C k /C ) log(1 + 1 /a (1)) + ( C k − /C ) log a (1) + o ( C k − ) (as → ∞ ) ≤ C k − R/ ( C a (1)) + ( C k − /C ) log a (1) + o ( C k − ) (as → ∞ ) . Hence we havedim H W ≥ lim k →∞ ( C k − /C ) log a (1) + o ( C k − ) C k − R/ ( a (1) C ) + ( C k − /C ) log a (1) + o ( C k − )= 11 + Ra (1) log a (1) . (cid:3) Proof of Theorem 1.2
Theorem 6.1.
Let
R > . Let ( c k ) k ∈ N be any real sequence satisfying ≤ c k ≤ R forall k ∈ N . Let C k = c · · · c k for all k ∈ N . Then, there exists M = M ( R ) > suchthat for all M ≥ M , we can find a prime number p ∈ [ M, M ] satisfying dim H { A ∈ [ p /c , ( p + 1) /c ) : ⌊ A C k ⌋ is prime-representing } ≥
11 + R/ ( p log p ) . Proof of Theorem 6.1.
Let M = M ( R ) and M = M ( R, , , d ) be so as in Lemma 4.1and Proposition 5.1, respectively. Set M = max { M , M } . By Lemma 4.1, for ev-ery M ≥ M , we find a prime number p ∈ [ M, M ], and we construct I k ⊆ N k ( k ∈ N ), a : S ∞ k =1 I k → P with a (1) = p , and m : S ∞ k =1 I k → N satisfying (B1) to(B4). Then I k , a , m satisfy (C1) to (C4) with θ = 1, L = 1, Q = d , and B = P .Further, a (1) = p ≥ M holds. Therefore, applying Proposition 5.1, we concludeTheorem 6.1. (cid:3) Proof of Theorem 1.2.
Let ( c k ) k ∈ N be any bounded real sequence satisfying c k ≥ k ∈ N . Then there exists R > ≤ c k ≤ R for all k ∈ N . Therefore RIME-REPRESENTING FUNCTIONS AND HAUSDORFF DIMENSION 15 by Theorem 6.1 and the monotonicity of the Hausdorff dimension, there exists M > M ≥ M we find a prime number p ∈ [ M, M ] satisfyingdim H W ( c k ) ≥ dim H { A ∈ [ p /c , ( p + 1) /c ) : ⌊ A C k ⌋ is prime-representing }≥
11 + R/ ( p log p ) ≥
11 + R/ ( M log M ) . By taking M → ∞ , we have dim H W ( c k ) ≥
1. Therefore we conclude Theorem 1.2since dim H W ( c k ) ≤ dim H R = 1. (cid:3) Acknowledgement
The author was supported by JSPS KAKENHI Grant Number JP19J20878.
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