aa r X i v : . [ m a t h . N T ] J un On the Density of Ranges ofGeneralized Divisor Functions
Colin Defant Department of Mathematics University of FloridaUnited Statescdefant@ufl.edu
Abstract
The range of the divisor function σ − is dense in the interval [1 , ∞ ).However, the range of the function σ − is not dense in the interval (cid:20) , π (cid:19) . We begin by generalizing the divisor functions to a class offunctions σ t for all real t . We then define a constant η ≈ . r ∈ (1 , ∞ ), then the range of the function σ − r isdense in the interval [1 , ζ ( r )) if and only if r ≤ η . We end with anopen problem. Throughout this paper, we will let N denote the set of positive integers, andwe will let p i denote the i th prime number.For any integer t , the divisor function σ t is a multiplicative arithmeticfunction defined by σ t ( n ) = X d | nd> d t for all positive integers n . The value of This work was supported by National Science Foundation grant no. 1262930. ( n ) is the sum of the positive divisors of n , while the value of σ ( n ) is simplythe number of positive divisors of n . Another interesting divisor function is σ − , which is often known as the abundancy index. One may show [2] thatthe range of σ − is a subset of the interval [1 , ∞ ) that is dense in [1 , ∞ ). If t < −
1, then the range of σ t is a subset of the interval [1 , ζ ( − t )), where ζ denotes the Riemann zeta function. This is because, for any positive integer n , σ t ( n ) = X d | nd> d t < ∞ X i =1 i t = ζ ( − t ). For example, the range of the function σ − is a subset of the interval (cid:20) , π (cid:19) . However, it is interesting to notethat the range of the function σ − is not dense in the interval (cid:20) , π (cid:19) . Tosee this, let n be a positive integer. If 2 | n , then σ − ( n ) ≥ + 12 = 54 .On the other hand, if 2 ∤ n , then σ − ( n ) < X d ∈ N \ (2 N ) d = ζ (2) (cid:0) − − (cid:1) = π π <
54 , we see that there is a “gap” in the range of σ − . In other words,there are no positive integers n such that σ − ( n ) ∈ (cid:18) π , (cid:19) .Our first goal is to generalize the divisor functions to allow for nonintegralsubscripts. For example, we might consider the function σ −√ , defined by σ −√ ( n ) = X d | nd> d −√ . We formalize this idea in the following definition. Definition 1.1.
For a real number t , define the function σ t : N → R by σ t ( n ) = X d | nd> d t for all n ∈ N . Also, we will let log σ t = log ◦ σ t .In analyzing the ranges of these generalized divisor functions, we willfind a constant which serves as a “boundary” between divisor functions withdense ranges and divisor functions with ranges that have gaps. Note that, forany real number t , we may write σ t = I ∗ I t , where I and I t are arithmeticfunctions defined by I ( n ) = 1 and I t ( n ) = n t . As I and I t are multiplicative,we find that σ t is multiplicative. 2 The Ranges of Functions σ − r Theorem 2.1.
Let r be a real number greater than . The range of σ − r isdense in the interval [1 , ζ ( r )) if and only if p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! forall positive integers m .Proof. First, suppose that 1 + 1 p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! for all positive inte-gers m . We will show that the range of log σ − r is dense in the interval[0 , log( ζ ( r ))), which will imply that the range of σ − r is dense in [1 , ζ ( r )).Choose some arbitrary x ∈ (0 , log( ζ ( r ))), and define X = 0. For each posi-tive integer n , we define α n and X n in the following manner. If X n − + log ∞ X j =0 p jrn ! ≤ x , define α n = −
1. If X n − + log ∞ X j =0 p jrn ! > x ,define α n to be the largest nonnegative integer that satisfies X n − + log α n X j =0 p jrn ! ≤ x . Define X n by X n = X n − + log (cid:16)P α n j =0 1 p jrn (cid:17) , if α n ≥ X n − + log (cid:16)P ∞ j =0 1 p jrn (cid:17) , if α n = − . Also, for each n ∈ N , define D n by D n = ( log (cid:16)P ∞ j =0 1 p jrn (cid:17) − log (cid:16)P α n j =0 1 p jrn (cid:17) , if α n ≥ , if α n = − , and let E n = n X i =1 D i . Note thatlim n →∞ ( X n + E n ) = lim n →∞ X n + n X i =1 D i ! = lim n →∞ n X i =1 log ∞ X j =0 p jri ! = log( ζ ( r )) . X n ) ∞ n =1 is bounded and monotonic, we knowthat there exists some real number γ such that lim n →∞ X n = γ . We wish toshow that γ = x .Notice that we defined the sequence ( X n ) ∞ n =1 so that X n ≤ x for all n ∈ N .Hence, we know that γ ≤ x . Now, suppose γ < x . Then lim n →∞ E n = log( ζ ( r )) − γ > log( ζ ( r )) − x . This implies that there exists some positive integer N such that E n > log( ζ ( r )) − x for all integers n ≥ N . Let m be the smallestpositive integer that satisfies E m > log( ζ ( r )) − x . If α m = − m > D m = 0, so E m − = E m > log( ζ ( r )) − x . However, this contradicts theminimality of m . If α m = − m = 1, then 0 = D m = E m > log( ζ ( r )) − x ,which is also a contradiction. Thus, we conclude that α m ≥
0. This meansthat X m + D m = X m − + log ∞ X j =0 p jrm ! > x , so D m > x − X m . Furthermore,log ∞ Y i = m +1 ∞ X j =0 p jri !! = ∞ X i = m +1 log ∞ X j =0 p jri ! = log( ζ ( r )) − m X i =1 log ∞ X j =0 p jri ! = log( ζ ( r )) − E m − X m < x − X m < D m , (1)and we originally assumed that 1 + 1 p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! . This means thatlog (cid:18) p rm (cid:19) < D m = log ∞ X j =0 p jrm ! − log α m X j =0 p jrm ! , or,equivalently, log (cid:18) p rm (cid:19) + log α m X j =0 p jrm ! < log (cid:18) p rm p rm − (cid:19) . If α m >
0, wehavelog (cid:18) p rm (cid:19) ! ≤ log (cid:18) p rm (cid:19) + log α m X j =0 p jrm ! < log (cid:18) p rm p rm − (cid:19) , so (cid:18) p rm (cid:19) < p rm p rm − p rm + 1 p rm < p rm − < p rm p rm − p rm − p rm >
2, this is a contradiction. Hence, α m = 0. By the definitions of α m and X m , this implies that X m − + log (cid:18) p rm (cid:19) > x and that X m = X m − . Therefore,log (cid:18) p rm (cid:19) > x − X m − = x − X m . However, recalling from (1) that ∞ X i = m +1 log ∞ X j =0 p jri ! < x − X m , we find that ∞ X i = m +1 log ∞ X j =0 p jri ! < log (cid:18) p rm (cid:19) , which is a contradiction because we originally assumed that1 + 1 p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! . Therefore, γ = x .We now know that lim n →∞ X n = x . To show that the range of log σ − r isdense in [0 , log( ζ ( r ))), we need to construct a sequence ( C n ) ∞ n =1 of elementsof the range of log σ − r that satisfies lim n →∞ C n = x . We do so in the followingfashion. For each positive integer n , write Y n = ( , if α n ≥ , if α n = − ,Z n = ( , if α n ≥ , if α n = − , and β n = ( α n , if α n ≥ , if α n = − . Now, for each positive integer n , define C n by C n = n X k =1 Y k log β k X j =0 p jrk ! + Z k log n X j =0 p jrk !! . X n , we have X n = n X k =1 Y k log β k X j =0 p jrk ! + Z k log ∞ X j =0 p jrk !! . Therefore, lim n →∞ C n = lim n →∞ X n = x . All we need to do now is show that each C n is in the range of log σ − r . We have C n = n X k =1 Y k log β k X j =0 p jrk ! + Z k log n X j =0 p jrk !! = X k ∈ N k ≤ nα k ≥ log α k X j =0 p jrk ! + X k ∈ N k ≤ nα k = − log n X j =0 p jrk ! = log Y k ∈ N k ≤ nα k ≥ σ − r ( p α k k ) Y k ∈ N k ≤ nα k = − σ − r ( p nk ) = log σ − r Y k ∈ N k ≤ nα k ≥ p α k k Y k ∈ N k ≤ nα k ≥ p nk . We finally conclude that if 1 + 1 p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! for all positive integers m , then the range of σ − r is dense in the interval [1 , ζ ( r )).Conversely, suppose that there exists some positive integer m such that1 + 1 p rm > ∞ Y i = m +1 ∞ X j =0 p jri ! . Fix some N ∈ N , and let N = v Y i =1 q γ i i be thecanonical prime factorization of N . If p s | N for some s ∈ { , , . . . , m } , then σ − r ( N ) ≥ p rs ≥ p rm .
6n the other hand, if p s ∤ N for all s ∈ { , , . . . , m } , then σ − r ( N ) = v Y i =1 σ − r ( q γ i i ) = v Y i =1 γ i X j =0 q jri ! < v Y i =1 ∞ X j =0 q jri ! < ∞ Y i = m +1 ∞ X j =0 p jri ! . Because N was arbitrary, this shows that there is no element of the range of σ − r in the interval " ∞ Y i = m +1 ∞ X j =0 p jri ! , p rm ! , which means that the rangeof σ − r is not dense in [1 , ζ ( r )).Theorem 2.1 provides us with a method to determine values of r > σ − r is dense in [1 , ζ ( r )). However, doing so isstill a somewhat difficult task. Luckily, for r ∈ (1 , Lemma 2.1. If j ∈ N \{ , , } , then p j +1 p j < √ .Proof. Pierre Dusart [1] has shown that, for x ≥
396 738, there must beat least one prime in the interval (cid:20) x, x + x
25 log x (cid:21) . Therefore, whenever p j >
396 738, we may set x = p j + 1 to get p j +1 ≤ ( p j + 1) + p j + 125 log ( p j + 1) < √ p j . Using Mathematica 9.0 [3], we may quickly search through all theprimes less than 396 738 to conclude the desired result. Theorem 2.2.
Let r be a real number in the interval (1 , . The range of σ − r is dense in the interval [1 , ζ ( r )) if and only if p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! for all m ∈ { , , } . roof. Let F ( m, r ) = (cid:18) p rm (cid:19) m Y i =1 ∞ X j =0 p jri ! so that the inequality1 + 1 p rm ≤ ∞ Y i = m +1 ∞ X j =0 p jri ! is equivalent to F ( m, r ) ≤ ζ ( r ). In light of The-orem 2.1, it suffices to show that if F ( m, r ) ≤ ζ ( r ) for all m ∈ { , , } ,then F ( m, r ) ≤ ζ ( r ) for all m ∈ N . Thus, let us assume that r is such that F ( m, r ) ≤ ζ ( r ) for all m ∈ { , , } . If m ∈ N \{ , , } , then Lemma 2.1 tellsus that p m +1 p m < √ ≤ r √
2, which implies that 2 p rm +1 > p rm . We then have F ( m + 1 , r ) = (cid:18) p rm +1 (cid:19) m +1 Y i =1 ∞ X j =0 p jri ! > (cid:18) p rm +1 (cid:19) m Y i =1 ∞ X j =0 p jri ! > (cid:18) p rm +1 (cid:19) m Y i =1 ∞ X j =0 p jri ! > (cid:18) p rm (cid:19) m Y i =1 ∞ X j =0 p jri ! = F ( m, r )for all m ∈ N \{ , , } . This means that F (3 , r ) < F (4 , r ) ≤ ζ ( r ). Fur-thermore, F ( m, r ) < ζ ( r ) for all integers m ≥ F ( m, r )) ∞ m =5 is astrictly increasing sequence and lim m →∞ F ( m, r ) = ζ ( r ).We have seen that, for r ∈ (1 , σ − r is dense in [1 , ζ ( r ))if and only if F ( m, r ) ≤ ζ ( r ) for all m ∈ { , , } . Using Mathematica 9.0,one may plot a function g m ( r ) = F ( m, r ) − ζ ( r ) for each m ∈ { , , } . It isthen easy to verify that g has precisely one root, say η , in the interval (1 , g ′ ( r ) > r ∈ (1 , g ( r ) , g ( r ) , g ( r ) ≤ r ∈ (1 , η ] and that g ( r ) > r ∈ ( η, Theorem 2.3.
Let η be the unique number in the interval (1 , that satisfiesthe equation (cid:18) η η − (cid:19) (cid:18) η + 13 η − (cid:19) = ζ ( η ) . If r ∈ (1 , ∞ ) , then the range of the function σ − r is dense in the interval [1 , ζ ( r )) if and only if r ≤ η . roof. In virtue of the preceding paragraph, we know from the fact that g ( η ) = F (2 , η ) − ζ ( η ) = (cid:18) η η − (cid:19) (cid:18) η + 13 η − (cid:19) − ζ ( η ) = 0that if r ∈ (1 , σ − r is dense in [1 , ζ ( r )) if and only if r ≤ η . We now show that the range of σ − r is not dense in [1 , ζ ( r )) if r > F (1 , r ) > ζ ( r ) for all r >
3. For r >
3, we have F (1 , r ) = (cid:18) r (cid:19) ∞ X j =0 jr > (cid:18) r (cid:19) = 1 + 12 r + 34 (cid:18) r − (cid:19) > r + 1( r − r − = 1 + 12 r + Z ∞ x r dx > ζ ( r ) . We end by acknowledging that it might be of interest to consider the numberof “gaps” in the range of σ − r for various r . For example, for which values of r ∈ (1 , ∞ ) is there precisely one gap in the range of σ − r ? More generally, ifwe are given a positive integer L , then, for what values of r > σ − r a union of exactly L disjoint subintervals of [1 , ζ ( r )]? References [1] Dusart, Pierre. Estimates of some functions over primes without R.H.,arXiv:1002.0442 (2010).[2] Laatsch, Richard. Measuring the abundancy of integers. Math. Mag. 59(1986), no. 2, 84–92.[3] Wolfram Research, Inc., Mathematica, Version 9.0, Champaign, IL(2012). 9010
Mathematics Subject Classification : Primary 11B05; Secondary 11A25.