A15 Integers 18 (2018) Ranges of Unitary Divisor

#A15 INTEGERS 18 (2018)

RANGES OF UNITARY DIVISOR FUNCTIONS

Colin Defant1 Department of Mathematics, University of Florida, Gainesville, Florida2 Current email: [email protected]

Received: 6/22/17, Revised: 11/26/17, Accepted: 2/20/18, Published: 3/9/18

Abstract

For any real t, the unitary divisor function t⇤ is the multiplicative arithmetic func- ↵ ↵t tion deﬁned by t⇤(p ) = 1 + p for all primes p and positive integers ↵. Let t⇤(N) denote the topological closure of the range of t⇤. We calculate an explicit constant ⌘⇤ 1.9742550 and show that ⇤ r(N) is connected if and only if r (0, ⌘⇤]. We end⇡with some open problems. 2

1. Introduction

c For each c C, the divisor function c is defined by c(n) = d n d . Divisor 2 | functions, especially 1, 0, and 1, are among the most extensively-studied arith- P metic functions [2, 10, 12]. For example, two very classical number-theoretic topics are the study of perfect numbers and the study of friendly numbers. A positive integer n is said to be perfect if 1(n) = 2, and n is said to be friendly if there exists m = n with 1(m) = 1(n) [14]. Motivated by the very dicult problems 6 related to perfect and friendly numbers, Laatsch [11] studied 1(N), the range of 1. He showed that 1(N) is a dense subset of the interval [1, ) and asked if 1 1(N) is in fact equal to the set Q [1, ). Weiner [16] answered this question in \ 1 the negative, showing that (Q [1, )) 1(N) is also dense in [1, ). \ 1 \ 1 The author has studied ranges of divisor functions in a variety of contexts [4, 5, 6, 7, 8]. For example, it is shown in [4] that (c) as (c) , where N ! 1 < ! 1 (c) denotes the number of connected components of c(N). Here, the overline N denotes the topological closure. In [15], Sanna develops an algorithm that can be used to calculate r(N) when r > 1 is real and is known with sucient precision. In addition, he proves that ( r) is finite for such r. The author [5] has since N extended this result, showing that (c) is finite whenever (c) 0 and c = 0. N <  6 Very recently, Zubrilina [17] has obtained asymptotic estimates for ( r) when N 1This work was supported by National Science Foundation grant no. 1262930. 2Current Address: Department of Mathematics, Princeton University INTEGERS: 18 (2018) 2 r > 1. She has also shown that there is no real number r such that (r) = 4. N In this paper, we study the close relatives of the divisor functions known as unitary divisor functions. A unitary divisor of an integer n is a divisor d of n such that gcd(d, n/d) = 1. The unitary divisor function c⇤ is defined by

c c⇤(n) = d d n gcd(d,n/dX| )=1

(see, for example, [1], [3], or [9]). The function c⇤ is multiplicative and satisﬁes ↵ ↵c c⇤(p ) = 1 + p for all primes p and positive integers ↵. If t [ 1, 0), then one may use the same argument that Laatsch employed in 2 [11] in order to show that ⇤(N) = [1, ). In particular, ⇤(N) is connected if t 1 t t [ 1, 0). On the other hand, ⇤(N) is a discrete disconnected set if t 0 (indeed, 2 t in this case, t(N) [0, s] is ﬁnite for every s > 0). The purpose of this paper is to \ prove the following theorem. Let ⇣ denote the Riemann zeta function.

Theorem 1. Let ⌘⇤ be the unique number in the interval (1, 2] that satisﬁes the equation ⌘⇤ ⌘⇤ 2 2 + 1 (3 + 1) ⇣(⌘⇤) ⌘ 2⌘ = . (1) 2 ⇤ · 3 ⇤ + 1 ⇣(2⌘⇤)

If r R, then r(N) is connected if and only if r (0, ⌘⇤]. 2 2 Remark 1.1. In the process of proving Theorem 1, we will show that there is indeed a unique solution to the equation (1) in the interval (1, 2].

In all that follows, we assume r > 1 and study ⇤ r(N). We ﬁrst observe that 1 v ⇤ r(N) [1, ⇣(r)/⇣(2r)). This is because if q1 qv is the prime factorization of ✓ · · · some positive integer, then v v v 1 v i ir r r ⇤ r(q1 qv ) = ⇤ r(qi ) = 1 + qi 1 + qi < 1 + p · · ·  i=1 i=1 i=1 p Y Y ⇣ ⌘ Y Y 2r 1 p ⇣(r) = = . 1 p r ⇣(2r) p Y ✓ ◆ It is straightforward to show that 1 and ⇣(r) are elements of ⇤ r(N). Therefore, Theorem 1 tells us that ⇤ r(N) = [1, ⇣(r)/⇣(2r)] if and only if r (0, ⌘⇤]. 2

2. Proofs

th In what follows, let pi denote the i prime number. Let ⌫p(x) denote the exponent of the prime p appearing in the prime factorization of the integer x. To start, we need the following technical yet simple lemma. INTEGERS: 18 (2018) 3

2r 2r ps + 1 pm + 1 Lemma 1. If s, m N and s m, then 2r r 2r r for all r > 1. 2  ps + ps  pm + pm x2r + 1 Proof. Fix some r > 1, and write h(x) = . Then x2r + xr

r r 1 h0(x) = x 2 . x(xr + 1)2 xr ✓ ◆ We see that h(x) is increasing when x 3. Hence, in order to complete the proof, it suces to show that h(2) h(3). Let f(s) = 2s32s + 22s + 2s (22s3s + 32s + 3s).  For s 1, we have s 2 s 2 s 2 s 2 s 2 s 2 f 00(s) = 18 log (18) + 4 log (4) + 2 log (2) 12 log (12) 9 log (9) 3 log (3) > 18s log2(18) 12s log2(12) 9s log2(9) > 18s log2(18) 2(12s log2(12)). It is easy to verify that 18s log2(18) 2(12s log2(12)) is increasing in s for s 1, so we obtain 2 2 f 00(s) > 18 log (18) 2(12 log (12)) > 0. A simple calculation shows that f 0(1) > 0, so it follows that f 0(s) > 0 for all s 1. Since f(1) = 0 and r > 1, we have f(r) > 0. Equivalently, 22r3r + 32r + 3r < 2r32r + 22r + 2r. It follows that (22r + 1)(32r + 3r) < (22r + 2r)(32r + 1). This shows 22r + 1 32r + 1 that < , which completes the proof. 22r + 2r 32r + 3r

The following theorem replaces the question of whether or not ⇤ r(N) is connected with a question concerning inﬁnitely many inequalities. The advantage in doing this is that we will further reduce this problem to the consideration of a ﬁ- nite list of inequalities in Theorem 3. Recall from the introduction that ⇤ r(N) is connected if and only if it is equal to the interval [1, ⇣(r)/⇣(2r)].

Theorem 2. If r > 1, then ⇤ r(N) = [1, ⇣(r)/⇣(2r)) if and only if

p2r + pr 1 1 m m 1 + p2r + 1  pr m i=m+1 i Y ✓ ◆ for all positive integers m.

p2r + pr 1 1 Proof. First, suppose that m m 1 + for all positive integers m. p2r + 1  pr m i=m+1 i Y ✓ ◆ We will show that the range of log ⇤ r is dense in [0, log (⇣(r)/⇣(2r))), which will then imply that the range of ⇤ r is dense in [1, ⇣(r)/⇣(2r)). Fix some x (0, log (⇣(r)/⇣(2r))) . 2 INTEGERS: 18 (2018) 4

We will construct a sequence (Ci)i1=1 of elements of the range of log ⇤ r that con- verges to x. First, let C0 = 0. For each positive integer n, if Cn 1 < x, let ↵nr Cn = Cn 1 + log 1 + pn , where ↵n is the smallest positive integer that satis- ↵nr ﬁes Cn 1 + log 1 + pn x. If Cn 1 = x, simply set Cn = Cn 1 = x. For each  n N, Cn log ⇤ r(N). Indeed, if Cn = Cn 1, then 2 2 6 n n n ↵ir ↵ir ↵i Cn = log 1 + pi = log 1 + pi = log ⇤ r pi . i=1 i=1 ! i=1 ! X Y Y If, however, Cn = Cn 1 = x, then we may let l be the smallest positive integer such that Cl = x and show, in the same manner as above, that l ↵i Cn = Cl = log ⇤ r pi . i=1 ! Y Let us write = lim Cn. Note that exists and that x because the sequence n !1  (Ci)i1=1 is nondecreasing and bounded above by x. If we can show that = x, then we will be done. Therefore, let us assume instead that < x. ↵nr We have Cn = Cn 1 + log(1 + pn ) for all positive integers n. Write Dn = n r ↵ r log(1 + p ) log(1 + p n ) and E = D . As n n n i i=1 X n n ↵ir x+ lim En > + lim En = lim (Cn+En) = lim log 1 + p + Di n n n n i !1 !1 !1 !1 i=1 i=1 ! X X n r = lim log 1 + p = log (⇣(r)/⇣(2r)) , n i !1 i=1 X we have lim En > log (⇣(r)/⇣(2r)) x. Therefore, we may let m be the smallest n positive in!1teger such that E > log (⇣(r)/⇣(2r)) x. If ↵ = 1 and m > 1, then m m Dm = 0. This forces Em 1 = Em > log (⇣(r)/⇣(2r)) x, contradicting the mini- mality of m. If ↵ = 1 and m = 1, then 0 = E > log (⇣(r)/⇣(2r)) x, which is m m also a contradiction since we originally chose x < log(⇣(r)/⇣(2r)). Consequently, ↵m > 1. Due to the way we deﬁned Cm and ↵m, we have

(↵m 1)r Cm 1 + log 1 + pn > x. Hence, ⇣ ⌘ (↵ 1)r ↵ r log 1 + p m log 1 + p m > x C . n n m ⇣ ⌘ p2r + pr 1 1 Using our original assumption that m m 1 + , we have p2r + 1  pr m i=m+1 i Y ✓ ◆ p2r + pr 1 1 ⇣(r) log m m log 1 + = log E C p2r + 1  pr ⇣(2r) m m m i=m+1 i ✓ ◆ X ✓ ◆ ✓ ◆ INTEGERS: 18 (2018) 5

↵mr r (↵m 1)r ↵mr pm + pm < x Cm < log 1 + pn log 1 + pn = log ↵mr . pm + 1 ⇣ ⌘ ✓ ◆ Thus, p2r + pr p↵mr + pr m m < m m . 2r ↵mr pm + 1 pm + 1

2r (↵m+1)r 3r ↵mr Rewriting this inequality, we get pm + pm < pm + pm . Dividing through by ↵ r (2 ↵m)r r (3 ↵m)r p m yields p + p < 1 + p , which is impossible since ↵ 2. This m m m m m contradiction proves that = x, so ⇤ r(N) = [1, ⇣(r)/⇣(2r)]. To prove the converse, suppose there exists some positive integer m such that

p2r + pr 1 1 m m > 1 + . p2r + 1 pr m i=m+1 i Y ✓ ◆ We may write this inequality as

1 p2r + 1 1 1 m < 1 + . (2) p2r + pr pr m m i=m+1 i Y ✓ ◆ Fix a positive integer N. If ⌫ (N) = 1 for all s 1, 2, . . . , m , then ps 2 { } m 1 1 ⇣(r) 1 1 ⇤ r(N) 1 + = 1 + r . pr ⇣(2r) p s=1 s i=m+1 i Y ✓ ◆ Y ✓ ◆

⌫ps (N) On the other hand, if ⌫ps (N) = 1 for some s 1, 2, . . . , m , then ⇤ r ps 6 2 { }  1 ⇣ ⌘ 1 + 2r . This implies that ps

2r 2r 1 1 1 ⇣(r) 1 + ps ⇣(r) ps + 1 ⇤ r(N) 1 + 2r 1 + r = r = 2r r  p p ⇣(2r) 1 + p ⇣(2r) p + p ✓ s ◆ i=1 ✓ i ◆ s s s Yi=s 6 in this case. Using Lemma 1, we have

2r ⇣(r) pm + 1 ⇤ r(N) 2r r .  ⇣(2r) pm + pm

As N was arbitrary, we have shown that there is no element of the range of ⇤ r in the interval 1 ⇣(r) p2r + 1 ⇣(r) 1 1 m , 1 + . ⇣(2r) p2r + pr ⇣(2r) pr m m i=m+1 i ! Y ✓ ◆ This interval is a gap in the range of ⇤ r because of the inequality (2). INTEGERS: 18 (2018) 6

As mentioned above, we wish to reduce the task of checking the inﬁnite collection of inequalities given in Theorem 2 to that of checking ﬁnitely many inequalities. We do so in Theorem 3, the proof of which requires the following lemma.

pj+1 3 Lemma 2. If j N 1, 2, 3, 4, 6, 9 , then < p2. 2 \ { } pj p 6 Proof. In [13], it is shown that j+1 < p3 2 for all j 10. We easily verify the pj  5 cases j = 5, 7, 8 by hand.

Theorem 3. If r (1, 3], then ⇤ r(N) = [1, ⇣(r)/⇣(2r)] if and only if 2 p2r + pr 1 1 m m 1 + p2r + 1  pr m i=m+1 i Y ✓ ◆ for all m 1, 2, 3, 4, 6, 9 . 2 { } Proof. Let p2r + pr m 1 F (m, r) = m m 1 + p2r + 1 pr m i=1 i Y ✓ ◆ p2r + pr 1 1 ⇣(r) so that the inequality m m 1 + is equivalent to F (m, r) . p2r + 1  pr  ⇣(2r) m i=m+1 ✓ i ◆ Y ⇣(r) Let r (1, 3]. By Theorem 2, it suces to show that if F (m, r) for all 2  ⇣(2r) ⇣(r) m 1, 2, 3, 4, 6, 9 , then F (m, r) for all m N. Therefore, assume that r 2 { }  ⇣(2r) 2 ⇣(r) is such that F (m, r) for all m 1, 2, 3, 4, 6, 9 .  ⇣(2r) 2 { } We will show that F (m + 1, r) > F (m, r) for all m N 1, 2, 3, 4, 6, 9 . This 2 \ { } ⇣(r) will show that (F (m, r))m1=10 is an increasing sequence. As lim F (m, r) = , m !1 ⇣(2r) ⇣(r) it will then follow that F (m, r) < for all integers m 10. Furthermore, we ⇣(2r) ⇣(r) ⇣(r) will see that F (5, r) < F (6, r) and F (7, r) < F (8, r) < F (9, r) ,  ⇣(2r)  ⇣(2r) which will complete the proof. pm+1 3 r Let m N 1, 2, 3, 4, 6, 9 . By Lemma 2, < p2 p2. This shows that 2 \{ } pm  r r 2r r r pm+1 < 2pm, implying that 2pm > pmpm+1. Therefore,

r r 2r 2r r r pm r 1 (pm 1)(pm+1 + 1) 2pm + 2 > pmpm+1 + r pm+1 r = r . pm+1 pm+1 pm+1 INTEGERS: 18 (2018) 7

r pm+1 Multiplying each side of this inequality by 2r 2r and adding 1 to (pm+1 + 1)(pm + 1) each side, we get r r 2pm+1 pm 1 1 + 2r > 1 + 2r , pm+1 + 1 pm + 1 which we may write as r 2 2r r (pm+1 + 1) pm + pm 2r > 2r . pm+1 + 1 pm + 1 Finally, we get

p2r + pr m+1 1 (pr + 1)2 m 1 F (m + 1, r) = m+1 m+1 1 + = m+1 1 + p2r + 1 pr p2r + 1 pr m+1 i=1 i m+1 i=1 i Y ✓ ◆ Y ✓ ◆ p2r + pr m 1 > m m 1 + = F (m, r). p2r + 1 pr m i=1 i Y ✓ ◆ Now, let p2r + pr 1 1 V (r) = log m m log 1 + . m p2r + 1 pr m i=m+1 i ✓ ◆ X ✓ ◆ ⇣(r) Equivalently, V (r) = log(F (m, r)) log , where F is the function deﬁned m ⇣(2r) in the proof of Theorem 3. Observe that ✓ ◆

p2r + pr 1 1 m m 1 + p2r + 1  pr m i=m+1 i Y ✓ ◆ m+6 1 p2r 2pr 1 if and only if V (r) 0. If we let J (r) = m m , then m  m pr + 1 (pr + 1)(p2r + 1) i=m+1 i m m we have X

@ pr ((pr 1)4 12p2r) log p m+6 pr log p J (r) = m m m m i i . @r m (pr + 1)2(p2r + 1)2 (pr + 1)2 m m i=m+1 i X r r 4 2r pm((pm 1) 12pm ) log pm It is not dicult to verify that r 2 2r 2 1 for all r [1, 2] (pm + 1) (pm + 1) 2 and m 1, 2, 3, 4, 6, 9 . Therefore, when r [1, 2] and m 1, 2, 3, 4, 6, 9 , we 2 { } 2 2 { } have @ m+6 pr log p m+6 log p J (r) 1 i i 1 i > 7. @r m (pr + 1)2 pr i=m+1 i i=m+1 i X X INTEGERS: 18 (2018) 8

1 Numerical calculations show that J (r) > for all m 1, 2, 3, 4, 6, 9 and m 400 2 { } n r 1 + : n 0, 1, 2, . . . , 2800 . 2 2800 2 { } Because each function J nis continuous in r for r [1, 2], woe see that m 2 1 1 J (r) > 7 = 0 m 400 2800 ✓ ◆ for all r [1, 2] and m 1, 2, 3, 4, 6, 9 . 2 2 { } We introduced the functions Jm so that we could write @ 1 log p (p2r 2pr 1) log p V (r) = i m m m > (log p )J (r) > 0 @r m pr + 1 (pr + 1)(p2r + 1) m m i=m+1 i m m X for all m 1, 2, 3, 4, 6, 9 and r [1, 2]. A quick numerical calculation shows 2 { } 2 that V2(1.5) < 0 < V2(2), so the function V2 has exactly one root, which we will call ⌘⇤, in the interval (1, 2]. Further calculations show that Vm(2) < 0 for all m 1, 3, 4, 6, 9 . Hence, V (r) 0 for all m 1, 2, 3, 4, 6, 9 and r (1, ⌘⇤]. By 2 { } m  2 { } 2 Theorem 3, this means that if r (1, 2], then ⇤ r(N) [1, ⇣(r)/⇣(2r)] if and only if 2 r ⌘⇤.  Next, note that

@ 1 log p (32r 2 3r 1) log 3 (32r 2 3r 1) log 3 V (r) = i · > · @r 2 pr + 1 (32r + 1)(3r + 1) (32r + 1)(3r + 1) i=3 i X (32r + 1) log 3 log 3 > > 1.1 (32r + 1)(3r + 1) 32 + 1 n for all r [2, 3]. Let A = 2 + : n 0, 1, 2, . . . , 400 . With a computer 2 400 2 { } program, one may verify thatnV2(r) > 0.003 for all r A. Becauseo V2 is continuous, 1 2 this shows that V (r) > 0.003 1.1 > 0 for all r [2, 3]. Consequently, 2 400 2 ✓ ◆ ⇤ r(N) = [1, ⇣(r)/⇣(2r)) if r [2, 3]. 6 2 We are now in a position to prove Theorem 1. Note that the equation deﬁning ⌘⇤ in the statement of this theorem is simply a rearrangement of the equation V (⌘⇤) = 0. Therefore, we have shown that the theorem is true for r (1, 3]. 2 2 In order to prove the theorem for r > 3, it suces (by Theorem 3) to show that ⇣(r) F (1, r) > for all r > 3. If r > 3, then ⇣(2r)

r+1 r 2 2r r+1 2r r 2 1 1 2 + 2 + 1 + r + r 1 (2 + 1) 2 + 2 + 1 r 1 2 (r 1)2 F (1, r) = 2r = 2r > 2r = 1 2 + 1 2 + 1 2 + 1 1 + 22r 1 1 1 r 1 + 2r + (r 1)2r 1 1 + r + 1 x dx ⇣(r) > = 2 2 > . ⇣(2r) ⇣(2Rr) ⇣(2r) INTEGERS: 18 (2018) 9

3. Future Directions

Let ⇤(t) denote the number of connected components of ⇤(N). It would be N t interesting to obtain analogues of Zubrilina’s results [17] by ﬁnding asymptotic estimates for ⇤( r) as r . Let N ! 1

E⇤ = t R: ⇤(t) = m . m { 2 N }

Theorem 1 tells us that E⇤ = [ ⌘⇤, 0). The sets E⇤ are the natural unitary 1 m analogues of the sets Em deﬁned in [5, Section 4]. Continuing the analogy, we say a positive integer m is a unitary Zubrilina number if E⇤ = (the name comes from m ; Zubrilina’s result that E = ). We do not have any speciﬁc examples of unitary 4 ; Zubrilina numbers, but we still make the following conjectures.

Conjecture 1. There are inﬁnitely many unitary Zubrilina numbers.

Conjecture 2. For r > 1, ⇤( r) is monotonically increasing as a function of r. N

Note that Conjecture 2 implies that the sets Em⇤ are intervals.

Acknowledgements. The author thanks the referee for carefully reading the manuscript and providing very helpful suggestions.

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