ON THE EQUATION τ(λ(n)) = ω(n) + k

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1 ON THE EQUATION τ(λ(n)) = ω(n) + k A. GLIBICHUK, F. LUCA AND F. PAPPALARDI Abstract. We investigate some properties of the positive integers n that satisfy the equation τ(λ(n)) = ω(n)+k providing a complete description for the solutions when k = 0,, 2, and giving some properties of the solutions in the other cases.. Introduction For every positive integer n, the function τ(n) counts the number of divisors of n, the function ω(n) counts the number of distinct prime divisors of n, while the Carmichael function λ(n) is the eponent of the multiplicative group of the invertible congruence classes modulo n. The value of the function λ(n) can be computed as follows: if n = ; 2 α 2 if n = 2 α, α > 2; λ(n) = p α (p ) if n = p α and p 3 or p = 2, α 2; [λ(p α ),..., λ(p αs s )] if n = p α... p αs In [?], Erdős, Pomerance and Schmutz proved a number of fundamental properties of λ. In the process of proving the lowerbound λ(n) > (log n) c 0 log log log n for all large n, provided c 0 < / log 2, they proved the inequality n (4λ(n)) 3τ(λ(n)). Numerical calculations suggest that the stronger inequality () n λ(n) τ(λ(n)) holds with the only eceptions of n = 2, 6, 8, 2, 24, 80, 20, 240. This will be proved in Corollary??. One of the tools for proving (??) is the inequality τ(λ(n)) > ω(n) which holds with the only eceptions F. L. and F. P. were supported in part by the Italian-Meican Agreement of Scientific and Technological Cooperation : Kleinian Groups and Egyptian Fractions. s.

2 2 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI n = 2, 6, 2, 24, 30, 60, 20, 240 as we will prove in Proposition?? and Proposition??. This motivates us to compare τ(λ(n)) with ω(n). Since τ(λ(n)) ω(n) holds for all positive integers n (see Proposition??), we can write τ(λ(n)) = ω(n) + k, where k is some nonnegative integer depending on n. We then fi k 0 and investigate the positive integers n such that τ(λ(n)) = ω(n) + k. Throughout this paper, we use to denote a positive real number. We also use the Landau symbols O and o and the Vinogradov symbols and with their usual meanings. We write log for the maimum between and the natural logarithm of. For a set A of positive integers we write A() = A [, ]. We write p and q with or without subscripts for prime numbers. Let us set A k = {n : τ(λ(n)) = ω(n) + k}. We will show in Theorem?? that if k is a positive integer and b k = 2(k + ) log 2 (2(k + ) 2 + k + ), then the upperbound #A k () k (log log ) b k holds as. Furthermore, in Theorem??, we will show that if k > 4, then the lowerbound #A k () k holds as. We will also give complete description on the sets A 0, A and A 2 (Proposition??, Proposition?? and Proposition??). We will show that A 0 contains 8 integers while the infiniteness of A and A 2 would follow if it were known that there eist infinitely many primes of the form 2q + with q also prime. Finally, in Proposition?? we deal with the cases k = 3, 4 proving that if either A 3 or A 4 are infinite then there eists an even positive integer c such that the set of primes of the form p = cq β +, with q prime and β 4 is infinite. This eplains the difficulty of proving the infiniteness of A k for k =, 2, 3, 4. Acknowledgements. This paper started during very enjoyable visits of A. G. to the Mathematics Department of the University Roma Tre and of F. P. to the Mathematical Institute of the UNAM in Morelia. These authors would like to thank these departments for their hospitality and support. The authors would also like to thank Sergei Konyagin for some useful comments.

3 THE EQUATION τ(λ(n)) = ω(n) + k 3 2. Determining A k for small values of k Proposition. For any positive integer n, we have that More precisely, τ(λ(n)) ω(n). τ(λ(n)) ω(n/(2, n)) + τ(λ o (n )), where n is the product of the primes dividing n, and λ o (m) denote the the odd part of λ(m). That is, λ o (m) = λ(m)/(2, λ(m)). Proof. Let us first note that if n m, then λ(n) λ(m), and therefore τ(λ(n)) τ(λ(m)). Thus, we can assume that n is square-free (indeed, if n is the product of the distinct primes dividing n, then ω(n) = ω(n ) and τ(λ(n)) τ(λ(n ))). Suppose that n is odd and n = p p 2 p r, where p < < p r are primes. Let 2 < q 2 < < q s be all the odd prime factors of λ(n) and write p = 2 α q α 2 2 q α s s ; p 2 = 2 α 2 q α q α 2s s ;. p r = 2 α r q α r2 2 q αrs s. If A i = ma{α i,..., α ri } for i =,..., s, then τ(λ(n)) = τ([p,..., p r ]) = (A + )(A 2 + ) (A s + ). Consider now the matri α... α s... α r... α rs We know that the entries of the matri consist of nonnegative integers. The elements in the first column are positive and less than or equal than A. For each i =,..., r, the elements of the i th column are nonnegative integers less than or equal to A i. Furthermore, for each fied natural number s, we have that the number of rows r is less or equal than the maimum number of distinct s tuples (a,..., a s ) with a ([, A ] and a i [0, A i ] for i = 2,..., s. This follows from the fact that 2 ) α i s j=2 qα ij j are distinct positive integers. i=,...,s Hence, r A (A 2 + ) (A s + ).

4 4 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI From the above discussion, we deduce that τ(λ(n)) = (A + )(A 2 + ) (A s + ) r + τ(λ o (n)) = ω(n) + τ(λ o (n)), where λ o (n) = λ(n)/(2, λ(n)) is largest odd divisor of λ(n). So, if n is square-free and odd, then τ(λ(n)) ω(n) +, while if n is square-free and even, then which concludes the proof. τ(λ(n)) = τ(λ(n/2)) ω(n/2) + = ω(n), Lemma?? is the main tool to determine the set A k for k 2. Proposition 2. A 0 = {2, 6, 2, 24, 30, 60, 20, 240}. Proof. Let n A 0. We apply Lemma?? and we obtain that if n is odd, then τ(λ(n)) > ω(n), which is impossible. If n is even, the condition τ(λ(n)) = ω(n), implies by Lemma?? that τ(λ o (n )) =. This is only possible if λ(n ) = 2 α for some α N. If n = 2 γ and τ(λ(2 γ )) =, then γ = so that n = 2. Assume now that n is not a power of 2 and write n = 2 γ 0 (2 2α + ) γ (2 2αr + ) γr, where γ j for j = 0,..., r, 0 α < < α r, and the numbers 2 2α i + are primes for each i =,..., r. Plugging the epression above for n in the identity τ(λ(n)) = ω(n), we obtain ma{τ(λ(2 γ 0 )), 2 αr + } γ γ r = r +, which is satisfied only for r = or r = 2 since from the above we gather that r + 2 αr + 2 r +. If r = 2, then necessary α 2 =. This forces α = 0, γ = γ 2 =, and γ 0 4, which correspond to the four values 30, 60, 20 and 240 for n. Finally, if r =, then α = 0, and this forces γ = and γ 0 3, which correspond to the three values 6, 2 and 24 for n. We are now ready to prove the motivating inequality (??):

5 THE EQUATION τ(λ(n)) = ω(n) + k 5 Corollary. Let ϕ denote the Euler function. With the only eceptions n = 2, 6, 8, 2, 24, 80, 20, 240, we have n λ(n) τ(λ(n)). Furthermore, ϕ(n) λ(n) τ(λ(n)) with the only eception n = 24. Finally, the inequality ϕ(n) λ(n) ω(n) holds unless n is a power of 2 times a product of distinct Fermat primes. Proof. Let v p (m) be the eponent of the prime p in the factorization of the positive integer m. We know that λ(n) divides ϕ(n) and if p odd, then v p (ϕ(n)) = l β n v p (l β (l )) ( ) ω(n) ma{v p (l β (l ))} l β n v p (λ(n) ω(n) ), while v 2 (ϕ(n)) = v 2 (n) + l n v 2(l ) + ω(n)v 2 (λ(n)). So, necessarily ϕ(n) 2λ(n) ω(n). Furthermore, the only circumstances in which ϕ(n) = 2λ(n) ω(n) is when ϕ(n) is a power of 2. If this happens, then n is necessarily a power of 2 times a product of distinct Fermat primes. In all other cases, we have ϕ(n) λ(n) ω(n) and this proves the third inequality. In order to prove the second, it is enough to notice that τ(λ(n)) ω(n) by Proposition??, therefore we only need to show that ϕ(n) λ(n) τ(λ(n)) when ϕ(n) = 2 a and n 24. Observe that the latter is certainly true when n is a power of 2 since for α > 2, ϕ(2 α ) = 2 α 2 (α 2)(α ) = λ(2 α ) τ(λ(2α )). In the other cases, if we write with α < < α r, then n = 2 α0 (2 2α + ) (2 2αr + ), ϕ(n) = 2 2α + +2 αr +ma{α 0,0} 2 2αr (+/2+ +/2 r )+ma{α 0,0} 2 3M+, where M = ma{log 2 (λ(2 α 0 ), 2 αr }. Here, we use log 2 for the logarithm in base 2. Similarly, λ(n) τ(λ(n)) = 2 M(M+). Finally 3M + M(M +) for M > 2 while the case when M 2 leads to r 2 so that n {3, 6, 2, 24, 48, 5, 0, 20, 40, 80, 5, 30, 60, 20, 240}, and the only value of n from the above set that does not satisfy the inequality ϕ(n) λ(n) τ(λ(n)) is n = 24. This completes the proof of the second statement.

6 6 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI As for the first statement, note that if n A 0 the statement holds if and only if n {30, 60}. So, we can assume that n A 0 and thus τ(λ(n)) ω(n) +. This implies that λ(n) τ(λ(n)) λ(n)ϕ(n) unless ϕ(n) is a power of 2. In order to conclude the proof we need to verify that the statement holds when ϕ(n) is a power of 2 and n 2, 8, and we need to show that λ(n)ϕ(n) n. We claim that the inequality above holds unless n {2, 3, 6, 2, 24} (values for which the statement is verified directly). Indeed, let p be the greatest prime divisor of n. If p 5, then n ϕ(n) = l n l l 3 4 p p λ(n). Similarly, if p = 3, then n/ϕ(n) 3 λ(n) unless n {3, 6, 2, 24}. Finally, if p = 2, then n/ϕ(n) = 2 λ(n) unless n = 2. If ϕ(n) is a power of 2, then we proceed as in the proof of the second inequality. Observe that if n = 2 α 0, then n λ(n) τ(λ(n)) unless α 0 =, 3. If n = 2 α0 (2 2α + ) (2 2αr + ) with α < < α r and if M = ma{log 2 (λ(2 α 0 ), 2 αr } so that 2 M(M+) = λ(n) τ(λ(n)), then n 2 2(2α + +2 αr )+α 0 2 5M+2. Since 5M + 2 M(M + ) for M > 5, we are left with checking the statement for integers that divide and this is done by a short calculation. Proposition 3. A = {, 3, 4, 8, 0, 5, 20, 40, 48, 80, 26, 252, 480, 504, 50, 020, 2040, 2730, 4080, 5460, 860, 890, 0920, 6320, 6380, 2840, 32760, 65520, 6q, 2q, 24q}, where q = 2p + is prime with p > 2 also prime. Proof. We follow the same method as in the proof of Proposition??. If n > is odd, then, from Lemma??, we obtain that λ o (n ) =. This implies that λ(n ) = 2 α for some α 0. Thus, n = (2 2α + ) γ (2 2αr + ) γr, where γ j for j =,..., r, 0 α < < α r, and again 2 2α i + is prime for i =,..., r.

7 THE EQUATION τ(λ(n)) = ω(n) + k 7 The equation τ(λ(n)) = ω(n) + is equivalent to (2 αr + )γ γ r = r +. Since α r r, the above is satisfied only if r = or r = 2. In the first case, we have necessarily α = 0 and γ =, so that n = 3. In the second case, we have α = 0, α 2 = and γ = γ 2 =, so that n = 5. Assume now that n is even. If n = 2 γ, then the equation τ(λ(n)) = 2 is only satisfied for n = 4 and for n = 8. If n is not a power of 2, then, from Lemma??, we get τ(λ o (n )) 2. This can only happen if either λ(n ) = 2 a, or λ(n ) = 2 a p, with p an odd prime. In the first case, we have that n = 2 γ0 (2 2α + ) γ (2 2αr + ) γr, where γ j for j = 0,..., r, 0 α < < α r, and again 2 2α i + is prime for i =,..., r. If we plug the above epression for n in the identity τ(λ(n)) = ω(n)+, we obtain ma{τ(λ(2 γ 0 )), 2 αr + } γ γ r = r + 2, which can only be satisfied for r 3 since r αr + 2 r +. A quick computation shows that γ j = for all j and we have only the following possibilities: r (α,..., α r ) n (0) 48 () 0, 20, 40, (0,, 2) 50, 020, 2040, 4080, 860, 6320 The net case to consider is when λ(n ) = 2 a p so that each odd prime dividing n is either of the form 2 2α +, or of the form 2 β p +. Hence, n = 2 γ0 (2 2α + ) γ (2 2αr + ) γr (2 β p + ) γ r+ (2 βs p + ) γ r+s, where γ j for j = 0,..., r + s, 0 α < < α r, 2 2α i + is prime for i =,..., r, < β < < β s, and 2 β k p + is prime for k =,..., s. We distinguish two more sub-cases: p 2 n and p 2 n. If p 2 n, then the equation τ(λ(n)) = ω(n) + translates into (2) ma{τ(λ(2 γ 0 )), 2 αr +, β s + } γ γ r+s = r + s + 2. In this case, there eists j r such that γ j 2 and since ma{a, b} (a + b)/2, we have that the left hand side (??) is greater or equal than 2 αr + + β s +. Using the fact that α r r and that β s s, we

8 8 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI obtain once again that 2 r + r +, which implies that r = or r = 2. If r =, then necessarily α = 0, γ = 2, s = and β = γ 2 =. This implies n = 2 γ and γ 0 =, 2, 3. If r = 2, then necessarily α = 0, α 2 = and s 2 since the left hand side of (??) is greater or equal of 2s+2. Checking all possibilities, we find that n = 2 γ and γ 0 =, 2, 3, 4. If p 2 n, then the equation τ(λ(n)) = ω(n) + translates into (3) 2 ma{τ(λ(2 γ 0 )), 2 αr +, β s + } γ γ r+s = r + s + 2. For the same reason as above, we have r = or r = 2 and s = or s = 2. If r = s =, then we have the family of solutions n = 2 γ0 3 (2p+), where γ 0 =, 2, 3 and 2p + is prime with p 3. If r = s = 2, then we have the solutions n = 2 γ , where γ 0 =, 2, 3, 4. The remaining cases r =, s = 2 and r = 2, s = produce a value of the right hand side of (??) equal to 5 and therefore do not lead to any more solutions. Proposition 4. We have that A 2 = F F 2 F 3 F 4 I I 2 I 3, where 5, 2 4, 2 5 3, 2 5 5, 2 β 3 2, , F = 2 α 3 7, 2 α 5 7, 3 5 7, α 6, β 3 ; 2 α 3 β 5 7, 3 β 7, 3 β F 2 = 2 α 3 β 5 3, 2 α 3 β 7 3, α 4, 2 α β =, 2 ; F 3 = { 2 α } α 4 ; { 2δ 3 F 4 = β 7 9, 2 α 3 β α 4, β, δ 3 } ; I = {2 α (2p + ) 2p +, p 3 primes, α 3} ; I 2 = {3 (2p + ) 2p +, p 3 primes}; { I 3 = 2 α 3 5 (2 β p + ) 2 β p +, p 3 primes, α 4, β =, 2 }. Proof. Following the same approach as in the previous results, we obtain that in order for n to satisfy τ(λ(n)) = ω(n) + 2, we need to have λ(n ) = 2 α p β, where α 0 and β = 0,, 2. This implies that n should be of the form n = 2 γ0 A B C,

9 THE EQUATION τ(λ(n)) = ω(n) + k 9 where A, B and C are either or of the respective forms: A = (2 2α + ) γ (2 2αr + ) γr, B = (2 β p + ) γ r+ (2 βs p + ) γ r+s, C = (2 δ p 2 + ) γ r+s+ (2 δt p 2 + ) γ r+s+t, where we assume the following conditions: γ j for j = 0,..., r+s+t, 0 α < < α r, 2 2α i + is prime for i =,..., r, < β < < β s, 2 β k p + is prime for k =,..., s, < δ < < δ t, and 2 δ l p 2 + is prime for l =,..., t. Here, we allow either one of r, s, t, γ 0 to be zero with the obvious meaning. The equation τ(λ(n)) = ω(n) + 2 is equivalent to (4) Θ Λ γ γ r+s+t = r + s + t + min{, γ 0 } + 2, where if either (s + t > 0 and p 3 n) or (s + t = 0) or (t = 0, s > 0 and p 2 n); Θ = 3/2 if t > 0 and p 2 n; 2 if t = 0, s > 0 and p 2 n; 3 if t > 0 and p 2 n; and Λ = ma{τ(λ(2 γ 0 )), 2 αr +, β s +, δ t + }. Here, the terms β s + (resp. δ t + ) are to be omitted if s = 0 (resp. t = 0). If s = t = 0, the above implies that r 3 and n = 2 δ0 3 δ 5 δ2 7 δ 3. In this case, all possible solutions of τ(λ(n)) = ω(n) + 2 are: r (δ 0, δ, δ 2, δ 3 ) n 0 (4,0,0,0) 2 4 (0,0,,0) 5 (δ, 2, 0, 0), δ =, 2, , , (5,, 0, 0) (5, 0,, 0) (6,,, 0) (δ,, 0, ), δ 6 2 δ 3 7, δ 6 (δ, 0,, ), δ 6 2 δ 5 7, δ 6 3 (0,,, ) (7,,, ) which are eactly the 22 elements of F. When t = 0, s 0, the equation (??) simplifies to (5) Θ Λ γ γ r+s = r + s + min{, γ 0 } + 2,

10 0 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI where Θ = { if p 2 n; 2 if p 2 n; Λ = ma{τ(λ(2 γ 0 )), 2 αr +, β s + }, and the middle term is omitted if r = 0. In such a case, we have that p n and s β s (s + min{, γ 0 })/2. This is only possible for n even and s = β s =. This implies that n = 2 γ 0 (2p + ) with γ 0 =, 2, 3, which are eactly the elements of I. Assuming r > 0, the left hand side of (??) is greater than or equal to 2 αr + β s + 2, which implies that 2 αr r + min{, γ 0 }. From the last inequality, it follows that r 2 + min{, γ 0 }. We distinguish the two sub-cases p = 3 and p > 3. In the first subcase, s r + min{, γ 0 } and β s (r + s + min{, γ 0 })/2. This implies that n = 2 δ0 3 δ 5 δ2 7 δ3 3 δ 4. In this sub-case, all possible solutions of τ(λ(n)) = ω(n) + 2 are: (r, s) (δ 0, δ, δ 2, δ 3, δ 4 ) n (, ) (0, δ, 0,, 0), δ =, 2 3 7, (, 2) (δ,, 0,, ), δ 4 2 δ (δ, 2, 0,, ), δ 4 2 δ (δ, 0,,, ), δ 4 2 δ (2, 2) (0,,,, ) (0, 2,,, ) (2, ) (δ,,,, 0), δ 4 2 δ (δ, 2,,, 0), δ 4 2 δ (δ,,, 0, ), δ 4 2 δ (δ, 2,, 0, ), δ 4 2 δ which are eactly the 32 elements of F 2. In the sub-case r > 0, s > 0, t = 0, p > 3, we have β s 2s. Thus, 2 αr + 2s + 2 αr + β s + 2 r + s + min{γ 0, } + 2, and s r + + min{γ 0, } 2 αr, which implies that s = and β 2. Note also that α r and r cannot be 3 since this would imply s =, 2 αr + 5, τ(λ(n)) 0, and ω(n) 8, which is impossible because ω(n) r + s Therefore, n = 2 δ0 3 δ 5 δ2 (2 β p + ) δ 3,

11 THE EQUATION τ(λ(n)) = ω(n) + k with p > 3. If 5 2 n, then we have the solutions n = 2 α 3 5 2, α =, 2, 3, 4, which are eactly the elements of F 3, while if 5 2 n, then we have the solutions n = 3 (2p + ), which are elements of I 2, and n = 2 α 3 5 (2 β + ), α =, 2, 3, 4, and β =, 2, which are elements of I 3. The last case to consider is when t > 0, so that there is a prime dividing n of the form 2 β p 2 +. Now equation τ(λ(n)) = ω(n) + 2 is equivalent to (6) Θ Λ γ γ r+s+t = r + s + t + min{, γ 0 } + 2, where if p 3 n; Θ = 3/2 if p 2 n; 3 if p 2 n; and Λ = ma{τ(λ(2 γ 0 )), 2 αr +, β s +, δ t + }. Here, the terms 2 αr + (resp. β s + ) are to be omitted if r = 0 (resp. s = 0). We claim that r, s 0, and we will show this later. Therefore, from (??), we deduce that 2 αr + β s + δ t + 3 r + s + t + min{, γ 0 } + 2. On one side, the above implies that 2 αr r + min{, γ 0 }, so that γ 0 and either r =, α = 0 or r = 2, α 2 =, α = 0. On another side, the above implies that s + t β s + δ t 2r. If r =, then s = t = β s = δ t =, and since 2p 2 + is prime, we necessarily have p = 3. Hence, n = 2 γ0 3 γ 7 γ2 9 γ 3, and the only solutions of τ(λ(n)) = 6 of the above form are the first 9 elements of F 4. If r = 2, then 4 s + t β s + δ t 4. This implies that s = t = 2 and (β, β 2, δ, δ 2 ) = (, 2,, 2), so that again p = 3, n = 2 γ0 3 γ 5 γ2 7 γ3 3 γ3 9 γ4 37 γ 5, and the only solutions of τ(λ(n)) = 9 of the above form are the last 2 elements of F 4. Finally, we need to prove the claim r, s 0. If r = 0, s 0, then from (??) we deduce that 3(s + t + 2)/2 3(β s + δ t + 2)/2 s + t + 3, which implies s + t 0, which is a contradiction. A similar argument rules out the possibility r = 0 and s = 0. Lastly, if r 0 and s = 0,

12 2 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI then from (??) and from δ t t we deduce that 3(2 αr + t + 2)/2 r + t + 3, which is again a contradiction and ends the proof of the proposition. 3. Lower bounds on the counting functions of A k Theorem. A k is nonempty for all nonnegative integers k. Proof. Let p = 3, p 2 = 5, p 3 = 3 and p 4 = 3. Then, for each m 3 and for each t {4, 5, 6, 7} the number n = 2 m+ 7 p p t 3 verifies ω(n) = t and τ(λ(n)) = τ(2 m 3 5) = 4m. Hence, τ(λ(n)) ω(n) = 4(m ) (t 4) can assume all possible values greater than or equal to 8. Finally, 3 A 0, 4 A, 5 A 2, 7 A 3, 7 A 4, 3 A 5, 62 A 6, and 3 A 7, which completes the proof. In what follows, we show that if k is sufficiently large, then A k contains many elements. Theorem 2. For all k 0,, 2, 3, 4, we have the lowerbound #A k () k as. Proof. The proof uses the famous Theorem of Chen that we state in the following form (see [?], or Lemma.2 in [?], or Chapter in [?]). Lemma. Let a N be an even number. There eists a constant c = c(a) such that if > 0 (a), then the number of primes p [/2, ] such that p (mod a) and (p )/a has at most two prime factors each of which eceeds /0 is at least c a /. We write k = 4s + r, with s, r {0,, 2, 3} and distinguish the two cases: Case. r 3; Case 2. r = 3. In Case, we apply Chen s Theorem with the choice a = 2 s and obtain that there are either at least M a a / primes p /42 with p = 2 s+2 q and q prime, or at least N a a / primes p /42 with p = 2 s q q 2, where q and q 2 are distinct primes which eceed /0.

13 THE EQUATION τ(λ(n)) = ω(n) + k 3 Assume that we are in the first instance. integers n of the form n = 7pT, where if r = 2; T = 2 if r = ; 6 if r = 0. Then consider the M a With these choices, we have that ω(n) = 4 r, λ(n) = 2 s 3 q and τ(λ(n)) = 4(s + ), therefore τ(λ(n)) ω(n) = 4s + r = k. Assume now that we are in the second instance. Then consider the N a integers n of the form n = 2pT where if r = 2; T = 3 if r = ; 5 if r = 0. For s 2 and for (s = and r 0), we have that ω(n) = 4 r, λ(n) = 2 s q q 2 and τ(λ(n)) = 4(s + ), so that again τ(λ(n)) ω(n) = k. In Case 2, we apply Chen s Theorem with the choice a = 2 s+ and obtain that either there are at least M a a / primes p /50 with p = 2 s+ q and q prime, or at least N a a / primes p /50 with p = 2 s+ q q 2, and q and q 2 distinct primes which eceed /0. Assume that we are in the first instance. Then consider the M a integers n of the form n = 20p. For s, we have that ω(n) = 5 and τ(λ(n)) = 4(s + 2), so that τ(λ(n)) ω(n) = 4s + 3 = k. Assume that we are in the second instance. Then consider the N a integers n of the form n = 50p. For s 3, we have that ω(n) = 5 and τ(λ(n)) = 4(s + 2), so that again τ(λ(n)) ω(n) = 4s + 3 = k. Net assume that k = 7. Then we apply Chen s Theorem with the choice a = 2 and obtain that either there are at least M a a / primes p /92 with p = 2q and q prime, or at least N a a / primes p /92 with p = 2q q 2, and q and q 2 distinct primes which eceed /0. Assume that we are in the first instance. Then consider the M a integers n of the form n = 2 6 3p. We have that ω(n) = 3 and τ(λ(n)) = τ(2 4 p), so that τ(λ(n)) ω(n) = 0 3 = 7. Assume that we are in the second instance. Then consider the N a integers n of the form n = p = (2q q 2 + ). We have that ω(n) = and τ(λ(n)) = 8, so that again τ(λ(n)) ω(n) = 8 = 7. Finally, we treat the case k =. Here, we apply Chen s Theorem with the choice a = 4 and deduce that either there eist M / primes p /450, such that p = 4q, with q prime, or

14 4 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI there eist N / primes p /450, such that p = 4q q 2, where q and q 2 are distinct primes which eceed /0. If we are in the instance when M /, then we note that for large the M positive integers n = p = 450p, where p is of the form 4q +, are all, have ω(n) = 5 and λ(n) = q, therefore τ(λ(n)) = 6 = ω(n) +. If we are in the instance when N /, then for large the N positive integers n = p, where p is such that p = 4q q 2, with distinct primes q and q 2 which eceed /0, have the property that τ(λ(n)) = τ(4q q 2 ) = 2 = ω(n) +. Thus, #A () ma{m, N} /, which completes the proof of this theorem. The remaining cases are k = 0,, 2, 3, 4, need to be treated separately. Propositions??,?? and?? address the first three cases and certainly there is no hope even to show that A k is infinite for k = 0,, 2. While the net result is not such a precise characterization of A k for k = 3, 4 as Propositions??,?? and?? for the smaller values of k, its aim is to show that it is beyond our reach to show that either one of these two sets is infinite. Proposition 5. Assume that A 3 A 4 is infinite. Then there eists an even positive integer c such that the set of primes of the form p = cq β +, with q prime and β 4 is infinite. Proof. Assume that n A 3 A 4. Then τ(λ(n)) ω(n) + 4. Write m = λ(n) and note that ω(n) is at most the number of divisors of m of the form p for some prime p. Hence, m can have at most four divisors d such that d + is composite. Write m = 2 α l, where l is odd. If α 9, then 2 3, 2 5, 2 6, 2 7 and 2 9 are five divisors of m none of the form p for some prime p. Thus, α 8. If τ(l) 6, then l (hence, m) has at least five odd divisors >, and certainly none of them is of the form p for some prime p. Thus, τ(l) 5, which shows that either l = q β for some prime q and some β 4, or l = q q 2, where q and q 2 are distinct primes. Assume that l = q β holds for infinitely many n. Then there eist infinitely many primes p of the form p = 2 α 0 q β for some α 0 {,..., 9}, and β {,..., 4}, which implies the conclusion of the proposition. Assume now that l = q q 2 holds for infinitely many n. Suppose further that q < q 2. We then distinguish two cases. The first case is when q remains bounded for infinitely many such n. Then 2 α q can take only finitely many values. Since we have infinitely many values for n, there must eist some fied even positive integer c (an even divisor

15 THE EQUATION τ(λ(n)) = ω(n) + k 5 of a number of the form 2 9 q over all the finitely many possibilities for q ), such that p = cq 2 holds for infinitely many primes p, which implies the conclusion of the proposition. The second case is when q tends to infinity as n tends to infinity in A 3 A 4. If for infinitely many such n we have that either 2q + or 2q 2 + is prime, then we get the conclusion of the proposition with c = 2. Assuming that this is not the case, we show that we get a contradiction. Note first that α 3, for if not 2 3, q, q 2, 2q and 2q 2 are five divisors of n none of which is of the form p for some odd prime p. Assume now that α =. Then τ(λ(n)) = τ(2q q 2 ) = 8, therefore ω(n) 4. Since the only prime factors of n are in {2, 3, 2q +, 2q 2 +, 2q q 2 +}, we deduct that one of 2q + and 2q 2 + must be prime, which is a contradiction. Finally, if α = 2, then τ(λ(n)) = τ(4q q 2 ) = 2, therefore ω(n) 8. Since all the prime factors of n belong to {2, 3, 5, 2q +, 2q 2 +, 4q +, 4q 2 +, 2q q 2 +, 4q q 2 + }, we get again that one of 2q + or 2q 2 + must be a prime, which is the final contradiction. 4. Upper bounds on the counting functions of A k Our first result here shows that numbers n A k have ω(n) bounded in terms of k. Proposition 6. If n A k, then ω(n) 2(k + ) 2 +. Proof. We use the same idea and notations as in the proof of Proposition??. Let n A k, and put m = λ(n) = 2 α l, where α is a nonnegative integer and l is odd. If α 2k + 3, then 2 3, 2 5,..., 2 2k+3 are k + divisors of m none of which is of the form p for some prime p, which is a contradiction. If τ(l) k + 2, then l (hence, m) has k + odd divisors >, and obviously none of them is of the form p for some prime p, which is again a contradiction. Hence, α 2k + 2 and τ(l) k +, therefore ω(n) = τ(λ(n)) k = τ(2 α l) k = (α + )τ(l) k (2k + 3)(k + ) k = 2(k + ) 2 +. An upperbound for the counting function #A k () of A k follows from Proposition?? with a little etra work. Let us set b k = 2(k + ) log 2 (2(k + ) 2 + k + ). We then have the following result.

16 6 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI Theorem 3. For all nonnegative integers k we have the upper bound #A k () k (log log ) b k as. Proof. Let K 2 be any fied positive integer. Let π K () be the number of primes p such that ω(p ) K. We begin with the following lemma. Lemma 2. There eists an absolute constant c 0 such that the following estimate holds π K () (log log + c 0) K+ (K )! as. Proof. Let P() = {p : ω(p ) K}. Put y = / log log and u = log / log y = log log. For a positive integer n we write P (n) for the largest prime factor of n. Let Ψ(, y) = {n : P (n) y}. By a result of de Bruijn (see [?], as well as Corollary.3 of [?], [?] and Chapter III.5 of [?]), the bound (7) #Ψ(, y) ep( ( + o())u log u) < holds as u, where u = log / log y, provided that u y /2, which is satisfied for the above choice of y. Therefore, if P () = P() Ψ(, y), then we have that #P (). Now let P 2 () = {p : q 2 p for some q y}. For a fied q y, the number of < n such that q 2 n and is /q 2. Thus, #P 2 () q dt 2 q y y t ( ) 2 y = o. Put P 3 () = P()\ (P () P 2 ()). Write p = P m, where P = P (p ). Since P > y and p P 2 (), we deduce that P (m) < P. Thus, ω(m) K. Fi m. By Brun s sieve (see, for eample, Theorem 2.3 in [?]), we have that the number of primes p such that p = mp for some prime P is ϕ(m) (log /m) 2 (log log )2 ϕ(m)(log y) 2 ϕ(m)(log ). 2

17 THE EQUATION τ(λ(n)) = ω(n) + k 7 Summing up over all the acceptable values of m, we get #P 3 () (log log )2 (log log )2 (log log )2 (log log )2 (log log )2 m ω(m) K K k= K k= K k= K k= m ω(m)=k k! k! ϕ(m) ( ϕ(m) p α ( p ) k p α (p ) p + O() k! (log log + c 0) k. It remains to note that in the above sum the last term dominates as tends to infinity. We are now ready to prove Theorem??. Assume that k 3, since otherwise the result follows immediately from Propositions??,??,?? and Brun s sieve even with a smaller b k (i.e., b 0 = 0, b = and b 2 = ). Now note that if p n and n A k, then 2 ω(p ) τ(p ) τ(λ(n)) = ω(n) + k 2(k + ) 2 + k +, (by Proposition??), therefore ω(p ) K = log 2 (2(k +) 2 +k +). Lemma?? shows that (8) #{p : ω(p ) K} K (log log ) K+. We put A k, () for the set of n A k () such that either P y = / log log, or P 2 divides n. As in the proof of Lemma??, (9) #A k,. Let A k,2 () stand for the complement of A k, () in A k (). Now write n A k, () as n = P m, where P = P (n). So, P > y = / log log, P 2 does not divide n, and ω(m) = ω(n) 2(k + ) 2. Fiing m, the number of values for P /m such that ω(p ) K is, by estimate ) k

18 8 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI (??), (log log(/m)) K+ (log log ) K+ π K (/m) k m(log(/m)) 2 k m(log y) 2 k (log log ) K+3 m. Summing up the above inequality over all the values of m with ω(m) 2(k + ) 2, we get that the number of possibilities is (log log ) K+3 #A k,2 () k m k (log log ) K+3 k m ω(m) 2(k+) 2 2(k+) 2 l=0 l! (log log ) K+3+2(k+)2, ( p α ) l p α which together with (??) completes the proof of this theorem. A more careful analysis (along the lines of the proof of Theorem 4. in [?]) shows that Theorem?? holds with a somewhat smaller b k. Furthermore, it is clear that one can write down a formula for the implied constant in terms of k. We do not enter into such details. 5. A more general Statement Let f() be any function which tends to infinity with n and which is monotonically decreasing for > 0. Let (0) B f = {n : τ(λ(n)) ω(n) < ep((log log n)/f(n))}. We then show the following result. Theorem 4. If B f estimate holds #B f () is the set appearing at (??), then the following +o() as. We start by proving the following lemma: Lemma 3. Let P f = {p : ω(p) < 2(log log p)/ f(p)}. Then the following estimate holds () #P f () as. +o()

19 THE EQUATION τ(λ(n)) = ω(n) + k 9 Proof. Let be large, put y = / log log and let P 2 () = {p P f () : p Ψ(, y)}. If p P 2 (), then p = Qm, where Q = P (p ) > y and m /y. Fi m. By Brun s method, the number of primes Q /m such that p = Qm + is also prime is ϕ(m)(log(/m)) 2 (log log )2 ϕ(m)(log y) 2 ϕ(m)(log ). 2 Using the minimal order ϕ(m)/m / log log of the Euler function in the interval [, ], we get that if m is fied, then the number of acceptable primes p P 2 () with (p )/P (p ) = m is (log log log )3 m. Let M() be the set of acceptable values for m. Since ω(p ) 2(log log p)/ f(p), f is increasing for large and p > y for all p P 2 (), it follows that z = ma{2(log log p)/ f(p) : p P 2 ()} 2 log log f(y) = o(log log ) as. Furthermore, M() {m : ω(m) z}. We then get (log log )3 (log log )3 (2) #P 2 () m m. Put m M() S k () = m ω(m)=k m. k z m ω(m)=k Clearly, by unique factorization, the multinomial formula and Stirling s formula, ( ) (3) S k () k ( ) k e log log + O(), k! p α k p α where we also used the obvious fact that p = O(), α p 2 α 2 together with Mertens s formula = log log + O(). p p

20 20 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI Since for every fied value of A > the function (ea/t) t is increasing for t < A, it follows that ( ) z e log log + O() S k () = ep (z log(e(log log + O())/z)) z ( 2 log log ep log(o( ) f(y))) = ep(o(log log )) f(y) (4) = (log ) o(), for k z. Hence, by inequalities (??) and (??) and estimate (??), we get (log log )3 #P 2 () S k () k z (log log )4 ma{s k () : k z} = +o(), which together with estimate (??) implies inequality (??) and completes the proof of the lemma. By partial summation, we immediately get Corollary 2. If P f is the set of primes appearing in Lemma??, then p = O(). p P f Proof of Theorem??. Let again y = / log log, w = / and It follows by inequality (??) that B () = {n w} Ψ(, y). (5) #B () 2 once is large. Let B 2 () = {n : ω(n) > 0 log log }. It follows from results of Norton [?] that #B 2 () (log ) λ, where λ = + 0 log(0/e) > 2, therefore (6) #B 2 () < (log ). 2 Now put B 3 () = B f ()\(B () B 2 ()),

21 THE EQUATION τ(λ(n)) = ω(n) + k 2 and assume that n B 3 (). Replacing f() by min{f(), log log log }, we may assume that f() log log log. Then p λ(n) for all prime factors p of n, therefore so 2 ω(p ) τ(λ(n)) ω(n) + ep((log log n)/f(n)) (7) ω(p ) < < 0 log log + ep((log log )/f(w)) ( ).(log log ) < ep, f(w).6(log log ) f(w), where we used the fact that./ log 2 <.6. Let B 4 () = {n B 3 () : P (n) > w}. Since w p/(log p) 2 holds for all p [w, ] once is large, it follows that if p = P (n) for n B 4 (), then the inequality ω(p ) <.6(log log ) f(p/(log p) 2 ) < 2(log log p) g(p), holds for large, where g is the function g(t) = (f(t/(log t) 2 )) 2, which is increasing for large t. Thus, p P g. Let us now write n = P m, where m < /p <, and let us fi m. Then p P g (/m) and, by Lemma??, the number of such choices for p is #P g (/m) m(log /m) 2+o() = m+o(). Summing up the above inequality for m, we get #B 4 () #P g (/m) m (8) = +o() +o(), m m because m log log = (log )o(). m From now on, we are assuming that n B 5 () = B 3 ()\B 4 (). Let n = P m, where P = P (n) [y, w]. Since.6 log log < 2 log log y 2 log log P for large, and f(w) f(p ), we get that ω(p ) <.6(log log ) f(w) < 2(log log P ). f(p )

22 22 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI In particular, P P f 2. By Lemma??, we get that if m /y is fied, then the number of choices for P is at most #P f 2(/m) m(log(/m)) 2+o() m(log y) 2+o() m(log ), 2+o() where we used the facts that /m y and log y = log / log log = (log ) +o(). Let M() be the set of acceptable values of m. Then (9) #B 5 () m M() m+o() +o() m M() m. Let Q() be the set of primes dividing some m M(). Note that Q() consists primes q satisfying the inequality (??). We put v = ep(ep((log log )/f(w))) and split the primes in Q into two subsets as follows: Q = {q v} Q. Q 2 = Q [v, w]. Note that if q Q 2, then 2 log log q f(q) 2 log log f(q)f(w) 2 log log f(w) > ω(q ), therefore Q 2 P f. This argument shows that ( ) m q α m M() q Q Q 2 α 0 ( ep q + ( )) q + O (20). q α q Q q Q 2 q 2 α 2 Since q Q q q v (by Mertens s formula), v = log log v + O() = o(log log ) q Q 2 q q P f q = O() (by Corollary??), and q 2 α 2 q α = O(),

23 THE EQUATION τ(λ(n)) = ω(n) + k 23 we get from estimate (??) that m ep(o(log log )) = (log )o(), m M() which together with (??) gives (2) #B 5 () +o(). Since B 3 () B 4 () B 5 (), we get, by estimates (??) and (??), that (22) #B 3 () < (log ), 2+o() which together with estimates (??) and (??) completes the proof of this theorem. 6. Average and normal orders of τ(λ(n)) ω(n) Our last result addresses average and normal orders of the function h(n) = τ(λ(n)) ω(n). Theorem 5. (i) There eist positive constants c 0, c such that the inequalities ) ) log (23) ep (c 0 log h(n) ep (c log log log log hold for all. (ii) The inequality n 0.5(+o())(log log n)2 h(n) = 2 holds for almost all positive integers n. Proof. (i) In [?], it is shown that inequalities (??) hold with some constants c 0 and c for the function τ(λ(n)) = h(n) + ω(n). Since the average value of ω(n) is log log = ep(o( log / log log )), the required inequality follows. (ii) In [?], it is shown that the normal order of both ω(ϕ(n)) and Ω(ϕ(n)) is 0.5(log log n) 2. Since ω(λ(n)) = ω(ϕ(n)), while Ω(λ(n)) Ω(ϕ(n)), it follows that the normal order of both ω(λ(n)) and Ω(λ(n)) is also 0.5(log log n) 2. Finally, since 2 ω(λ(n)) τ(λ(n)) 2 Ω(λ(n)), and since the normal order of ω(n) is log log n = 2 o((log log n)2), the desired inequalities follow.

24 24 A. GLIBICHUK, F. LUCA AND F. PAPPALARDI 7. Comments and Remarks We suspect that for every k there eist constants a k > 0 and c k 0 such that (24) #A k () = a k ( + o()) (log log )c k as. Widely believed conjectures concerning the distribution of Sophie Germain primes p together with Proposition?? seem to support the above conjecture (??) at k = (with c = 0 and some a > 0). Note that an upper bound of the above shape is given in Theorem??. We would like to leave this conjecture as an open problem. References [] W. D. Banks, K. Ford, F. Luca, F. Pappalardi, I. E. Shparlinski, Values of the Euler function in various sequences, Monatsh. Math. 46 (2005), no., 9. [2] N. G. de Bruijn, On the number of positive integers and free of primes > y, II, Indag. Math., 28 (966), [3] E. R. Canfield, P. Erdős and C. Pomerance, On a problem of Oppenheim concerning Factorisatio Numerorum, J. Number Theory, 7 (983), 28. [4] J. R. Chen, On the representation of a large even integer as the sum of a prime and a product of at most two primes, Sci. Sinica 6 (973), [5] P. Erdős and C. Pomerance, On the normal number of prime factors of ϕ(n), Rocky Mtn. J. of Math. 5 (985), [6] P. Erdős, C. Pomerance and E. Schmutz, Carmichael s lambda function, Acta Arith. 58 (99), no. 4, [7] K. Ford, The number of solutions of ϕ() = m, Ann. Math. 50 (999), [8] H. Halberstam and H.-E. Rickert, Sieve Methods, Academic Press, London, 974. [9] A. Hildebrand and G. Tenenbaum, Integers without large prime factors, J. Théor. Nombres Bordeau, 5 (993), [0] F. Luca and C. Pomerance, On the average number of divisors of the Euler function, Publ. Math. (Debrecen), to appear. [] K. K. Norton, The number of restricted prime factors of an integer. I, Illinois J. Math. 20 (976), ; II, Acta Math. 43 (979), [2] G. Tenenbaum, Introduction to analytic and probabilistic number theory, Cambridge Univ. Press, 995. (Glibichuk) Department of Mechanics and Mathematics, Moskow State University, Vorobevy gory,, MSU Main Building, Moskow, 9992, Russia address: aanatol@mail.ru

25 THE EQUATION τ(λ(n)) = ω(n) + k 25 (Luca) Instituto de Matemáticas, Universidad Nacional Autónoma de Méico, C.P , Morelia, Michoacán, Méico address: fluca@matmor.unam.m (Pappalardi) Dipartimento di Matematica, Università degli Studi Roma Tre, Largo S. L. Murialdo,, Roma, I 0046, Italy address: pappa@mat.uniroma3.it