Annales Univ. Sci. Budapest., Sect. Comp. 45 (06) 0 0 ITERATES OF THE SUM OF THE UNITARY DIVISORS OF AN INTEGER Jean-Marie De Koninck (Québec, Canada) Imre Kátai (Budapest, Hungary) Dedicated to Professor Pavel D. Varbanets on the occasion of his 80-th birthday Communicated by Karl-Heinz Indlekofer (Received March 6, 06; accepted July 9, 06) Abstract. Given an integer k 0, let σk(n) stand for the k-fold iterate of σ (n), the sum of the unitary divisors of n. We show that σ (p+) σ tends (p+) to for almost all primes p.. Introduction and notation Let σ (n) be the sum of the unitary divisors of n, that is, σ (n) := d n (d,n/d)= Given an integer k 0, let σ k (n) stand for the k-fold iterate of σ (n), that is, σ 0(n) = n, σ (n) = σ (n), σ (n) = σ (σ (n)), and so on. The function σ (n) is easily checked to be multiplicative with σ (p α ) = p α + for each prime p and integer α. Key words and phrases: Sum of unitary divisors, shifted primes. 00 Mathematics Subject Classification: N37. Research supported in part by a grant from NSERC. d.
0 J.-M. De Koninck and I. Kátai In 97, Erdős and Subbarao [3] proved that (.) σ(n) for almost all n. (n) σ This is uite a contrast with the easily proven estimate σ (n) for almost all n. σ (n) In 99, Kátai and Wisjmuller [6] proved that σ 3(n) σ (n) for almost all n and conjectured that, given an arbitrary integer k 0, This remains unproven. σ k+ (n) σ k (n) for almost all n. Here, we consider similar uotients, namely those where the arguments of the functions σk+ and σ k are running over shifted primes.. Main result Theorem. We have σ(p + ) for almost all primes p. (p + ) σ Of course, the above statement is euivalent to the following one. Given any ε > 0, we have } lim π() # p : σ (p + ) σ > + ε (p + ) = 0. In the following, we denote by p(n) and P (n) the smallest and largest prime factors of n, respectively. We let µ(n) stand for the Moebius function and φ(n) for the Euler totient function. For each integer n, we let ω(n) stand for the number of distinct prime factors of n and set ω() = 0. The letters p,, π and
Iterates of the sum of the unitary divisors of an integer 03 Q, with or without subscript, will stand eclusively for primes. On the other hand, the letters c and C, with or without subscript, will stand for absolute constants but not necessarily the same at each occurrence. We will let stand for the set of all primes. Moreover, we shall at times use the abbreviations = log, = log log, and so on. We denote the logarithmic integral dt by li(). Finally, we let π() stand for the number of primes p log t and we write π(; k, l) for #p : p l (mod k)}. 3. Preliminary results For the proof of our main result, we will need the following lemmas. Our first lemma is a classical result. Lemma. (Brun-Titchmarsh Theorem) For every positive integer k <, we have π(; k, l) φ(k) log(/k). A proof of the following result follows from Theorem 3. in the book of Halberstam and Richert [4]. Lemma. There eists a positive constant C such that #p, : p + = a } < C φ(a) log (/a). The following can be obtained from Theorem 4. in the book of Halberstam and Richert [4]. Lemma 3. Given an arbitrary positive number δ <, there eists an absolute constant C > 0 such that lim #p : P (p + ) < δ or P (p + ) > δ } < C δ li(). It was proved by the first author [5] that the distribution of the numbers ω(σ(p + )) (log log p) 3 (log log p) 3/ as p runs through the primes obeys the Gaussian Law. It is easy to show that the same statement holds if σ(p + ) is replaced by σ (p + ). The following result is an immediate conseuence of this observation.
04 J.-M. De Koninck and I. Kátai Lemma 4. We have π() #p : ω(σ (p + )) > } 0 as. 4. Proof of the main result First let t(n) := It is clear that, for each integer n, σ (n) = ( + ) n α = ep α n and therefore that σ(n) σ (n) = σ (σ (n)) σ (n) α σ (n) α n ep α. ( log + ) α α α σ (n) ep = et(n). α α n From this observation, it follows that the claim in Theorem is euivalent to the assertion that (4.) t(p + ) 0 for almost all primes p. Let δ > 0 be any small number, let := p : p } and consider the set () := p : P (p + ) < δ or P (p + ) > δ }. In light of Lemma 3, (4.) # () < C δ li(). This is why we only need to work with the set () := \ (). So, let us assume that p (), let T be a large integer and consider the set D T made up of all those primes p such that π p + for some prime π > T.
Iterates of the sum of the unitary divisors of an integer 05 Using Lemma, for some constant C 3 > 0, we then have #p : p D T } (4.3) T <π< C 3 li() p p+ 0 (mod π ) T <π< li() T log T + O( 3/4). = T <π< φ(π ) = C 3li() π(; π, ) T <π< π(π ) Hence, in light of (4.3), we may now discard those primes p for which p D T, since their number is O(li()/(T log T )). This is why, for each prime number, we now focus our attention on the set (4.4) E () := p () : p D T and T σ (p + )} and the sum S T () := T #E (). Moreover, for each prime, we let B be the semigroup generated by those primes Q such that Q +. Let us now consider a fied prime T and those integers (4.5) K = π α παr r for which π αj j + for j =,..., r with α j = if π j > T. In order to estimate #E (), we first introduce the set H K,R :=p : p + = KRmP, P (R) T, p(mp ) > T, µ (m) =, (m, B ) =, P (p + ) = P }. Writing each prime p H K,R as p + = KRmP = ap, then, since P was assumed to be such that P > δ, it follows that a < δ, and therefore, using
06 J.-M. De Koninck and I. Kátai Lemma, that (4.6) #H K,R C δ C log m C C δ log φ(kr) C C δ log φ(kr) µ (m) φ(krm) m/kr (m,b )= µ (m) φ(m) T <π π+ 0 (mod ) ( + ) π } C C δ log φ(kr) ep = = C C δ log log ep } φ(kr) C C δ li() ep }. φ(kr) (4.7) On the one hand, observe that R P (R)T φ(r) for some positive constant C 4. pt ( + φ(p) + ) φ(p ) + = ( + ) p + p(p ) + = pt = ( + p ( )) p + O = pt = ep p + O() C 4 log T pt On the other hand, writing each K as K = K K, where P (K ) T and p(k ) > T. Then, by the nature of K (see (4.5)), it is clear that K is suarefree and that ω(k ) r T. From this, it follows that, for some constant C 5 > 0, (4.8) K φ(k) T φ(k ) K j=0 j! π π+ 0 (mod ) j π = < C 5 T (T )!.
Iterates of the sum of the unitary divisors of an integer 07 Recalling that for all K, R N, we have φ(kr) φ(k)φ(r), and combining (4.7) and (4.8) in (4.6), it follows that (4.9) #E () T H K,R C 6 δli() (T )! ep } log T K,R for some constant C 6 > 0, from which it follows that (4.0) S T () = o(li()) ( ). We now move to estimate t(p+) when the primes such that α σ (p+) satisfy > T. We first consider the case where α =, and for this we introduce the sum p () T < ε s(p + ) := T < ε σ (p+) where ε > 0 is an arbitrarily small number. We then have, using Lemmas, and 3, s(p + ) = #p () : p + = QHmP, P (H) T, p(m) > T, + T < ε C δ C δ C δ, P > δ, Q + 0 (mod ), (m, B ) = } + #p () : p + = HmP, P (H) T, p(m) > T, P > δ, P + 0 T < ε T < ε T < ε from which it follows that (4.) s(p + ) δ p () Q (mod ) (m,b )= P (H)T, p(m)>t Q (mod ) π (mod ) (mod )} φ(hmq) Q φ(h) P (H)T, p(m)>t ( + ) π + π(π ) + log T ep }, log T T = o(li()) ( ).
08 J.-M. De Koninck and I. Kátai To account for those [ +ε, +ε ], we will consider the sum r(p + ) := Proceeding as above, we obtain that p () r(p + ) +ε +ε C δli() C δli() from which it clearly follows that (4.) +ε +ε σ (p+). Q Q+ 0 (mod ) +ε +ε +ε +ε C δ C 6 li() π(; Q, ) Q Q+ 0 (mod ) C 6 +ε +ε C δ C 6 li() +ε 3, p () r(p + ) = o(li()). We now split t(p + ) into five sums as follows. Q (4.3) t(p + ) = t (p + ) + t (p + ) + t 3 (p + ) + t 4 (p + ) + t 5 (p + ), where t j (p + ) = α σ (p+) I j α (j =,..., 5), with the five intervals I j being defined as I = [, T ], I = (T, ε ], I 3 = ( ε, +ε ), I 4 = [ +ε, +ε ], I 5 = ( +ε, ). We will show that the net five ineualities hold for almost all primes p ().
Iterates of the sum of the unitary divisors of an integer 09 First of all, in light of (4.0), we have (4.4) t (p + ) T (p ), ( ) li() with the eception of at most O In light of (4.), we have that T (4.5) t (p + ) s(p + ) + T < ε primes p (). o() + T log T (p ). Clearly, (4.6) t 3 (p + ) ε << +ε log ( ) + ε < ε. ε In light of (4.), we have (4.7) t 4 (p + ) r(p + ) + > +ε ( ) o() + O +ε 3 (p ). Finally, using Lemma 4, it follows that (4.8) t 5 (p + ) ε (p ). Gathering ineualities (4.4) through (4.8) in (4.3), we have thus established that π() # p : t(p + ) > } δ T and since this is true for every δ > 0 and for every large number T, our claim (4.) is established and the proof of Theorem is complete. 5. Final remark The unitary analog of φ(n) denoted by φ (n) and introduced by Cohen [] is defined by φ (n) := (p α ). p α n
0 J.-M. De Koninck and I. Kátai Denote by φ k (n) the k-fold iterate of φ (n). Erdős and Subbarao [3] claimed that φ (n) (n ) (n) φ ecept on a set of integers n of zero density. In fact, they mentioned that they could prove this result by using the same methods as those used to prove (.). With the approach we used to prove Theorem, we can also prove the following. Theorem. Given any ε > 0, } φ π() # (p + ) φ < ε 0 (p + ) ( ). References [] Cohen, E., Arithmetic functions associated with the unitary divisors of an integer, Math. Zeitschrift, 74 (960), 66 80. [] De Koninck, J.-M. and I. Kátai, On the distribution of the number of prime factors of the k-fold iterate of various arithmetic functions, preprint. [3] Erdős, P. and M.V. Subbarao, On the iterates of some arithmetic functions, in The theory of arithmetic functions (Proc. Conf., Western Michigan Univ., Kalamazoo, Mich., 97), pp. 9 5. Lecture Notes in Math., Vol. 5, Springer, Berlin, 97. [4] Halberstam, H.H. and H.E. Richert, Sieve Methods, Academic Press, London, 974. [5] Kátai, I., Distribution of ω(σ(p+)), Annales Univ. Sci. Budapest., Sect. Math., 34 (99), 7 5. [6] Kátai, I. and M. Wijsmuller, On the iterates of the sum of unitary divisors, Acta Math. Hungar., 79 (998), no., 49 67. J.-M. De Koninck Dép. math. et statistiue Université Laval Québec Québec GV 0A6 Canada jmdk@mat.ulaval.ca I. Kátai Computer Algebra Department Eötvös Loránd University 7 Budapest Pázmány Péter Sétány I/C Hungary katai@inf.elte.hu