Positive Integers n Such That σ(φ(n)) = σ(n)

2 3 47 6 23 Journal of Integer Sequences, Vol. (2008), Article 08..5 Positive Integers n Such That σ(φ(n)) = σ(n) Jean-Marie De Koninck Départment de Mathématiques Université Laval Québec, PQ GK 7P4 Canada jmdk@mat.ulaval.ca Florian Luca Instituto de Matemáticas Universidad Nacional Autónoma de Méico C.P. 58089 Morelia, Michoacán Méico fluca@matmor.unam.m Abstract In this paper, we investigate those positive integers n for which the equality σ(φ(n)) = σ(n) holds, where σ is the sum of the divisors function and φ is the Euler function. Introduction For a positive integer n we write σ(n) and φ(n) for the sum of divisors function and for the Euler function of n, respectively. In this note, we study those positive integers n such that σ(φ(n)) = σ(n) holds. This is sequence A03363 in Sloane s Online Encylopedia of Integer Sequences. Let A be the set of all such positive integers n and for a positive real number we put A() = A [,]. Our result is the following.

Theorem. The estimate holds for all real numbers >. ( ) #A() = O (log ) 2 The above upper bound might actually be the correct order of magnitude of #A(). Indeed, note that if m is such that σ(2φ(m)) = 2σ(m) () and if q > m is a Sophie Germain prime, that is a prime number q such that p = 2q + is also prime, then n = mp A. The numbers m = 238, 2806, 5734, 5937, 798, 8097,... all satisfy relation (), and form Sloane s sequence A37733. More generally, if k and m are positive integers such that σ(2 k φ(m)) = 2 k σ(m) (2) and q <... < q k are primes with p i = 2q i + also primes for i =,...,k and q > m, then n = p...p k m A. Now recall that the Prime K-tuplets Conjecture of Dickson (see, for instance, [2, 4, 8]) asserts that, ecept in cases ruled out by obvious congruence conditions, K linear forms a i n+b i, i =,...,K, take prime values simultaneously for about c/(log ) K integers n, where c is a positive constant which depends only on the given linear forms. Under this conjecture (applied with K = 2 and the linear forms n and 2n+), we obtain that there should be /(log ) 2 Sophie Germain primes q, which suggests that #A() /(log ) 2. We will come back to the Sophie Germain primes later. Throughout, we use the Vinogradov symbols and and the Landau symbols O and o with their regular meanings. We use log for the natural logarithm and p, q and r with or without subscripts for prime numbers. 2 Preliminary Results In this section, we point out a subset B() of all the positive integers n of cardinality O(/(log ) 2 ). For the proof of Theorem we will work only with the positive integers n A()\B(). Further, 0 is a sufficiently large positive real number, where the meaning of sufficiently large may change from a line to the net. We put ( ) log y = ep. log log For a positive integer n we write P(n) for the largest prime factor of n. It is well known that Ψ(,y) = #{n P(n) y} = ep( ( + o())u log u) (u ), (3) where u = log / log y, provided that u y /2 (see [], Corollary.3 of [6], or Chapter III.5 of [9]). In our case, u = log log, so, in particular, the condition u y /2 is satisfied for > 0. We deduce that u log u = (log log )(log log log ). 2

Thus, if we set B () = {n P(n) y}, then We now let #B () = Ψ(,y) = ep ( ( + o())(log log )(log log log )) < for > (log ) 0 0. (4) z = (log ) 26 and for a positive integer n we write ρ(n) for its largest squarefull divisor. Recall that a positive integer m is squarefull if p 2 m whenever p is a prime factor of m. It is well known that if we write S(t) = {m t m is squarefull}, then #S(t) = ζ(3/2) ζ(3) t/2 + O(t /3 ), where ζ is the Riemann Zeta-Function (see, for eample, Theorem 4.4 in [7]). By partial summation, we easily get that m t/2. (5) m t m squarefull We now let B 2 () be the set of positive integers n such that one of the following conditions holds: (i) ρ(n) z, (ii) p n for some prime p such that ρ(p ± ) z, (iii) there eist primes r and p such that p n, p ± (mod r) and ρ(r ± ) z. We will find an upper bound for #B 2 (). Let B 2, () be the set of those n B 2 () for which (i) holds. We note that for every n B 2, () there eists a squarefull positive integer d z such that d n. For a fied d, the number of such n does not eceed /d. Hence, #B 2, () d z d squarefull d (log ) 3, (6) where we have used estimate (5) with t = z. Now let B 2,2 () be the set of those n B 2 () for which (ii) holds. We note that each n B 2,2 () has a prime divisor p such that p ± (mod d), where d is as above. Given d and p, the number of such n does not eceed /p. Summing up over all choices of p and d we get that #B 2,2 () d z d squarefull p ± (mod d) p (log log ) 2 d z d squarefull 3 p d d z d squarefull log log φ(d) (log log )2 (log ) 3, (7)

where in the above estimates we used aside from estimate (5), the fact that the estimate ( ) p log log p (a,b) + O (8) φ(b) p a (mod b) p holds uniformly in a, b and when b, where a and b are coprime and p (a,b) is the smallest prime number p a (mod b) (note that p (,b) b+ and p (,b) = p (b,b) b φ(b) for all b 2), together with the well known minimal order φ(n)/n / log log, valid for n in the interval [,]. Let B 2,3 () be the set of those n B 2 () for which (iii) holds. Then there eists r such that r ± (mod d) for some d as above, as well as p n such that r p or r p +. Given d, r and p, the number of such n does not eceed /p, and now summing up over all choices of d, r and p, we get that #B 2,3 () d z r ± (mod d) d squarefull r d z r ± (mod d) d squarefull r (log log ) 2 d z d squarefull (log log ) 3 d z d squarefull (log log ) 4 d z d squarefull p ± (mod r) p log log φ(r) r ± (mod d) r φ(d) d r p (log log )4 (log ) 3, (9) where in the above estimates we used again estimate (5), estimate (8) twice as well as the minimal order of the Euler function on the interval [,]. Hence, using estimates (6) (9), we get #B 2 () #B 2, () + #B 2,2 () + #B 2,3 () < (log ) 0 for > 0. (0) We now put and set w = 0 log log S(w,) = ω(m) w m m, 4

where ω(m) denotes the number of distinct prime factors of the positive integer m. Note that, using the fact that p = log log t + O() and Stirling s formula k! = (+o())kk e k 2πk, p t we have S(w,) = k w ω(m)=k m m = k w k! ( p α ) k p α = ( ( )) k k! p + O p 2 k w p p 2 k w ( ) k e log log + O() ( e log log + O() k w k w k w ( ) w e log log + O() w for > 0 because 0 log(0/e) >. (log log + O()))k k! ) k (log ) 0 log(0/e) < (log ) () We now let B 3 () be the set of positive integers n such that one of the following conditions holds: (i) ω(n) w, (ii) p n for some prime p for which ω(p ± ) w, (iii) there eist primes r and p such that p n, p ± (mod r) and ω(r ± ) w. Let B 3, (), B 3,2 () and B 3,3 () be the sets of n B 3 () for which (i), (ii) and (iii) hold, respectively. To bound the cardinality of B 3, (), note that, using (), we have #B 3, () = ω(n) w n ω(n) w n n = S(w,) < (log ) (2) for > 0. To bound the cardinality of B 3,2 (), note that each n B 3,2 () admits a prime divisor p such that ω(p ±) w. Fiing such a p, the number of such n does not eceed /p. Summing up over all such p we have, again in light of (), #B 3,2 () ω(p±) w p p ω(p+) w p+ + 2 p + + ω(p ) w p p (2S(w, + ) + S(w,)) < 3S(w,) + 2 (log ) (3) 5

for > 0. To bound the cardinality of B 3,3 (), note that for each n B 3,3 () there eist some prime r with ω(r ± ) w and some prime p n such that p ± (mod r). Given such r and p, the number of such n does not eceed /p. Summing up over all choices of r and p given above we get, again using (), for > 0. #B 3,3 () ω(r±) w r = (log log ) (log log ) p ± (mod r) p ω(r±) w r p ω(r ) w r r = (log log ) r + ω(r+) w r+ (log log )(S(w,) + 3S(w, + )) 4(log log )S(w,) + O(log log ) Hence, using estimates (2) to (4), we get We now let #B 3 () #B 3, () + #B 3,2 () + #B 3,3 () < ω(r±) w r 3 r + φ(r) (log log ) (log ) (4) (log ) 0 for > 0. (5) B 4 () = {n n (B () B 2 ()) and p 2 φ(n) for some p > z}. Let n B 4 () and let p 2 φ(n) for some prime p. Then it is not possible that p 2 n (because n B 2 ()), nor is it possible that p 2 q for some prime factor q of n (again because n B 2 ()). Thus, there must eist distinct primes q and r dividing n such that q (mod p) and r (mod p). Fiing such p, q and r, the number of acceptable values of such n does not eceed /(qr). Summing up over all the possible values of p, q and r we arrive at #B 4 () z p z p q (mod p) r (mod p) q<r, qr (log log )2 (log ) 3, qr z p 2 (log log ) 2 (p ) 2 (log log ) 2 6 q (mod p) q z p p 2 q 2

where we used estimates (8) and (5). Hence, We now let #B 4 () < (log ) 0 for all > 0. (6) τ = ep ( (log log ) 2) and let B 5 () stand for the set of n which are multiples of a prime p for which either p or p + has a divisor d > τ with P(d) < z. Fi such a pair d and p. Then the number of n divisible by p is at most /p. This shows that #B 5 () P(d)<z τ<d P(d)<z τ<d p ± (mod d) p p log log P(d)<z τ<d d. (7) It follows easily by partial summation from the estimates (3) for Ψ(,v), that if we write v = log τ/ log z, then S = d log ep(( + o())v log v). Since v = (log log )/26, we get that and therefore that v log v = (/26 + o())(log log )(log log log ) S for all > 0, which together with estimate (7) gives Thus, setting log ep(( + o())v log v) < (8) (log ) #B 5 () < (log ) 0 for > 0. (9) B() = we get, from estimates (4), (0), (5), (6) and (9) that 5 B i (), (20) i= #B() 5 i= #B i () (log ) 0 for all > 0. (2) 3 The Proof of Theorem We find it convenient to prove a stronger theorem. 7

Theorem 2. Let a and b be any fied positive integers. Setting A a,b = {n σ(aφ(n)) = bσ(n)}, then the estimate holds for all 3. #A a,b () a,b (log ) 2 Proof. Let be large and let B() be as in (20). We assume that n is a positive integer not in B(). We let A () be the set of n A a,b ()\B() for which (P(n) )/2 is not prime but such that there eists a prime number r > z and another prime number q P(n) for which r gcd(p(n) +,q + ). To count the number of such positive integers n, let r and q be fied primes such that r q +, and let P be a prime such that q P and r P +. The number of positive integers n such that P(n) = P does not eceed /P. Note that the congruences P (mod r) and P (mod q) are equivalent to P a q,r (mod qr), where a q,r is the smallest positive integer m satisfying m (mod r) and m (mod q). We distinguish two instances: Case : qr < P. Let A () be the set of such integers n A (). Then #A () z<r q (mod r) q log log z<r log log z<r P a q,r (mod qr) qr<p q (mod r) q r (log log ) 2 z<r (log log ) 2 z<r q (mod r) q rφ(r) where in the above inequalities we used estimates (8) and (5). Case 2: qr P. P φ(qr) q (log log )2, (22) r2 (log ) Let A () be the set of such integers n A (). Here we write n = Pm. Note that P > P(m) because y > z for large and n B () B 2 (). Furthermore, since r q +, we may write q = rl. Since q P, we may write P = sq + = s(rl )+ = srl+ s. Since r P +, we get that s (mod r), and therefore that s 2 (mod r). Hence, there eists a nonnegative integer λ such that s = λr + 2. If λ = 0, then s = 2 leading to 8

P = 2q +, which is impossible. Thus, λ > 0. Let us fi the value of λ as well as that of r. Then rl = q and (λr 2 + 2r)l (λr + ) = P (23) are two linear forms in the variable l which are simultaneously primes. Note that P /m and since P lr(λr + 2), we get that l /(mr(λr + 2)). In particular, mr(λr + 2). Recall that a typical consequence of Brun s sieve (see for eample Theorem 2.3 in [3]), is that if L (m) = Am + B and L 2 (m) = Cm + D are linear forms in m with integer coefficients such that AD BC 0 and if we write E for the product of all primes p dividing ABCD(AD BC), then the number of positive integers m y such that L (m) and L 2 (m) are simultaneously primes is Ky ( E ) 2 (log y) 2 φ(e) for some absolute constant K. Applying this result for our linear forms in l shown at (23) for which A = r, B =, C = λr 2 + 2r and D = (λr + ), we get that the number of acceptable values for l does not eceed K mr(λr + 2) (log(/(mr(λr + 2))) 2 (log log )2 mr(λr + 2), ( ) 2 (λr + 2)(λr + )r φ((λr + 2)(λr + )r) where for the rightmost inequality we used again the minimal order of the Euler function on the interval [,]. Here, K is some absolute constant. Summing up over all possible values of λ, r and m, we get #A () (log log ) 2 mr(λr + 2) z<r λ m ( ) ( < (log log ) 2 r 2 where we used again estimate (5). z<r (log log )2 (log ) 2 (log ) 3 = From estimates (22) and (24), we get #A () #A () + A () < λ ) ( λ m ) m (log log )2 (log ), (24) (log ) 0 for all > 0. (25) Now let A 2 () be the set of those n A a,b ()\(A () B()) and such that (P(n) )/2 is not prime. With n A 2 (), we get that n = Pm, where P > P(m) for > 0 because 9

y > z for large and n B () B 2 (). Then bσ(n) = bσ(m)(p + ). Let d be the largest divisor of P + such that P(d ) z. Then P + = d l, and bσ(n) = bσ(m)d l. Furthermore, φ(n) = φ(m)(p ), so that aφ(n) = aφ(m)(p ). Let d be the largest divisor of P which is z-smooth; that is, with P(d) z. Then P = dl. Note that l is squarefree (because n B 2 ()) and l and φ(m) are coprime (for if not, there would eist a prime r > z such that r l and r φ(m), so that r 2 φ(n), which is impossible because n B 4 ()). Since z > a for > 0, we get that aφ(m)d and l are coprime, and therefore that σ(φ(n)) = σ(aφ(m)d)σ(l). We thus get the equation σ(aφ(m)d)σ(l) = bσ(m)d l. Now note that l and σ(l) are coprime. Indeed, if not, since l is squarefree, there would eist a prime factor r of l (necessarily eceeding z) dividing q + for some prime factor q of l, that is of P. But this is impossible because n A (). Thus, l σ(aφ(m)d). Note now that Pm, so that m /P /y. Furthermore, ma{d,d } τ because n B 5 (). Let us now fi m, d and d. Then l σ(aφ(m)d), and therefore the number of choices for l does not eceed τ(σ(aφ(m)d)), where τ(k) is the number of divisors of the positive integer k. The above argument shows that if we write M for the set of such acceptable values for m, then #A 2 () τ2 y ma{τ(σ(aφ(m)d)) m M, d τ, d τ}. (26) To get an upper bound on τ(σ(aφ(m)d)), we write aφ(m)d as aφ(m)d = AB, where A is squarefull, B is squarefree, and A and B are coprime. Clearly, so that Since B is squarefree, it is clear that σ(b) Furthermore, σ(aφ(m)d) = σ(ab) = σ(a)σ(b), τ(σ(aφ(m)d)) τ(σ(a))τ(σ(b)). q aφ(m)d (q + ). ω(aφ(m)d) ω(aφ(n)) ω(a) + ω(n) + p n ω(p ) w 2 + w + O(), (27) so that we have ω(aφ(m)d) 2w 2 if > 0. The above inequalities follow from the fact that n B 3 (). Also, for each one of the at most 2w 2 prime factors q of aφ(m)d, the number q+ has at most w prime factors itself and its squarefull part does not eceed z, again because 0

n B 3 (). In conclusion, σ(b) C, where C is a number with at most 2w 3 prime factors whose squarefull part does not eceed z 2w2. Thus, where ω(c) 2w 3 and ρ(c) z 2w2. Clearly, and since also we finally get that τ(σ(b)) τ(c) 2 ω(c) τ(ρ(c)), τ(ρ(c)) ρ(c) z w2 = ep(w 2 log z) = ep(o(log log ) 3 ), 2 ω(c) 2 2w3 = ep(o(log log ) 3 ), τ(σ(b)) = ep(o(log log ) 3 ). (28) We now deal with σ(a). We first note that P(A) z for > 0. Indeed, if q > z divides A, then q 2 aφ(n). If > 0, then z > a, in which case the above divisibility relation forces q 2 φ(n) which is not possible because n B 3 (). By looking at the multiplicities of the prime factors appearing in A, we easily see that A ρ(a)ρ(d) ρ(p ) p m q aφ(m)d q z q ω(aφ(m)d) As we have seen at estimate (27), ω(aφ(m)d) 2w 2, in which case the above relation shows that A ρ(a)z ω(aφ(m)d)+ω(aφ(m)d)2 z 4w4 +2w 2 = ep(o(log log ) 5 ). But since σ(a) < A 2 and τ(σ(a)) σ(a) A 2, we get that which together with estimate (28) gives τ(σ(a)) = ep(o(log log ) 5 ), (29) τ(σ(aφ(m)d)) τ(σ(a))τ(σ(b)) = ep(o(log log ) 5 ). Returning to estimate (26), we get that ( #A 2 () τ2 y ep(o(log log )5 ) = ep log ) log log + O((log log )5 ) < for all > (log ) 0 0. (30) Thus, writing C() = A a,b ()\(A () A 2 () B()), we get that ( A a,b () = #C() + O (log ) 0. ). (3)

Moreover, if n C(), then n = Pm, with P > P(m) for > 0, and (P )/2 is a prime. Hence, P = 2q, where q is a Sophie Germain prime. Since also P /m, it follows by Brun s method that the number of such values for P is m(log /m) 2 (log log )2 = m(log y) 2 m(log ). 2 In the above inequalities we used the fact that /m P y. Summing up over all m, we get #C() (log log ) 2 (log log )2. (32) m(log ) 2 log m In particular, #A a,b () (log log )2. (33) log This is weaker than the bound claimed by our Theorem 2. However, it implies, by partial summation, that n + 2 t d(#a a,b(t)) n A a,b () = + #A a,b(t) t ( t= + O t=2 t(log log t) 2 dt 2 t 2 log t = O((log log ) 3 ). (34) To get some improvement, we return to C(). Let n C() and write it as n = Pm, where P > P(m). Write also P = 2q +. Then φ(n) = 2φ(m)q. Moreover, q > (y )/2 > z for > 0 so that q does not divide aφ(m) (because q > a and n B 3 ()). Hence, σ(aφ(n)) = σ(2aφ(m))(q + ). On the other hand, bσ(n) = bσ(m)(p + ) = 2bσ(m)(q + ). Thus, the equation σ(aφ(n)) = σ(bn) forces σ(2aφ(m)) = 2bσ(m), implying that m A 2a,2b (/P) A 2a,2b (/y). The argument which leads to estimate (32) now provides the better estimate In particular, we get #C() m A 2a,2b () #A a,b () (log log ) 2 m(log ) 2 (log log )5 (log ) 2. (log log )5 (log ) 2. (35) This is still somewhat weaker than what Theorem 2 claims. However, it implies that the sum of the reciprocals of the numbers in A a,b is convergent. In fact, by partial summation, ) 2

it follows, as in estimates (34), that n A a,b n y n y = #A a,b(t) t t d(#a a,b(t)) t= t=y ( + O y ) t(log log t) 5 t 2 (log t) dt 2 (log log y)5 + (log y) 2 y t(log t) 3/2dt (log y) /2. (36) We now take another look at C(). Let again n C() and write n = Pm. Fiing m and using the fact that P /m is a prime such that (P )/2 is also a prime, we get that the number of choices for P is m(log(/m)) 2. Hence, #C() m A 2a,2b (/y) m(log(/m)) 2. We now split the above sum at m = /2 and use estimate (36) to get #C() m A 2a,2b ( /2 ) (log /2 ) 2 m(log(/m)) 2 + m A 2a,2b m + (log y) 2 m A 2a,2b /2 m /y m A 2a,2b m /2 m(log(/m)) 2 (log ) + 2 (log y) = (log log )5/2 + 5/2 (log ) 2 (log ) 5/2 (log ) 2, (37) which together with estimate (3) leads to the desired conclusion of Theorem 2. m 4 Acknowledgments This work was done in April of 2006, while the second author was in residence at the Centre de recherches mathématiques of the Université de Montréal for the thematic year Analysis and Number Theory. This author thanks the organizers for the opportunity of participating in this program. He was also supported in part by grants SEP-CONACyT 46755, PAPIIT IN04505 and a Guggenheim Fellowship. The first author was supported in part by a grant from NSERC. 3

References [] E. R. Canfield, P. Erdős and C. Pomerance, On a problem of Oppenheim concerning Factorisatio Numerorum, J. Number Theory 7 (983), 28. [2] L. E. Dickson, A new etension of Dirichlet s theorem on prime numbers, Messenger of Math. 33 (904), 55 6. [3] H. Halberstam and H.-E. Richert, Sieve Methods, Academic Press, London, 974. [4] G. H. Hardy and J. E. Littlewood, Some problems on partitio numerorum III. On the epression of a number as a sum of primes, Acta Math. 44 (923), 70. [5] A. Hildebrand, On the number of positive integers and free of prime factors > y, J. Number Theory 22 (986), 289 307. [6] A. Hildebrand and G. Tenenbaum, Integers without large prime factors, J. de Théorie des Nombres de Bordeau 5 (983), 4 484. [7] A. Ivić, The Riemann Zeta-Function, Theory and Applications, Dover Publications, 2003. [8] A. Schinzel and W. Sierpiński, Sur certaines hypothèses concernant les nombres premiers, Acta Arith. 4 (958), 85 208. Erratum, Acta Arith. 5 (959), 259. [9] G. Tenenbaum, Introduction to analytic and probabilistic number theory, Cambridge Univ. Press, 995. 2000 Mathematics Subject Classification: Primary A25; Secondary N37. Keywords: Euler function, sum of divisors. (Concerned with sequences A03363 and A37733.) Received June 26 2007; revised version received February 9 2008. Published in Journal of Integer Sequences, February 9 2008. Return to Journal of Integer Sequences home page. 4