HARMONIOUS PAIRS MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE

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1 HARMONIOUS PAIRS MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE Dedicated to the memory of Felice Bateman, Paul Bateman, and Heini Halberstam ABSTRACT. Let σ be the usual sum-of-divisors function. We say that a and b form a harmonious pair if a σ(a) = ; equivalently, the harmonic mean of is 2. For example, 4 and 2 form a σ(a) + b σ(b) a and σ(b) b 4 harmonious pair, since = 4 and 2 = 3. Every amicable pair is harmonious, but there are many σ(4) 7 σ(2) 7 others. We show that the count of numbers that belong to a harmonious pair having both members in [, x] is at most x/ exp((log x) 2 +o() ), as x.. INTRODUCTION Let σ(n) denote the sum of the divisors of the natural number n. Recall that m and n are said to form an amicable pair if σ(m) = σ(n) = m + n. The study of amicable pairs dates back to antiquity, with the smallest such pair 220 and 284 known already to Pythagoras. While amicable pairs have been of interest for 2500 years, many of the most natural questions remain unsolved. For example, although we know about 2 million amicable pairs, we have no proof that there are infinitely many. In the opposite direction, there has been some success in showing that amicable pairs are not so numerous. In 955, Erdős showed that the set of n belonging to an amicable pair has asymptotic density zero [4]. This result has been subject to steady improvement over the past 60 years [, 5, 8, 9, 0]. We now know that the count of numbers not exceeding x that belong to an amicable pair is smaller than (.) x/ exp((log x) /2 ) for all large x. If m and n form an amicable pair, then σ(m) = σ(n) = m + n. From this, one sees immediately that =. In this paper, we study solutions to this latter equation. m σ(m) + n σ(n) Definition. We say a and b form a harmonious pair if a σ(a) + b σ(b) =. The terminology here stems from the following simple observation: a and b form a harmonious pair precisely when σ(a)/a and σ(b)/b have harmonic mean 2. While every amicable pair is harmonious, there are many examples not of this kind, for instance 2 and 20, or 3 and Our main theorem is an upper bound on the count of numbers belonging to a harmonious pair. While harmonious pairs certainly appear to be more thick on the ground than amicable pairs, we are able to get an upper estimate of the same general shape as (.). Theorem. Let ɛ > 0. The number of integers belonging to a harmonious pair a, b with max{a, b} x is at most x/ exp((log x) 2 ɛ ), for all x > x 0 (ɛ). As a corollary of Theorem, the reciprocal sum of those integers that are the larger member of a harmonious pair is convergent. Note that Theorem does not give a reasonable bound on the number of harmonious pairs lying in [, x]. Date: October 9, Mathematics Subject Classification. Primary A25, Secondary N37.

2 2 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE We are not aware of any previous work on harmonious pairs, as such. However, the following result can be read out of a paper of Borho [2]: If a, b form a harmonious pair and Ω(ab) = K, then ab K 2K. Borho states this for amicable pairs, but only the harmonious property of the pair is used in the proof. We also discuss discordant numbers, being those numbers that are not a member of a harmonious pair. We show there are infinitely many discordant numbers, in fact, more than x/(log x) ɛ of them in [, x], when ɛ > 0 is fixed and x is sufficiently large. Probably a positive proportion of numbers are discordant, but we have not been able to prove this. A weaker assertion that seems to escape us: it is not the case that the numbers that belong to some harmonious pair form a set of asymptotic density. At the end of the paper we use harmonious pairs to disprove a conjecture in [7]. Notation. Throughout this paper, we use the Bachmann Landau symbols O and o as well as the Vinogradov symbols and with their regular meanings. Recall that A = O(B) and A B are both equivalent to the fact that the inequality A < cb holds with some constant c > 0. Further, A B is equivalent to B A, while A = o(b) means that A/B 0. We write log k x for the iteration of the natural log function, with the undertanding that x will be big enough to have log k x. We let P + (n) denote the largest prime factor of n, with the convention that P + () =. We write s(n) for the sum of the proper divisors of n, so that s(n) = σ(n) n. If p is prime, we write v p (n) for the exponent of p appearing in the prime factorization of n. We let τ(n) denote the number of positive divisors of n and let ω(n) denote the number of these divisors which are prime. 2. PROOF OF THEOREM The following proposition, whose proof constitutes the main part of the argument, establishes half of Theorem. This proof largely follows the plan in [9, 0], though here we have more cases. Proposition 2. The number of integers b x that are members of a harmonious pair a, b with max{a, b} x and P + (b) P + (a) is x/ exp((log x) 2 ) for all x 3. Proof. We may assume that x > x 0 where x 0 is some large, absolute constant. For α in (0, ) and x 3 we put L α = exp((log x) α ). We aim to bound the count of b-values by O(x/L α ) with some fixed α (0, ), whose size we will detect from our arguments. We will pile various conditions on b and keep track of the counting function of those b x failing the given conditions.. We eliminate numbers b x having a square full divisor d > 2 L2 α. The counting functions of those is bounded above by x d x, L α d L 2 α /2 d squarefull where the above estimate follows from the Abel summation formula using the fact that the counting function of the number of square full numbers m t is O(t /2 ). 2. We eliminate numbers b x for which P + (b) L α. Putting y = L α, we have u := log x/ log y = (log x) α. By known estimates from the theory of smooth numbers (e.g., [3]), we have that the number of integers b x with P + (b) x /u is x (2.) as x, exp(( + o())u log u) when u > (log x) ɛ, so certainly the above count is < x/l α once x is sufficiently large.

3 HARMONIOUS PAIRS 3 3. We assume that α < /2. We eliminate numbers b x having a divisor d > L 2α with P + (d) L 2 α. Put y = L 2 α. For each t L 2α, we have u = log t/ log y 0.5(log x) α. Thus, u log u (log x) α log 2 x, and in particular u log u > 3(log x) α for x > x 0. So the number of such d t is at most t/l 2 α, uniformly for t [L 2α, x], assuming x > x 0. Fixing d, the number of b x divisible by d is x/d. Using the Abel summation formula to sum the reciprocals of such d, we get that the number of such b x is bounded by L 2α <d x P + (d) L 2 α x d x L 2 α x L 2α dt t x log x L 2 α x L α. 4. We eliminate numbers b x having a prime factor p > L 2 α such that p gcd(b, σ(b)). Let us take a closer look at such numbers. Suppose that p > L 2 α and p σ(b). Then there is a prime power q l dividing b such that p σ(q l ). If l 2, then 2q l > σ(q l ) p > L 2 α, so q l > L 2 α/2 and q l b with l 2, but such b s have been eliminated at. So, l =, therefore q (mod p). Thus, b is divisible by pq for some prime q (mod p). The number of such numbers up to x is at most x pq. Summing up the above bound over all primes q x with q (mod p) while keeping p fixed, then over all primes p (L 2 α, x] gives us a count of L 2 α<p x q (mod p) q x x pq x(log 2 x) L 2 α<p x p 2 x L 2 α x L α. 5. We eliminate the numbers b x/l α, since obviously there are only at most x/l α such values of b. Let d = gcd(b, σ(b)). Then P + (d) L 2 α by 4, so by 3 we have d L 2α. Write b = P m, where P = P + (b). By 2, we can assume that P > L α. 6. We eliminate b x corresponding to some a x/l 2 2α. Indeed, let b have a corresponding a with the above property. With c = gcd(a, σ(a)), we have an equality of reduced fractions b/d σ(b)/d (σ(a) a)/c =. σ(a)/c Notice that c is determined uniquely in terms of a. Thus, b/d = (σ(a) a)/c is also determined by a. Since d L 2α, the number b is determined in at most L 2α ways from a. So the number of b corresponding to some a x/l 2 2α is at most x/l 2α < x/l α. 7. Similar to 6, we eliminate a bounded number of subsets of b x which have some corresponding a x with a counting function of size O(x/L 2 2α ). In particular, by an argument similar to the one at, we may assume that a has no divisor which is squarefull and larger than L 4 2α /2. In particular, if p2 a, then p < L 2 2α. We may further assume that P + (a) > L 2α by an argument similar to the one at 2, and that if d is the largest divisor of a such that P + (d ) L 4 2α, then d L 4α, again by an argument similar to the one used at 3. Assuming α 6, we then have P + (a) > L 2α L 4α d. Further, P + (a) > L 2α > L 2 2α. Hence, P + (a) 2 a. Thus, a = Q n, where Q = P + (a) and Q n.

4 4 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE 8. Recall that c = gcd(a, σ(a)). By an argument similar to 4, we may eliminate numbers b x with some corresponding a having the property that there exists a prime factor p c such that p > L 4 2α. Indeed, in this case p a. Further, p σ(a) so there is a prime power q l dividing a such that p σ(q l ). If l 2, then 2q l > σ(q l ) p > L 4 2α, contradicting 7. So, l =, q (mod p), and pq divides a, so the number of such a x is at most x/pq. Summing up the above bound over all primes q (mod p) with q x, then over all primes p (L 4 2α, x], we get a count on the number of such a of x pq x(log 2 x) x, L 4 2α <p x q (mod p) q x and we are in a situation described at the beginning of 7. L 4 2α <p x p 2 By 8, we have that if p c, then p L 4 2α. So from 7, c d L 4α. 9. We eliminate b x for which P + (P + ) L 2α. Assume that b satisfies this condition. Then P + x/m + 2x/m is a number having P + (P + ) y = L 2α, and P + > P > L α, by 2. Thus, u := log(2x/m )/ log y (log x) α, so that u log u > 3(log x) α for x > x 0. Fixing m, the number of such P (even ignoring the fact that they are prime) is, again by (2.), at most x L 2 αm. Summing over all m x, we get at most O(x(log x)/l 2 α) = O(x/L α ) such b. 0. We may eliminate those b x corresponding to an a with P + (Q + ) L 4α. Indeed, assume that b satisfies the above property. Then Q + x/n + 2x/n. Further, Q + > Q > L 2α by 7 and P + (Q + ) L 4α, so that with y = L 4α, we have u := log(2x/n )/ log y > (log x) 2α. This shows that u log u > 4(log x) 2α for x > x 0. Thus, the number of possible numbers of the form Q + (even ignoring the fact that Q is prime), is, again by (2.), at most x L 3 2α n. Summing up the above bound for n x, we see there are at most O(x(log x)/l 3 2α ) possible a. So we are in the situation described at the beginning of 7.. Reducing the left and right-hand sides of the equation a Hence, σ(a) = σ(b) b σ(b) (2.2) Q n = a = (c/d)(σ(b) b) = (c/d)(p s(m ) + σ(m )), L 2 2α gives that a/c = (σ(b) b)/d. and so Q n d = c(p s(m ) + σ(m )). Since c L 4α and Q > L 2α, it follows that Q c. (Recall our assumption that α 6.) Hence, gcd(q, c) =, and c n d. Thus, P s(m ) + σ(m ) = Q λ, where λ = n d/c. Further, since d L 2α < Q and Q n, it follows that Q λ. We now break symmetry and make crucial use of our assumption that P Q. We claim that P a. Assume for a contradiction that P a. Recalling (2.2), and using the fact that P > L α > max{c, d}, we get that P σ(b) b, therefore P σ(b), so P d, which is false. Let R = P + (P + ). We note that R a. Indeed, the argument is similar to the argument that P a. To see the details, assume that R a. Since R > L 2α max{c, d}, it follows from (2.2) that R σ(b) b. But R P + σ(b), therefore R b. Thus, R d, which is impossible since R > d.

5 HARMONIOUS PAIRS 5 Now R σ(b)/d = σ(a)/c. Thus, there is some prime power Q l 2 dividing a such that R σ(q l 2 ). Hence, L 2α < R σ(q l 2 ) < 2Ql 2. If l 2, we then get L 2α < 2Q l 2 L4 2α by 7, which is false for x > x 0. Thus, l =, and we have that R Q 2 +. In particular, Q 2 > R > L 2α. Since Q 2 Q (the case Q 2 = Q is possible), it follows that Q 2 P. We write a = Q 2 n 2 and going back to relation (2.2), we get σ(b) b = Q 2 n 2 d c. Note that Q 2 > L 2α max{c, d}, so indeed c n 2 d. Write λ 2 = n 2 d/c. We then have (2.3) P s(m ) + σ(m ) = σ(b) b = Q 2 λ 2. Note that Q 2 s(m ), for if not, then we would also get that Q 2 σ(m ). Thus, Q 2 m b and Q 2 σ(m ) σ(b), therefore Q 2 d, which is false since Q 2 > d. Reduce now equation (2.3) with respect to R, using P Q 2 (mod R ), to get that This shows that m + λ 2 0 (mod R ). (2.4) either m R /2 or λ 2 R /2. So the situation splits into two cases. We treat an instance a bit stronger then the first case above throughout steps 2 5, and the second situation in the subsequent steps We first assume that (2.5) m > L /4 6α. Note that the left inequality (2.4) implies (2.5) for x > x 0. (The negation of the weak inequality (2.5) will be useful in 7.) 2. We eliminate the numbers b x for which P + (m ) L 7α. Fix P and count the number of corresponding m (L /4 6α, x/p ]. If there are any such m, then with y = L 7α, we have u := log(x/p )/ log y 0.25(log x) α. Hence, u log u > 3(log x) α for x > x 0. By (2.), the number of these m is at most x/(l 2 αp ) for x > x 0. Summing over all primes P x, we get an upper bound of O(x(log 2 x)/l 2 α) = O(x/L α ) on the number of such b. 3. Let P 2 = P + (m ) and put m = P 2 m 2. Note that P 2 x/(p m 2 ) and P 2 > L 7α. Clearly, if α, then P 2 does not divide cd for large x because P 2 > L 7α max{c, d}. Also, since L 7α > L 2 α/2 for x > x 0, it follows that P 2 b. Thus, P 2 + σ(b). 4. We eliminate b x such that P + (P 2 + ) L 8α. Since P 2 + > L 7α, for fixed P, m 2, by arguments similar to the preceding ones, we get that the number of such P 2 is at most x/(l 2 αp m 2 ). Summing up the above inequality over all the primes P x and all positive integers m 2 x, we get a bound of O(x(log x)(log 2 x)/l 2 α) = O(x/L α ) on the number of such b. 5. Now we put R 2 = P + (P 2 + ). Then R 2 > d if α < 0, because R 2 > L 8α. Thus, R 2 σ(b)/d = σ(a)/c, therefore there exists Q l 3 dividing a such that R 2 σ(q l 3 ). Thus, 2Ql 3 > σ(ql 3 ) R 2. Since α < 0, we have R 2 > L 8α > L 4 2α for x > x 0, so, by 7, we get that l =. Thus, Q 3 a and Q 3 (mod R 2 ) (the case Q 3 = Q is possible). Now assume α 2. Then Q 3 > R 2 > L 8α c. Since a = (σ(b) b)c/d, it follows that Q 3 σ(b) b. Hence, P s(m ) + σ(m ) 0 (mod Q 3 ). Since Q 3 > L 8α, arguments similar to previous ones show that Q 3 s(m ). This puts P x/m in an arithmetic progression modulo Q 3. Since Q 3 Q P, it follows that x/m Q 3, so that the

6 6 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE number of such integers P (even ignoring the fact that P is prime) is O(x/(m Q 3 )). But m = P 2 m 2, where R 2 gcd(p 2 +, Q 3 + ). Fixing R 2, P 2, Q 3 and summing up over m 2 x, we get a count of O(x(log x)/(p 2 Q 3 )). Now we sum up over primes P 2 and Q 3 both at most x and both congruent to (mod R 2 ) getting a count of O(x(log x)(log 2 x) 2 /R 2 2 ). We finally sum over primes R 2 (L 8α, x] getting a bound of O(x(log x)(log 2 x) 2 /L 8α ) = O(x/L α ) on the number of such b. The steps 2 5 apply when m satisfies (2.5). Now assume that m fails (2.5). In this case, m < R /2 (assuming x > x 0 ). So by (2.4), we must have λ 2 R /2 > 0.5L 2α. Note that λ 2 = n 2 (d/c); therefore n 2 = λ 2 (c/d) > 0.5L 2α /d > L 0.5 2α for x > x 0, since d L 2α. 6. We eliminate b x such that their corresponding a has the property that P + (n 2 ) L 4α. Put y = L 4α. We fix Q 2 and count n 2 x/q 2, with n 2 > L 2α 0.5, such that P + (n 2 ) y. Since u := log(x/q 2 )/ log y 0.5(log x) 2α, it follows that u log u > 4(log x) 2α for x > x 0. Thus, for large x the number of corresponding a x is at most x/(l 3 2α Q 2). Summing up the above bound over primes Q 2 x, we get a count of order x(log 2 x)/l 3 2α. This is smaller than x/l2 2α for x > x 0, and so we are fine by 7. Now we put n 2 = Q 3 n 3, where Q 3 = P + (n 2 ) > L 4α. 7. We eliminate b x corresponding to a such that P + (Q 3 + ) L 6α. Fix Q 2 and n 3. Then Q 3 x/(q 2 n 3 ) and Q 3 > L 4α. Assuming that P + (Q 3 + ) L 6α, we get by previous arguments involving (2.) that the count of such Q 3 is smaller than x/(l 3 2α Q 2n 3 ), once x > x 0. Summing up the above bound over primes Q 2 x and all positive integers n 3 x, we get an upper bound of order x(log x)(log 2 x)/l 3 2α on the number of these a. Again, we are fine by 7. Write R 2 = P + (Q 3 + ). Then R 2 > L 6α. Assume now that α < 0. Then R 2 > max{c, d} and R 2 > L 4 2α for x > x 0. Since Q 3 > L 4α > L 2 2α (for x > x 0), 7 gives that Q 3 a. Thus, R 2 Q 3 + σ(a). Since R 2 does not divide c, we get that R 2 divides σ(a)/c = σ(b)/d. Hence, R 2 σ(b). Since b has no squarefull divisors exceeding L 2 α/2, there is a prime P 2 b such that R 2 P 2 +. In fact, we can take P 2 = P, in other words, R 2 P +. Suppose otherwise. Then R 2 since m fails (2.5), σ(m ) m 2 (L 6α) /2 < R 2. Hence, R P + and R 2 P +. σ(b) P + = σ(m ). However, 8. Consider the case when R = R 2. Then a = Q 2 Q 3 n 3 and both Q 2 and Q 3 are congruent to modulo R. Fixing Q 2, Q 3, the number of such a is at most x/q 2 Q 3. Summing this bound over all pairs of distinct primes Q 2, Q 3 up to x and congruent to modulo R, we get a bound of O(x(log 2 x) 2 /R 2 ). Now summing over all primes R (L 2α, x], we get a count that is < x/l 2α < x/l 2 2α for x > x 0, and we are fine by 7. From now on, we assume that R R 2, so that P (mod R R 2 ). 9. We eliminate numbers b x such that either m Q 2 R x or m Q 3 R 2 x. Suppose we are in the first case. Then P (mod R ), and P s(m ) + σ(m ) 0 (mod Q 2 ). Since Q 2 s(m ), this puts P into an arithmetic progression modulo Q 2. By the Chinese remainder theorem, P x/m is in an arithmetic progression modulo Q 2 R, and the number of such numbers (ignoring the condition that P is prime) is at most + x/(m Q 2 R ) 2x/(m Q 2 R ). Here is where we use the condition that m Q 2 R x. We keep R fixed and sum over all m x, and primes Q 2 (mod R ), getting a count of O(x(log x)(log 2 x)/r 2). Then we sum over all primes R (L 2α, x], getting a count of O(x(log x)(log 2 x)/l 2α ). This count of b values is < x/l α once x > x 0. The same applies when m Q 3 R 2 x. There, P (mod R 2 ) and the congruence P s(m ) + σ(m ) 0 (mod Q 3 ) together with the fact that Q 3 does not divide s(m ) puts P x/m in an

7 HARMONIOUS PAIRS 7 arithmetic progression modulo Q 3. By the Chinese remainder theorem, P x/m is in an arithmetic progression modulo Q 3 R 2, and the number of such possibilities (ignoring the fact that P is prime) is at most + x/(m Q 3 R 2 ) 2x/(m Q 3 R 2 ). Here we used that m Q 3 R 2 x. Summing up the above bound over all m x and primes Q 3 x in the arithmetic progression (mod R 2 ), we get a count of O(x(log x)(log 2 x)/r 2 2 ). Summing up the above bound over all R 2 > L 6α, we get a count of O(x(log x)(log 2 x)/l 6α ). So the number of these b is smaller than x/l α for x > x We now look at the instance m Q 2 R > x and m Q 3 R 2 > x. We will show that this set is empty for x > x 0. Indeed, write Q 2 = R l, Q 3 = R 2 l 2 for some even integers l, l 2 > 0. The inequalities m Q 2 R > x and m Q 3 R 2 > x yield m Q2 Q 3 R R 2 > x. Since R R 2 = (Q 2 + )(Q 3 + )/(l l 2 ) Q 2 Q 3 /l l 2, we get (2.6) m Q 2 Q 3 l l 2 x. Now P s(m ) + σ(m ) = σ(b) b = a(d/c) = Q 2 Q 3 (n 3 d/c). Since min{q 2, Q 3 } > max{d, c}, we have c n 3 d. Put l 3 = n 3 d/c. Then Q 2 Q 3 l 3 = σ(b) b < σ(b) x log 2 x. Thus, Q 2 Q 3 x(log 2 x)/l 3. Hence, using (2.6), In particular, xm log 2 x l 3 l l 2 m Q Q 2 l l 2 x giving l 3 l l 2 m log 2 x. (2.7) l m 2 (log 2 x) 2, l 2 m 2 (log 2 x) 2, l 3 m log 2 x. Write P = R R 2 l. We then have which is equivalent to (R R 2 l )s(m ) + σ(m ) = (R l )(R 2 l 2 )l 3, (2.8) (ls(m ) l l 2 l 3 )R R 2 + m = R l l 3 R 2 l 2 l 3 + l 3. Moving m to the other side, dividing by R R 2 and using (2.7), we get ( m ls(m ) l l 2 l 3 = O + l l 3 + l ) ( 2l 3 m 3 = O (log 2 x) 3 ) = o() (x ), R R 2 R 2 R L 6α where the last estimate above comes from the fact that m fails (2.5). Since the left hand side above is an integer, it must be 0 for x > x 0. Returning to (2.8), we get m = R l l 3 R 2 l 2 l 3 + l 3 < 0, a contradiction. Hence, this case cannot occur once x > x 0. Denouement. Glancing back through the argument, we find that every step can be carried out with α = 2. Hence, the total count of b-values is O(x/ exp((log x) /2 )). To count values of a paired with b having P + (a) P + (b), we use a method introduced by Wirsing [6]. Wirsing showed that the number of solutions n x to an equation of the form σ(n)/n = β is at most exp(o(log x/ log log x)), uniformly in β. The next lemma provides a sharper bound if the number of primes dividing n is not too large. Lemma 3. Let k be a positive integer and let x 0 5. Let β be a rational number. The number of integers n x with ω(n) k and σ(n) n = β is at most (2 log x)3k.

8 8 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE Proof. Write β = λ/µ, where the right-hand fraction is in lowest terms. If σ(n)/n = λ/µ, then µ n. So we may assume that ω(µ) k. Given an n with σ(n)/n = β, put P = {p 2k} {p µ}, and write n = AB, where A = p vp(n), B = p vp(n). Note that µ B. The main idea of the proof is to show that B nearly determines its cofactor A. Specifically, we will show that for any given B, the number of corresponding A is at most (log x) k. Since gcd(a, B) =, we have p P (2.9) σ(a)σ(b) = λ µ AB. Moreover, As a consequence, (2.0) A/σ(A) > p A ( /p) p A p P p k 2k + > 2. ( ) λ 2 µ B < σ(b) = A ( ) λ σ(a) µ B λ µ B. Thus, σ(b) λ µ B unless the final inequality is an equality, which occurs only if A = and σ(b) = λ µ B. Suppose that σ(b) λ µ B. Then there is a prime dividing σ(b) to a higher power than λ µ B. Let p be the least such prime and observe that p is entirely determined by β and B. By (2.9), p A. Suppose that p e A. Set A = A/p e and B = Bp e. Then (2.9) (2.0) hold with A and B replaced by A and B, respectively. From the analogue of (2.0), we find that if σ(b ) λ µ B, then A =, so that A = p e. Suppose that σ(b ) λ µ B. There is a prime dividing σ(b ) to a higher power than it divides λ µ B. Let p 2 be the smallest such prime. Then p 2 is entirely determined by β, B, and e, and p 2 A. If p e 2 2 A, we set A 2 = A /p e 2 2 and B 2 = B p e 2. If σ(b 2) λ µ B 2, then A = p e 2 pe 2 2. If not, there is a prime dividing B 2 to a higher power than λ µ B 2, which allows us to continue the argument. We carry out this process until A r =, which happens in r k steps. Then A = p e per r. Here each prime p i+ is entirely determined by β, B, and e,..., e i. Thus, A is entirely determined by β, B, and the exponent sequence e,..., e r. Clearly, 3 e i (2k + ) e i r i= p e i i = A x, and so each e i [, log x/ log 3]. Extend e,..., e r to a sequence e,..., e k by putting e i = 0 for r < i k. Since each e i [0, log x log 3 ], the number of possibilities for e,..., e r is at most ( + log x log 3 )k (log x) k, using in the last step that x 0 5. To bound the number of possibilities for n = AB, it now suffices to estimate the number of possibilities for B. We have B = p P pfp, where each f p [0, log x log 2 ]. Thus, B belongs to a set of size at most ( + log x ) #P (2 log x) #P (2 log x) π(2k)+ω(µ) (2 log x) k+k = (2 log x) 2k. log 2 Putting everything together gives a final upper bound of (log x) k (2 log x) 2k (2 log x) 3k. Proof of Theorem. In view of Proposition 2, it is enough to estimate the number of a involved in a pair a, b with max{a, b} x and P + (b) P + (a). With K to be specified shortly, we partition these a according to whether or not ω(a) K. Since σ(a) a is determined by b, Lemma 3 and Proposition 2 show that the number of a with ω(a) K is (2 log x) 3K x/ exp((log x) /2 ).

9 HARMONIOUS PAIRS 9 On the other hand, since τ(a) 2 ω(a), we have trivially that the number of a with ω(a) > K is x log x 2 K. Adding these two estimates and taking K = (log x) /2 (log log x) 2 finishes the proof. Remark. We have shown that there are not many integers which are the member of some harmonious pair contained in [, x]. It would be interesting to show that there are not too many such harmonious pairs. Note that the upper bound x exp(o(log x/ log log x)) follows trivially from Wirsing s theorem. One cannot immediately derive a sharper estimate from Theorem, since a single integer may be shared among many pairs. However, Theorem and Lemma 3 imply (arguing similarly to the proof just given) that the number of pairs with max{a, b} x and min{ω(a), ω(b)} (log x) 2 δ is at most x/ exp((log x) 2 +o() ), for any fixed δ > DISCORDANT NUMBERS Given a number a, is there a number b for which the pair a, b is harmonious? If not, we say that a is discordant. Since a and b form a harmonious pair exactly when σ(b)/b = σ(a)/s(a), deciding whether a is discordant amounts to solving a special case of the following problem: Problem (Recognition problem for σ(n)/n). Decide whether a given rational number belongs to the image of the function σ(n)/n. Rational numbers not in the range of σ(n)/n have been termed abundancy outlaws. In the early 970s, C.W. Anderson [] conjectured that the set {σ(n)/n} is recursive: In other words, an algorithm exists for deciding whether or not a given rational number is an outlaw. This conjecture is still open, but some partial results can be found in [8]. See also [5, 2, 6, 4]. Difficulties arise when trying to decide discordance even for small values of a. The smallest number whose status is unresolved seems to be a = ; to answer this, we would need to know whether or not 2 is an abundancy outlaw. Anderson noted that σ(b) b = 5 3 if and only if 5b is an odd perfect number with 5 b. Since σ(24) s(24) = 5 3, it follows that a = 24 is a member of a harmonious pair if and only if there is an odd perfect number precisely divisible by 5. It is perhaps not immediately clear that there are infinitely many discordant numbers. Here we prove the following modest lower bound. Proposition 4. The number of discordant integers n x is at least x/(log x) (e γ +o())/ log 3 x as x. The following simple lemma can be found in [] and [5]. Lemma 5. Suppose v and u are coprime positive integers. If v < σ(u), then v/u is an abundancy outlaw. Proof. Suppose v u = σ(n) n. Then u n, so that v u = σ(n) n Lemma 5 implies the following criterion for discordance. σ(u) u. Hence, v σ(u). Lemma 6. If n, u, v are positive integers with s(n)/σ(n) = u/v, gcd(u, v) =, and n (3.) σ(n) + u σ(u) <, then n is discordant. u Proof. Our assumptions imply that σ(u) < n σ(n) σ(n)/s(n) is an outlaw; hence, n is discordant. We now are ready to prove Proposition 4. = s(n) σ(n) = u v, so that v < σ(u). From Lemma 5,

10 0 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE Proof. Let ɛ > 0 be arbitrary but fixed and let π be the largest prime number smaller than e γ ɛ log 3 x. Thus, for large enough x we have π 5. Let B = (log 2 x)/(log 3 x) 2 and let A 0 denote the least common multiple of integers in [, B] coprime to π. Further, let A be the product of A 0 and all primes r with the property that π σ(r vr(a 0) ). That is, if r is a prime and r α A 0 with π σ(r α ), we multiply by r. We have π Aσ(A), A = exp(( + o())b), σ(a)/a = (e γ + o()) log 3 x, as x. Let k run over integers to x /4 such that A k and π kσ(k). We would like a lower bound for /k. For this, we restrict our attention to numbers of the form Aj, where j x /5 is squarefree with no prime factors below B and no prime factors in the residue class (mod π). Let i 0 = 3 log log(x /5 ) and let S denote the set of primes r in (B, x /i 0 ] with r (mod π). For an integer i i 0, the sum S i of reciprocals of squarefree numbers j x /5 composed solely of primes in S satisfies S i i! ( r ) i (i 2)! r 2 ( r ) i 2 > (i 2)! ( r ) i 2 By the prime number theorem for residue classes (cf. [8, Theorem ]), ( (3.2) r = ) log log x log log B + O(). π k i 2 ( ) 2 r. Thus, since /r2 /B, this sum is small compared with (/i 2 )( /r)2, so that ( ) i ( ) i k A j S i = A A i! r A e r = T T 2, A i! r j i i 0 i i 0 i>i 0 say. By (3.2), T (log x) /(π ) /(A log B). Also note that by (3.2), T 2 ( ) i ( ) i0 ( ) i0 e = o() i! r i i>i 0! r i 0 r 0 as x. Thus, k A log B (log x) (+o())/π = (log x) (+o())/π, x. k Next, for each k chosen, let q run over primes to x /2 /k, with q k, π q(q + ), and π qs(k) + σ(k) = s(qk). To arrange for this last condition, note that if π s(k), it is true automatically, and if π s(k), then there are at least π 3 allowable residue classes for q modulo π (we discard the classes 0,, σ(k)/s(k)). For m = qk so chosen, we have A m, π ms(m)σ(m), and by the prime number theorem for residue classes, m = m k k q q k k (log x) (+o())/π, x. Finally, for each m, we let p run over primes to x/m where p m and (3.3) π ps(m) + σ(m) = s(pm). For (3.3), we take p in the residue class σ(m)/s(m) modulo π. Since π σ(m)s(m), this is a nonzero residue class. Further, it is not the class since π m implies that σ(m) s(m) (mod π). Thus, if we r 2

11 HARMONIOUS PAIRS choose p satisfying (3.3), then π p +. For n = pm so chosen we have by the prime number theorem for residue classes (3.4) = x/m π log(x/m) x π log x m x (log x) n x m p m m (+o())/π as x. It remains to note that for each number n constructed we have n x, A n, and if s(n)/σ(n) = u/v with u, v coprime, then π u. Thus, as x, n σ(n) + u σ(u) A σ(a) + π π + (e γ + o()) log 3 x + e γ ɛ log 3 x +, and this expression is smaller than for all large x. Thus, by (3.), n is discordant. Since ɛ > 0 is arbitrary, (3.4) implies the proposition. The criterion (3.) is sufficient for discordance but not necessary; there are abundancy outlaws of the form σ(a)/s(a) not captured by Lemma 5. In order to detect (some) of these, we combine Lemma 5 with a bootstrapping procedure described in the following result. Lemma 7 (Recursive criterion for outlaws). Let v and u be coprime positive integers. Let P be the product of any finite set of primes p for which pvp(u) v σ(up ) σ(p vp(u) ) u is known to be an outlaw. If up > v u, then v u is an outlaw. Proof. If σ(n)/n = v/u, then u n. Let p be a prime dividing P. If p vp(u) n, then σ(n/p vp(u) ) n/p vp(u) = pvp(u) σ(p vp(u) ) v u, contradicting that the right-hand side is an outlaw. Thus, p vp(u)+ n for all p dividing P, and so up n. Hence, v u = σ(n), contrary to assumption. n σ(up ) up Example. As an illustration, let us show that 888 is discordant. We have σ(888) 2 σ(2) = σ(2 58) 2 58 = > Since σ(87) = 20 > 95, the fractional So Lemma 7, with P = p = 2, implies that s(888) = Then 2 58, and is a known outlaw by Lemma 5. Moreover, is an outlaw. Table displays the counts of numbers up to 2 k, for k =, 2,..., 26, that belong to a harmonious pair with other member at most 0 j, where j = 8, 9,..., 8. To collect this data, we modified a gp script posted by Michael Marcus to the Online Encyclopedia of Integer Sequences (see [3]). Given a rational number v u and a search limit L, the script (provably) finds all b L with σ(b) b = v u. For each < a 2k, we used this script to determine whether or not the equation σ(b) b = σ(a) s(a) has any solutions b 0j. Table 2 summarizes three counts: Numbers known to be harmonious because they are members of a pair contained in [, 0 8 ], numbers proved to be discordant, and numbers which fall into neither camp. The count of discordant numbers was obtained by tallying those a > for which σ(a)/s(a) could be determined to be an abundancy outlaw by at most five iterations of Lemma 7. (Here the 0th iteration corresponds to outlaws detected by Lemma 5.) We then added to the counts, since is discordant but not detected in this fashion. It would be interesting to prove (or disprove) that the set of discordant integers has positive lower density. It seems possible that we could detect further classes of discordant numbers by developing some of the ideas introduced in [4] for finding abundancy outlaws; this deserves further study. In the opposite direction, we do not know how to show that there are infinitely many non-discordant integers, i.e., that there are infinitely many harmonious pairs.

12 2 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE TABLE. Number of positive integers up to 2 k belonging to a harmonious pair with other member at most 0 j Harmonious Discordant Not classified Harmonious Discordant Not classified TABLE 2. Counts up to various heights of numbers belonging to a harmonious pair in [, 0 8 ], numbers known to be discordant, and numbers fitting neither classification. 4. CONCLUDING REMARKS Harmonious pairs have a surprising connection with a different generalization of amicable pairs recently studied by two of us [7]. Say that m and n form a δ-amicable pair if σ(m) = σ(n) = m + n + δ. When δ = 0, this reduces to the usual notion of an amicable pair. It was shown in [7] that for each fixed δ 0, the set of numbers in [, x] belonging to a δ-amicable pair has size O δ (x(log 2 x) 4 /(log x) /2 ). The same authors conjectured that for arbitrary B, this count is (4.) δ,b x/(log x) B.

13 HARMONIOUS PAIRS H single (x) H pair (x) (x) H single (x) H pair (x) (x) TABLE 3. Values of H single (x) = # of n involved in a harmonious pair a b x, H pair (x) = # of pairs a b x, and (x) = # of values of δ = a + b x. The conjectured upper bound (4.) turns out to be too optimistic. To explain why, we first describe how to associate to a harmonious pair a, b a family of δ-amicable numbers with δ = a+b. Since a/σ(a)+b/σ(b) =, the fractions a/σ(a) and b/σ(b) have the same denominator in lowest terms, say d. Thus, we can write a/σ(a) = u/d and b/σ(b) = v/d, where both right-hand fractions are reduced and u + v = d. Write a = ua 0, b = vb 0, σ(a) = da 0, σ(b) = db 0. Put n = ap and m = bq, where p a, q b are primes. Then the equation σ(n) = σ(m) amounts to requiring σ(a)(p + ) = σ(b)(q + ), or equivalently, a 0 (p + ) = b 0 (q + ). This holds precisely when (4.2) p = b 0 (a 0, b 0 ) t and q = a 0 (a 0, b 0 ) t for some positive integer t. In that case, n + m + δ = ap + bq + δ = ua 0 ( ) b0 (a 0, b 0 ) t = a 0b 0 t (a 0, b 0 ) (u + v) = da 0b 0 t (a 0, b 0 ). ( ) a0 + vb 0 (a 0, b 0 ) t + (a + b) But this last fraction is equal to both σ(m) and σ(n), and thus m and n form a δ-amicable pair. We have constructed a pair of δ-amicable numbers from each pair of primes p, q satisfying (4.2), as long as p a and q b. One expects that there are always infinitely many such pairs. When b 0 = a 0, which corresponds to the case when a, b form an amicable pair, this follows immediately from the prime number theorem for arithmetic progressions. In that case, the above construction produces x/ log x members of a δ-amicable pair not exceeding x, which is much larger than allowed by (4.). If b 0 a 0, we cannot rigorously prove the existence of infinitely many prime pairs satisfying (4.2), but this follows from the prime k-tuples conjecture. Here we expect the construction to produce x/(log x) 2 numbers in [, x] that belong to a δ-amicable pair. Again, this contradicts the conjectured bound (4.). The following related questions seem attractive but difficult. Question. Does the bound (4.) hold if δ cannot be written as a + b for any harmonious pair a, b? Question. Let (x) be the number of δ x that can be written as a sum of two members of a harmonious pair. Can one show that (x) = o(x), as x? Of course this would follow if we could show that the count H pair (x) of harmonious pairs in [, x] is o(x). Perhaps (x) H pair (x) 2 H single(x), where H single (x) is the quantity bounded in Theorem. See Table 3.

14 4 MARK KOZEK, FLORIAN LUCA, PAUL POLLACK, AND CARL POMERANCE 5. ACKNOWLEDGEMENTS Research of F. L. on this project started during the meeting on Unlikely Intersections at CIRM, Luminy, France in February 204 and continued when he visited the Mathematics Department of Dartmouth College in Spring 204. F. L. thanks the organizers of the Luminy meeting for the invitation, and CIRM and the Mathematics Department of Dartmouth College for their hospitality. M. K. s participation took place while on sabbatical visit at UGA to work with P. P. with support from the UGA Mathematics Department, the UGA Office of the Vice President for Research, and Danny Krashen s NSF CAREER grant DMS P. P. is supported by NSF award DMS The computations reported on were performed with Magma and PARI/gp. We thank the Magma support staff for their detailed assistance, and we thank Matti Klock for her help in getting gp2c up and running. REFERENCES [] C. W. Anderson, The solutions of Σ(n) = σ(n)/n = a/b, Φ(n) = φ(n)/n = a/b, and related considerations, unpublished manuscript. [2] W. Borho, Eine Schranke für befreundete Zahlen mit gegebener Teileranzahl, Math. Nachr. 63 (974), [3] N. G. de Bruijn, On the number of positive integers x and free prime factors > y. II, Indag. Math. 28 (966), [4] P. Erdős, On amicable numbers, Publ. Math. Debrecen 4 (955), 08. [5] P. Erdős and G. J. Rieger, Ein Nachtrag über befreundete Zahlen, J. Reine Angew. Math. 273 (975), 220. [6] J. A. Holdener, Conditions equivalent to the existence of odd perfect numbers, Math. Mag. 79 (2006), [7] P. Pollack and C. Pomerance, On the distribution of some integers related to perfect and amicable numbers, Colloq. Math. 30 (203), [8] C. Pomerance, On the distribution of amicable numbers, J. Reine Angew. Math. 293/294 (977), [9], On the distribution of amicable numbers. II, J. Reine Angew. Math. 325 (98), [0], On amicable numbers (204), to appear in a Springer volume in honor of Helmut Maier. [] G. J. Rieger, Bemerkung zu einem Ergebnis von Erdős über befreundete Zahlen, J. Reine Angew. Math. 26 (973), [2] R. F. Ryan, Results concerning uniqueness for σ(x)/x = σ(p n q m )/(p n q m ) and related topics, Int. Math. J. 2 (2002), [3] N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences, Sequence A00970; gp script by Michael Marcus online at [4] W. G. Stanton and J. A. Holdener, Abundancy outlaws of the form σ(n)+t N 9 pages. [5] P. A. Weiner, The abundancy ratio, a measure of perfection, Math. Mag. 73 (2000), [6] E. Wirsing, Bemerkung zu der Arbeit über vollkommene Zahlen, Math. Ann. 37 (959), DEPARTMENT OF MATHEMATICS, WHITTIER COLLEGE, WHITTIER, CA , USA address: mkozek@whittier.edu, J. Integer Seq. 0 (2007), no. 9, Article , MATHEMATICAL INSTITUTE, UNAM JURIQUILLA, SANTIAGO DE QUERÉTARO, QUERÉTARO DE ARTEAGA, MÉXICO AND SCHOOL OF MATHEMATICS, UNIVERSITY OF THE WITWATERSRAND, P. O. BOX WITS 2050, SOUTH AFRICA address: fluca@matmor.unam.mx DEPARTMENT OF MATHEMATICS, UNIVERSITY OF GEORGIA, ATHENS, GA 30602, USA address: pollack@math.uga.edu DEPARTMENT OF MATHEMATICS, DARTMOUTH COLLEGE, HANOVER, NH , USA address: carl.pomerance@dartmouth.edu