JARED DUKER LICHTMAN AND CARL POMERANCE

Transcription

1 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS JARED DUKER LICHTMAN AND CARL POMERANCE Abstract. A subset of the integers larger than is rimitive if no member divides another. Erdős roved in 935 that the sum of /a log a for a running over a rimitive set A is universally bounded over all choices for A. In 988 he asked if this universal bound is attained for the set of rime numbers. In this aer we make some rogress on several fronts, and show a connection to certain rime number races such as the race between πx and lix.. Introduction A set of ositive integers > is called rimitive if no element divides any other for convenience, we exclude the singleton set {}. There are a number of interesting and sometimes unexected theorems about rimitive sets. After Besicovitch [4], we know that the uer asymtotic density of a rimitive set can be arbitrarily close to /2, whereas the lower asymtotic density is always 0. Using the fact that if a rimitive set has a finite recirocal sum, then the set of multiles of members of the set has an asymtotic density, Erdős gave an elementary roof that the set of nondeficient numbers i.e., σn/n 2, where σ is the sum-of-divisors function has an asymtotic density. Though the recirocal sum of a rimitive set can ossibly diverge, Erdős [9] showed that for a rimitive set A, a log a <. a A In fact, the roof shows that these sums are uniformly bounded as A varies over rimitive sets. Some years later in a 988 seminar in Limoges, Erdős suggested that in fact we always have. fa := a log a log, a A P where P is the set of rime numbers. The assertion. is now known as the Erdős conjecture for rimitive sets. In 99, Zhang [9] roved the Erdős conjecture for rimitive sets A with no member having more than 4 rime factors counted with multilicity. After Cohen [6], we have.2 C := P log = , Date: June 6, Mathematics Subject Classification. Primary B83; Secondary A05, N05. Key words and hrases. rimitive set, rimitive sequence, Mertens roduct formula.

2 2 JARED DUKER LICHTMAN AND CARL POMERANCE the sum over rimes in.. Using the original Erdős argument in [9], Erdős and Zhang showed that fa < for a rimitive set A, which was later imroved by Robin to These unublished estimates are reorted in Erdős Zhang [] who used another method to show that fa <.84. Shortly after, Clark [5] claimed that fa e γ = However, his brief argument aears to be incomlete. Our rincial results are the following. Theorem.. For any rimitive set A we have fa < e γ. Theorem.2. For any rimitive set A with no element divisible by 8, we have fa < C Say a rime is Erdős strong if for any rimitive set A with the roerty that each element of A has least rime factor, we have fa / log. We conjecture that every rime is Erdős strong. Note that the Erdős conjecture. would immediately follow, though it is not clear that the Erdős conjecture imlies our conjecture. Just roving our conjecture for the case of = 2 would give the inequality in Theorem.2 for all rimitive sets A. Currently the best we can do for a rimitive set A of even numbers is that fa < e γ /2, see Proosition 2. below. For art of the next result, we assume the Riemann hyothesis RH and the Linear Indeendence hyothesis LI, which asserts that the sequence of numbers γ n > 0 such that ζ 2 + iγ n = 0 is linearly indeendent over Q. Theorem.3. Unconditionally, all of the odd rimes among the first 0 8 rimes are Erdős strong. Assuming RH and LI, the Erdős strong rimes have relative lower logarithmic density > The roof deends strongly on a recent result of Lamzouri [3] who was interested in the Mertens race between / and /eγ log x. For a rimitive set A, let PA denote the suort of A, i.e., the set of rime numbers that divide some member of A. It is clear that the Erdős conjecture. is equivalent to the same assertion where the rime sum is over PA. Theorem.4. If A is a rimitive set with PA [3, ex0 6 ], then fa log. PA If some rimitive set A of odd numbers exists with fa > PA / log, Theorem.4 suggests that it will be very difficult indeed to give a concrete examle! For a ositive integer n, let Ωn denote the number of rime factors of n counted with multilicity. Let N k denote the set of integers n with Ωn = k. Zhang [20] roved a result that imlies fn k < fn for each k 2, so that the Erdős conjecture holds for the rimitive sets N k. More recently, Banks and Martin [2] conjectured that fn > fn 2 > fn 3 >. The inequality fn 2 > fn 3 was just established by Bayless, Kinlaw, and Klyve [3]. We rove the following result. Theorem.5. There is a ositive constant c such that fn k c for all k. We let the letters, q, r reresent rimes. In addition, we let n reresent the nth rime. For an integer a >, we let P a and a denote the largest and smallest

3 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS 3 rime factors of a. Modifying the notation introduced in [], for a rimitive set A let A = {a A : a }, A = {a A : a = }, A = {a/ : a A }. We let fa = /a log a and so fa = a A fa. In this language, Zhang s full result [20] states that fn k f for all rimes, k. We also, let ga = a <P a, ha = with ga = a A ga and ha = a A ha. 2. The Erdős aroach a log P a, In this section we will rove Theorem.. We begin with an argument insired by the original 935 aer of Erdős [9]. Proosition 2.. For any rimitive set A, if q / A then fa q < e γ gq = eγ. q Proof. For each a A q, let S a = {ba : b P a}. Note that S a has asymtotic density ga. Since A q is rimitive, we see that the sets S a are airwise disjoint. Further, the union of the sets S a is contained in the set of all natural numbers m with m = q, which has asymtotic density gq. Thus, the sum of densities for each S a is dominated by gq, that is, 2. ga q = ga gq. a A q <q By Theorem 7 in [7], we have for x 285, 2.2 > e γ log2x, which may be extended to all x by a calculation. Thus, since each a A q is comosite, ga = e γ > a a log 2P a e γ a log a = e γ fa. Hence by 2., <P a fa q/e γ < ga q gq. Remark 2.2. Let σ denote the sum-of-divisors function and let A be the set of n with σn/n 2 and σd/d < 2 for all roer divisors d of n, the set of rimitive nondeficient numbers. Then an aroriate analog of ga gives the density of nondeficient numbers, recently shown in [2] to lie in the tight interval , In [4], an analog of Proosition 2. is a key ingredient for shar bounds on the recirocal sum of the rimitive nondeficient numbers.

4 4 JARED DUKER LICHTMAN AND CARL POMERANCE Remark 2.3. We have gp =. It is easy to see by induction over rimes r that g = =. q r r q< r Letting r we get that gp =. There is also a holistic way of seeing this. Since g is the density of the set of integers with least rime factor, it would make sense that gp is the density of the set of integers which have a least rime factor, which is. To make this rigorous, one notes that the density of the set of integers whose least rime factor is > y tends to 0 as y. As a consequence of gp =, we have 2.3 g = 2, an identity we will find to be useful. For a rimitive set A, let A k = {a A : 2 k a}, B k = {a/2 k : a A k }. The next result will hel us rove Theorem.. Lemma 2.4. For a rimitive set A, let k be such that 2 k / A. Then we have fa k < eγ 2 k g. Proof. The hyothesis 2 k / A imlies that / B k, so that B k is a rimitive set. If 2 k / A for a rime > 2, then B k is a rimitive set of odd comosite numbers, so by Proosition 2., fb k < e γ g. Now if 2 k A for some odd rime, then B k = {} and note / A by rimitivity. We have f2 k < 2 k e γ g since / A 2 k log2 k 2 k log2 < eγ 2 k g, which follows from 2.2. Since B k imlies / A, we have fa k = f2 k B k f2 k + 2 k fb k / A B k, / A / B k, / A < eγ 2 k g. / A With Lemma 2.4 in hand, we rove fa < e γ. Proof of Theorem.. From Erdős Zhang [], we have that fa 3 < If 2 A, then A 2 = {2}, so that fa = fa 3 + fa 2 < /2 log 2 < e γ. Hence we may assume that 2 / A. If A contains every odd rime, then A 2 consists of at most one ower of 2, and the calculation just concluded shows we may assume

5 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS 5 this is not the case. Hence there is at least one odd rime 0 / A. By Proosition 2., we have 2.4 fa = fa = A f + / A fa < A f + e γ / A First suose A contains no owers of 2. Then by Lemma 2.4, fa 2 = fa k < e γ 2 k g = e γ g. k k / A / A Substituting into 2.4, we conclude, using 2.3, 2.5 fa < A f + 2e γ / A For the last inequality we used that for every rime, 2.6 f e γ g <.082, g 2e γ g = e γ. g + fa 2. which follows after a short calculation using [7, Theorem 7]. Now if 2 K A for some ositive integer K, then K is unique and K 2. Also A K = {2 K } and A k = for all k > K, so again by Lemma 2.4, fa 2 = K k= K fa k < f2 K + Substituting into 2.4 gives fa < A 2.7 k= e γ 2 k / A f + f2 K K e γ / A g = f2 K + 2 K e γ / A g. g f2 K e γ g f e γ < e γ, using K 2, the identity 2.3, inequality 2.6, and f2 2 < 2 2 e γ. This comletes the roof. <x 3. Mertens rimes In this section we will rove Theorems.3 and Theorem.4. Mertens theorem, e γ log x, x, where γ is Euler s constant. We say a rime q is Mertens if 3. e γ log q, <q Note that by and let P Mert denote the set of Mertens rimes. We are interested in Mertens rimes because of the following consequence of Proosition 2., which shows that every Mertens rime is Erdős strong. Corollary Let A be a rimitive set. If q P Mert, then fa q fq. Hence if A q {q} for all q / P Mert, then A satisfies the Erdős conjecture.

6 6 JARED DUKER LICHTMAN AND CARL POMERANCE Proof. By Proosition 2. we have fa q max{e γ gq, fq}. If q P Mert, then e γ gq = eγ q q log q = fq, so fa q fq. <q Now, one would hoe that the Mertens inequality 3. holds for all rimes q. However, 3. fails for q = 2 since e γ > / log 2. We have comuted that q is indeed a Mertens rime for all 2 < q 0 8 = 2,038,074,743, thus roving the unconditional art of Theorem Proof of Theorem.3. To comlete the roof, we use a result of Lamzouri [3] relating the Mertens inequality to the race between πx and lix, studied by Rubinstein and Sarnak [8]. Under the assumtion of RH and LI, he roved that the set N of real numbers x satisfying e γ > log x, has logarithmic density δn equal to the logarithmic density of numbers x with πx > lix, and in articular 3.2 dt δn = lim x log x t = t N [2,x] We note that if a rime = n N, then for = n+ we have [, N because the rime roduct on the left-hand side is constant on [,, while / log x is decreasing for x [,. The set of rimes Q in N is recisely the set of non-mertens rimes, so Q = P \ P Mert. From the above observation, we may leverage knowledge of the continuous logarithmic density δn to obtain an uer bound on the relative uer logarithmic density of non-mertens rimes 3.3 δq := lim su x From the above observation, we have δn lim su x log x Q dt t log x Q = lim su x log. log /. log x Q Then letting d = be the ga between consecutive rimes, we have d δn lim su x log x, since log / = d / + O. The average ga is roughly log, so we may consider the rimes for which d < ɛ log, for a small ositive constant ɛ to be determined. We claim 3.4 lim su x log x d <ɛ log Q log 6ɛ,

7 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS 7 from which it follows δq = lim su x log x δn /ɛ + 6ɛ. Q log lim su x d /ɛ + log x Q d <ɛ log d ɛ log log Hence to rove Theorem.3 it suffices to rove 3.4, since taking ɛ = δn /4 gives 3.5 δq < 8 δn < By Riesel-Vaughan [6, Lemma 5], the number of rimes u to x with + d also rime is at most 8c 2x log 2 x 2, d +d rime where c 2 is for the twin-rime constant 2 2/ 2 = Denote the rime roduct by F d = d 2, and consider the multilicative function Hd = u d µuf d/u. We have H2k = 0 for all k, and for > 2 we have H = F, and H k = 0 if k 2. Thus, = y F d = Hu = d y d y u d u y + Hu / 2 y d y/u u y = y Hu u + Noting that c 2 := + /[ 2] = 2/c 2, we have d <ɛ log d ɛ log x +d rime 8c 2x log 2 x d ɛ log x y. 2 F d ɛ 8c 2c 2x log x + H = ɛ 6x log x. Thus, 3.4 now follows by artial summation, and the roof is comlete. Remark 3.. The concet of relative uer logarithmic density of the set of non- Mertens rimes in 3.3 can be relaced in the theorem with δ 0 Q := lim su x log log x. Indeed, δ 0 Q δq follows from the identity = log x + log x 2 Q Q Q tlog t 2 t Q log Remark 3.2. Greg Martin has indicated to us that one may be able to rove under RH and LI that the relative logarithmic density of Q exists and is equal to the logarithmic density of N. This toic will be addressed in a future aer. dt.

8 8 JARED DUKER LICHTMAN AND CARL POMERANCE 3.2. Proof of Theorem.4. We now use some numerical estimates of Dusart [8] to rove Theorem.4. We say a air of rimes q is a Mertens air if r<q r > log log q. We claim that every air of rimes, q with 2 < q < e 06 is a Mertens air. Assume this and let A be a rimitive set suorted on the odd rimes u to e 06. By 2., if / A, we have > log a r a log P a a A r<p a a A log a log a = fa log. a A Dividing by log we obtain fa f, which also holds if A. Thus, the claim about Mertens airs imlies the theorem. To rove the claim, first note that if is a Mertens rime, then, q is a Mertens air for all rimes q. Indeed, we have r<q r = r< r r< r<q > e γ log. r r r<q By 2.2, this last roduct exceeds e γ / log2q > e γ / logq, and using this in the above dislay shows that, q is indeed a Mertens air. Since all of the odd rimes u to 0 8 are Mertens, to comlete the roof of our assertion, it suffices to consider the case when > 0 8. Define E via the equation = + E r e γ log. Using [8, Theorem 5.9], we have for > 2,278,382, 3.6 E.2/log 3. A routine calculation shows that if q < e 4.999log 4, then = log r log q + E q > log + E log q. r<q It remains to note that 4.999log >,055,356. It seems interesting to record the rincile that we used in the roof. Corollary If A is a rimitive set such that a, P a is a Mertens air for each a A, then fa fpa. Remark 3.3. Kevin Ford has noted to us the remarkable similarity between the concet of Mertens rimes in this aer and the numbers γ n = γ + log k k k k n k n

9 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS 9 discussed in Diamond Ford [7]. In articular, while it may not be obvious from the definition, the analysis in [7] on whether the sequence γ, γ 2,... is monotone is quite similar to the analysis in [3] on the Mertens inequality. Though the numerical evidence seems to indicate we always have γ n+ < γ n, this is disroved in [7], and it is indicated there that the first time this fails may be near This may also be near where the first odd non-mertens rime exists. If this is the case, and under assumtion of RH, it may be that every air of rimes q is a Mertens air when > 2 and q < ex Odd rimitive sets We say a rimitive set is odd if every member of the set is an odd number. In this section we rove Theorem.2 and establish a curious result on arity for rimitive sets. Let ɛ 0 = / P Mert e γ g f. Lemma 4.. We have 0 ɛ 0 < Proof. By the definition of P Mert, the summands in the definition of ɛ 0 are nonnegative, so that ɛ 0 0. If > 2 is not Mertens, then > 0 8 > 2 0 9, so that 3.6 shows that 4. e γ g f < 5log 4. By [8, Proosition 5.6], we have n > nlog n + log log n + log log n 2./ log n, n 2. Using this we find that which with 4. comletes the roof. 5 n log n 4 < , n>0 8 Remark 4.2. Clearly, a smaller bound for ɛ 0 would follow by raising the search limit for Mertens rimes. Another small imrovement could be made using the estimate in [] for n. It follows from the ideas in Remark 3.2 that ɛ 0 > 0. Further, it may be rovable from the ideas in Remark 3.3 that ɛ 0 < 0 00 if the Riemann Hyothesis holds. We have the following result. Theorem 4.3. For any odd rimitive set A, we have 4.2 fa fpa + ɛ 0. Proof. Assume that A is an odd rimitive set. If PA is Mertens, Corollary 3.0. imlies that fa f, while if PA is not Mertens we have by Proosition 2. that fa max{f, e γ g} = e γ g. Thus, fa = fa f + e γ g ɛ 0 + f PA PA P Mert PA\P Mert PA by the definition of ɛ 0. This comletes the roof.

10 0 JARED DUKER LICHTMAN AND CARL POMERANCE This theorem yields the following corollary. Corollary If A is a rimitive set containing no multile of 8, then 4.2 holds. Proof. We have seen the corollary in the case that A is odd. Next, suose that A contains an even number, but no multile of 4. If 2 A, the result follows by alying Theorem 4.3 to A \ {2}, so assume 2 / A. Then A 2 is an odd rimitive set and fa 2 fa 2/2. We have by the odd case that 4.3 fa = fa 3 + fa 2 < fpa 3 + ɛ fpa 2 + ɛ 0. Since 2 fpa 2 fp \ {2} < and f2 = , 4.3 and Lemma 4. imly that fa < fpa, which is stronger than required. The case when A contains a multile of 4 but no multile of 8 follows in a similar fashion. Since a cube-free number cannot be divisible by 8, 4.2 holds for all rimitive sets A of cube-free numbers. Also, the roof of Corollary 4.3. can be adated to show that 4.2 holds for all rimitive sets A containing no number that is 4 mod 8. We close out this section with a curious result about those rimitive sets A where 4.2 does not hold. Namely, the Erdős conjecture must then hold for the set of odd members of A. Put another way, 4.2 holds for any rimitive set A for which the Erdős conjecture for the odd members of A fails. Theorem 4.4. If A is a rimitive set with fa > fpa + ɛ 0, then fa 3 < fpa 3. Proof Sketch. Without loss of generality, we may include in A all rimes not in PA, and so assume that PA = P and fa > C + ɛ 0. By Theorem 4.3 we may assume that A is not odd, and by Corollary 4.3. we may assume that 2 / A. By the roof of Theorem. see 2.5 and 2.7, if 3 A, we have fa < f eγ < C, a contradiction, so we may assume that 3 / A. We now aly the method of roof of Theorem. to A 3, where owers of 3 relace owers of 2. This leads to fa 3 < 2 eγ < C f2 = fpa 3. This comletes the argument. Note that 5. Zhang rimes and the Banks Martin conjecture x log log x, x. In Erdős Zhang [] and in Zhang [20], numerical aroximations to this asymtotic relation are exloited. Say a rime q is Zhang if log log q. q

11 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS Let P Zh denote the set of Zhang rimes. We are interested in Zhang rimes because of the following result. Theorem 5.. If PA P Zh, then fa f. Hence the Erdős conjecture holds for all rimitive sets A suorted on P Zh. Proof. As in [] it suffices to rove the theorem in the case that A is a finite set. By d A we mean the maximal value of Ωa for a A. We roceed by induction on d A. If d A, then fa f. If d A >, then fa fa /. The rimitive set B := A satisfies fb = fb = q fb q. Since d B q d B < d A, by induction we have fb q fq. Thus, since is Zhang, fa = fb = fb q q log q log, q q from which we obtain fa fa / / log. This comletes the roof. From this one might hoe that all rimes are Zhang. However, the rime 2 is not Zhang since C > / log 2, and the rime 3 is not Zhang since C /2 log 2 > / log 3. Nevertheless, as with Mertens rimes, it is true that the remaining rimes u to 0 8 are Zhang. Indeed, starting from.2, we comuted that 5. q log = C <q log log q for all 3 < q 0 8. The comutation stoed at 0 8 for convenience, and one could likely extend this further with some atience. It seems likely that there is also a race between q / log and / log q, as with Mertens rimes, and that a large logarithmic density of rimes q are Zhang, with a small logarithmic density of rimes failing to be Zhang. A related conjecture due to Banks and Martin [2] is the chain of inequalities, log > q q log q > q r qr log qr >, succinctly written as fn k > fn k+ for all k, where N k = {n : Ωn = k}. As mentioned in the introduction, we know only that fn > fn k for all k 2 and fn 2 > fn 3. More generally, for a subset Q of rimes, let N k Q denote the subset of N k suorted on Q. A result of Zhang [20] imies that fn Q > fn k Q for all k >, while Banks and Martin showed that fn k Q > fn k+ Q if Q / is not too large. We rove a similar result in the case where Q is a subset of the Zhang rimes and we relace fn k Q with hn k Q. Recall ha = a A /a log P a. Proosition 5.2. For all k and Q P Zh, we have hn k Q hn k+ Q.

12 2 JARED DUKER LICHTMAN AND CARL POMERANCE Proof. Since k is a Zhang rime, we have hn k+ Q = q q k q k+ log q k+ This comletes the roof. q q k+ q i Q = q q k q i Q q q k q i Q q q k q k+ log q k+ q k+ q k q q k log q k = hn k Q. It is interesting that if we do not in some way restrict the rimes used, the analogue of the Banks Martin conjecture for the function h fails. In articular, we have hn 2 > m 0 4 m n m n log n = C m m 0 4 k<m k log k >.638, while hn = C <.637. It is also interesting that the analogue of the Banks Martin conjecture for the function g is false since = gn = gn 2 = gn 3 =. We have already shown in 2. that ga q gq for any rimitive set A and rime q, so the analogue for g of the strong Erdős conjecture holds. 5.. Proof of Theorem.5. We now return to the function f and rove Theorem.5. We may assume that k is large. Let m = k and let Bn = e en. We have fn k = Ωa=k = j m a log a > Ωa=k e ek <a e ek+m Ωa=k Bk+j <a Bk+j Ωa=k Bk+j <a Bk+j a log a a log a > j m log Bk + j Ωa=k Bk+j <a Bk+j Thus it suffices to show that there is a ositive constant c such that for j m we have Bk + j 5.2 clog = c ek+j a m m, so that the roosition will follow. Let N k x denote the number of members of N k in [, x]. We use the Sathe Selberg theorem, see [5, Theorem 7.9], from which we have that uniformly for Bk < x Bk + m, as k, N k x x log log x k. k! log x a.

13 THE ERDŐS CONJECTURE FOR PRIMITIVE SETS 3 This result also follows from Erdős [0]. We have Thus, Ωa=k Bk+j <a Bk+j Ωa=k Bk+j <a Bk+j Bk+j a > N k x N k Bk + j Bk+j x 2 dx Bk+j 2Bk+j N k x x 2 dx. log log Bk + j Bk+j k a k! 2Bk+j dx x log x k + j k = log log Bk + j log log2bk + j k! k + j k ek+j, k! k the last estimate following from Stirling s formula. This roves 5.2, and so the theorem. The sets N k and Theorem.5 give us the following result. Corollary We have that lim su{fa : A [x,, A rimitive} > 0. x Acknowledgments The first-named author is grateful for suort from the office of undergraduate research at Dartmouth College. We thank Greg Martin for his thoughts in connection with Remark 3.2 and Kevin Ford for the content of Remark 3.3. We thank Paul Kinlaw, Zhenxiang Zhang, and the referee for some helful comments. References. C. Axler, New estimates for the n-th rime number, arxiv: v [math.nt]. 2. W. D. Banks and G. Martin, Otimal rimitive sets with restricted rimes, Integers 3 203, #A69, J. Bayless, P. Kinlaw, and D. Klyve, Sums over rimitive sets with a fixed number of rime factors, rerint, A. S. Besicovitch, On the density of certain sequences of integers, Math. Ann , D. A. Clark, An uer bound of /a i log a i for rimitive sequences, Proc. Amer. Math. Soc , H. Cohen, High recision comutation of Hardy-Littlewood constants, rerint htts:// hecohen/. 7. H. G. Diamond and K. Ford, Generalized Euler constants, Math. Proc. Cambridge Phil. Soc , P. Dusart, Exlicit estimates of some functions over rimes, Ramanujan J , P. Erdős, Note on sequences of integers no one of which is divisible by any other, J. London Math. Soc , , On the integers having exactly k rime factors, Annals Math ,

14 4 JARED DUKER LICHTMAN AND CARL POMERANCE. P. Erdős and Z. Zhang, Uer bound of /a i log a i for rimitive sequences, Proc. Amer. Math. Soc , M. Kobayashi, On the density of abundant numbers, PhD thesis, Dartmouth College, Y. Lamzouri, A bias in Mertens roduct formula, Int. J. Number Theory 2 206, J. D. Lichtman, The recirocal sum of rimitive nondeficient numbers, J. Number Theory 208, htts://doi.org/0.06/j.jnt H. L. Montgomery and R. C. Vaughan, Multilicative number theory I. Classical theory, Cambridge U. Press, Cambridge, H. Riesel and R. C. Vaughan, On sums of rimes, Ark. Mat , J. B. Rosser and L. Schoenfeld, Aroximate formulas for some functions of rime numbers, Illinois J. Math , M. Rubinstein and P. Sarnak, Chebyshev s bias, Exeriment. Math , Z. Zhang, On a conjecture of Erdős on the sum n / log, J. Number Theory 39 99, , On a roblem of Erdős concerning rimitive sequences, Math. Com , Deartment of Mathematics, Dartmouth College, Hanover, NH address: jdl.8@dartmouth.edu address: jared.d.lichtman@gmail.com Deartment of Mathematics, Dartmouth College, Hanover, NH address: carl.omerance@dartmouth.edu