NEW UPPER AND LOWER BOUND SIFTING ITERATIONS

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1 NEW UPPER AND LOWER BOUND SIFTING ITERATIONS ZARATHUSTRA BRADY. Introduction Let A be a oibl eighted et of hole number and for each oitive integer d et A d = {a A d a}. Suoe that κ z are uch that for ever quarefree integer d all of hoe rime factor are le than z e have A d κ ωd. d In articular e have A +. We ant to etimate the quantit SA z = {a A < z a = }. Suoe no that = z a contant z going to infinit. Define ifting function f κ F κ b + of κ κ SA z + of κ κ <z <z ith f κ a large a oible re. F κ a mall a oible given that the above inequalit hold for all choice of A atifing. Selberg [3] ha hon in a much more general context that the function f κ F κ are continuou monotone and comutable for > and that the tend to exonentiall a goe to infinit. Let β = βκ be the infimum of uch that f κ > 0. Selberg [3] ha hon that e have κ < β < κ e κ here the firt inequalit alie for all κ and the econd alie for κ ufficientl large. Further hen κ = e have β =. Our main goal i to get good etimate for βκ hen κ i lightl greater than. When κ the bet knon ieve arie from the identit SA z = A <z SA running over rime. Thi identit lead to the inequalitie κ f κ κ κ t κ F κ t dt t> κ F κ κ + κ t κ f κ t dt. t> The ue of thee inequalitie to roduce ne bound on f κ F κ from knon bound i knon a Buchtab iteration reference?. Infinite iteration of Buchtab inequalitie lead to hat i knon a the β-ieve. The current tate of the art for κ lightl greater than i due to Diamond Halbertam and Richert []. Their method run a follo. From Selberg uer bound ieve e have F κ σ κ

2 here σ κ olve the differential-difference equation { κ σ κ = e γ κ Γκ+ 0 < d d κ σ κ = κ κ σ. Uing thi a our tarting oint e iterate Buchtab inequalitie infinitel to obtain uer and loer bound fκ D Fκ D atifing Fκ D = σ κ 0 < ακ D d κ Fκ D = κ κ fκ D d ακ D fκ D = 0 0 < βκ D d d κ f D κ = κ κ F D κ β D κ for ome ακ D βκ D. In thi note e ll decribe variation on Buchtab inequalitie hich allo u to lightl imrove uon the function fκ D Fκ D above. Theorem. For an z e have here q run over rime. SA z SA 3. The ne iteration <z SA + 3 q<<z SA q Proof. Let a A. We need to ho that the number of time a i counted on the left hand ide of the above i at leat the number of time a i counted on the right. If a ha an rime factor belo then both quantitie are clearl zero o aume that a i ha no rime factor belo. Suoe a ha exactl k rime factor beteen and z. If k = 0 then both ide count a once. Thu e jut need to check that for an integer k e have 0 3 k + k 3 hich follo from the identit 3 k + 3 Corollar. For an real t e have κ F κ t κ F κ t 3 κ t κ f κ t x dx x + 3 κ t <x< k = k k. 3 t <<x< t κ F κ t x dx x Remark. The otimal in Theorem above aear to be = hich correond to taking z β t =. Thu thi uer bound iteration tend to be ueful onl for β +. β d.

3 Theorem. For an z e have SA z S A S A <z 3 here q r run over rime. r<q<<z qr< S A qr S q<<z S q r<q<<z A q A qr Proof. Let a A. Firt uoe that a ha no rime factor belo and ha exactl k rime factor beteen and z. If k i 0 then both ide count a once. Otherie e need to check that for an integer k e have and thi follo from the identit k k k k 3 k k = k k k No uoe that a ha mallet rime factor <. We grou together all of the ummand on the right hand ide ith a common a. In order for an uch ummand to be nonzero e mut have or equivalentl. Suoe that a ha exactl k rime factor trictl belo. Then the number of time a i counted in uch ummand i at mot k k = 3k k 4 3 and thi i at mot 0 unle k =. Thu the onl bad cae occur hen i the third mallet rime factor of a q i the econd mallet rime factor of a and r = i the mallet rime factor of a. If qr < then the contribution from thee ummand i jut = 0 o the bad cae onl occur hen qr. But then ince q r = e can combine thi bad grou of ummand ith the grou of ummand here i relaced b q and the total number of time a i counted in the to grou become = = 0. 3

4 Corollar. For an real t ith t 3 e have κ f κ t κ x dx f κ t κ t F xκ κ κ 3 κ3 t x<<x< t <x< t x<z<<x< + 6 κ3 t <z<<x< t xκ f κ t xκ F κ t xκ F κ x dx d x z dx x z dx Remark. A ith Theorem it eem that the otimal in Theorem i = correonding z β to t =. Thu thi loer bound iteration tend to be ueful onl hen β + β +. β d d z z. 3. To miracle at κ = When κ = the β-ieve roduce the otimal function f = f F = F ee Selberg [3]. Furthermore e have the more recie error term f e γ c + oh SA z F logz logz e γ + c + oh logz logz here c i a comutable contant in fact a more recie reult can be found in Ianiec []. The function f F h H are given b F = eγ 3 d F = f 3 d f = eγ log 4 d f = F d H = 3 d H = h 3 d h = + log 4 d h = H d It natural to ak hat haen to thee function hen e al the ne uer and loer bound iteration to them. 4

5 Theorem 3. If κ = 5 3 and t = are reciel equal that i F = tf t tft x dx 3 x + 3 t <x< Furthermore in thi cae even the error term match u: H = t Ht + t ht x dx 3 x + 3 t <x< then the to ide of the inequalit in Corollar t <<x< t <<x< tf t x dx x t Ht x dx x Proof. Conider the right hand ide of the firt claimed equalit a a function Φ t of and t. Since F = e γ i contant for 3 it enough to check that Φ = Φ t = 0 hen t =. We have Φ = t 3 f t t 3 F t dx x x t <x< and u to a multile of eγ thi i equal to 3 log t 3 = 3 log t + log 3 hich i indeed 0 hen t = t = 5 o for an x > t Thu e have Φ t = ft 3 ft 3 and u to a multile of eγ t log t log Φ. In order to calculate t firt note that ince 5 e have t x t 3 o tf t x = 0. t t <x< thi i equal to 3 logt log 3 = 3 log t t. F t x dx x + 3 t t <x< t log t d. d. e have F t x dx x + 0 t t hich i alo equal to 0 hen t = The econd claim i left a an involved exercie to the reader alternativel one can ue the method of roof of the next theorem. Since the loer bound iteration i much more comlicated e need a better method of checking that it ha the linear ieve a a fixed oint. For thi e ue the folloing eighted et introduced b Selberg [3] in order to exlain the arit roblem: let A + be the eighted et of integer beteen and ith the eight attached to n given b λn here λn = Ωn and let A be imilar ith the eight of n given b + λn. Set π ± z = SA ± z. 5

6 Thee function are invariant under Buchtab iteration: π ± z = π ± <<z and b the rime number theorem for < < 3 e have o for all > e have 3 π + z = π πz = eγ e γ logz + π / logz + O logz 3 π + z = F e γ logz + H logz + O logz 3 π z = f e γ logz h logz + O logz 3. Theorem 4. If κ = 7 4 and t = are equal that i f = tft t <x< t x<<x< t x<z<<x< t <z<<x< then the to ide of the inequalit in Corollar x dx t xf t xf t xf t xf x dx d x z dx x z dx d d z z. Furthermore the error term are equal a ell: h = t ht t x<<x< t <x< t x<z<<x< t <z<<x< x dx t H x x dx t h x d x z dx t H x x z dx t H x 6 d d z z.

7 Proof. B equation 3 it enough to check that for contant 7 < < 4 and = z SA z = S A S A <z 3 r<q<<z qr< S A + qr S q<<z S q r<q<<z A q A + qr + O logz 3. e have We have the ea inequalit z > > 3/4 and for < < z e have > z > /7 a 5 ell a > 5 >. Thu if n i a number belo hich i counted b either ide then z 4 ever rime factor of n mut be at leat /7 and Ωn mut be an even number trictl belo max = 6. We need to etimate the number of n belo hich contribute more to the left and ide than the right hand ide. Since the number of nonquarefree n hich can contribute to either ide i at mot 3 6/7 e can aume ithout lo that n i quare free. If n = q ith > q rime e mut have z > in order for n to contribute more to the left ide than the right ide. The number of uch n i at mot z < 4/7 o e ma aume ithout lo that n ha four ditinct rime factor > q > r > at leat one of hich i belo z o n in t counted on the left hand ide at all. Firt conider the cae. Since n z e have z > q. Then if n ha 3 k 4 rime factor belo z n i counted on the right hand ide ith multilicit k k 3 0 k 3 = k k 3 k 4 = 0 o e get the ame contribution to both ide. No uoe that < r. Since n z e have z > q. Then if n ha 3 k 4 rime factor belo z n i counted on the right hand ide ith multilicit 0 k + 5 k k 3 = k k 3 k 4 = 0 a before. Next uoe that > r and > z. We mut have z > q > in order to get an contribution from n. Then n i counted on the right hand ide ith multilicit = 0 o e get the ame contribution from both ide. Thu an bad n mut have z > q and > r r > > /7. The number of uch n i at mot O z z logz logz log = O. logz 3 4. Numerical reult hen κ = 3 When κ = 3 e have αd κ = βκ D = []. In articular e have ακ D < βκ D + o Corollar can be alied to in the range ακ D < < βκ D + ith t =. The imrovement to the value of F κ in thi range i nonzero but ver mall. Combining thi ith ordinar Buchtab iteration for the loer bound one can ho that β 3 < If e al the iteration from Corollar directl to F D κ f D κ then the value of t for hich the quantit κ f κ i imroved the mot are given b 4.85 t 5.5. Thi reult in the bound β 3 < Iterativel combining the imrovement from Corollarie and e get β 3 < β D κ 5. An infinite equence of iteration rule Here e ill decribe an infinite equence of iteration rule one for each k generalizing the uer and loer bound iteration rule decribed o far hich correond to the cae k = and k =. We ill alo rove an otimalit reult for thee iteration rule. 7

8 Theorem 5. If k and z k then k SA z k S A /k + k + + / k < < <z k k k+ < < <z k < < <z k /k <z S A k S A k S A /k + k k #{i k + i k+ } S A k+ Proof. It enough to rove thi hen A ha jut one element a A = {a}. We ma alo aume that a i quarefree and rite a = q q q m ith q < q < < q m and the q i rime. We ma aume alo that q < z ince otherie the reult i trivial. Thu e jut need to rove that the right hand ide i at leat 0. Note that ever nonzero ummand correond to ome divior d = j of a having j rime factor j k +. Our trateg i to combine the nonzero ummand into mall grou according to the combinatorial tructure of their rime factor uch that each grou of ummand ha a nonnegative um. The firt te i to combine the ummand correonding to d = j ith j k and j = q ith the ummand correonding to d/ j and to note that thee to ummand exactl cancel each other out. After thi te the onl ummand that remain are thoe hich have d = j ith j k and k > q. The next te i to grou the ummand correonding to d = j ith j k k = q l ith l > and k taking ome fixed value P ith P q. If l = then the total contribution from uch d i. If l = 3 then the total contribution from uch d i + #{ {... k q q } P q q } = #{i k i P q q }. If l = 4 then the total contribution from uch d i at leat 3 + k+ 3 = k Finall if l 5 then the total contribution from uch d i eail een to be oitive. In order to balance out the negative contribution coming from grou correonding to P = k P q k = q l ith l = 3 e ill aign ortion of the oitive exce from grou correonding to P ith l = or l = 4 to certain correonding P ith l = 3. If P = k P q k = q l ith l = and m 3 i minimal uch that q m doe not divide P then e grou the exce contribution from thi P ith the contribution correonding to P = P q m /q - note that the leat rime factor of P i then necearil equal to q k.

9 If P = k P q k = q l ith l = 4 then e take k of the exce contribution from thi P and divide it into k iece of ize... k and e aign the iece of i ize to P i = P q 3/ i+ noting once again that P i ha leat rime factor equal to q 3. To finih the argument e jut have to ho that for P = k P q k = q l ith l = 3 the total exce contribution that a aigned to P b the roce decribed in the lat to aragrah i at leat #{i k i P q q }. To ee thi let m 4 be minimal uch that q m doe not divide P or let m = k + if P q q = a. For an 3 j < m if e let P j = P q /q j then the leat rime factor of P j i q and a long a q j P q q e have P j q and the exce of correonding to P j i aigned to P. Additionall in the cae m < k + e let P = P q m /q 3 and e ee that the leat rime factor of P i q 4 that P q < P q and that k+ m of the exce correonding to P i aigned to P. Together e ee that the amount of exce hich a aigned to P i at leat #{3 j < m q j P q q } + k + m #{i k i P q q }. To ee that the kth iteration rule i otimal hen e et κ = = and = z ith z k + 3 < < k + e argue a in Theorem 4 to ee that e jut need to rove the folloing bound. Theorem 6. If A ± are eighted et of integer beteen and defined a in the dicuion before Theorem 4 then for an k if = z ith k + 3 < < k + and = e have z k SA k z = k S A k /k + k S A k /k / k < < <z k k+ < < <z k + O logz 3. k < < <z k /k <z k S A + k S A k k #{i k + i k+ } S A + k+ Proof. Suoe that a i counted a different number of time on both ide of the above. Then e necearil have λa = k and the leat rime dividing a i le than z. In order for the contribution of a to the right hand ide to be oitive there mut be rime > > k dividing a uch that < z and uch that the leat rime dividing a i at leat k o e conclude that an rime dividing a mut be at leat > k z k = z k+ = z k+ > z. In articular the number of uch a hich have a quare factor i O z o e ma aume ithout lo that a i quare free. If a ha at leat k + 4 rime factor then ince a ha ome collection of 9 k

10 k rime factor hoe roduct i at leat e have a > z 4 = a contradiction. Thu a ha trictl le than k + 4 rime factor and ince λa = k e ee that a ha either k or k + rime factor. If a ha exactl k rime factor then the mut all be le than z in order for the contribution z 3/. of a to the right hand ide to be oitive o a < z k < o the number of uch a i at mot z 3/ Thu e ma aume ithout lo that a ha exactl k + rime factor at leat k of hich are le than z. If to of the rime factor of a are z then the remaining rime factor of a mut have roduct at leat o a > z = a contradiction. If one of the rime factor of a i z and the remaining k + rime factor of a are all < z then the total contribution of a to the right hand ide i reciel 0. Thu e ma aume that all of the rime factor of a are le than z. If ever roduct of k rime factor of a i then the contribution of a i again reciel 0. Otherie e can rite a = q q k+ ith z < q < < q k+ q q k < q k+ < z and q k+ < z. Uing an uer bound ieve to bound the number of oible value for q q k b O logz z e ee that the number of uch a i O = O logz 3 Reference. logz 3 [] H. Diamond H. Halbertam and H.-E. Richert. Combinatorial ieve of dimenion exceeding one. J. Number Theor 83: [] H. Ianiec. On the error term in the linear ieve. Acta Arith. 9: [3] Atle Selberg. Collected aer. Vol. II. Sringer-Verlag Berlin 99. With a foreord b K. Chandraekharan. Deartment of Mathematic Stanford Univerit 450 Serra Mall Bldg. 380 Stanford CA addre: notzeb@tanford.edu 0