Coarse-grained integers - Smooth? Rough? Both!

are known, see for example the overview by Granville (2008). It is difficult, however, to compare our results to existing ones, since we consider integers with a fixed number of factors. This restriction is completely sufficient for the application we have in mind. To our knowledge these kind of investigations cannot be found in the literature. Our main result implies the following Corollary. Fix k ∈ N ≥2 , α > 0 and ε > 0. Then for large B and C = B 1+α we have uniformly for Actually, we can describe a piecewise smooth function that approximates π k B,C up to an additive error of order O x √ B and the hidden constant depends on ε, k and α only, see Theorem 9.2, Theorem 9.3. To avoid difficulties with non-squarefree numbers, instead of numbers we count lists of primes (0.2) It turns out that there are anyways only a few non-squarefree numbers counted by . This would actually be an equality if there were no non-squarefree numbers in the count. We defer a precise treatment until Section 8. We head for determining precise bounds κ k B,C (x) or π k B,C (x) that can be used in practical situations. However, to understand these bounds we additionally consider the asymptotical behaviours. This is tricky since we have to deal with the three parameters B, C and x simultaneously. To guide us in considering different asymptotics, we usually write B = x β , C = x γ and γ = β(1 + α). In particular, C = B 1+α . So we replace (B, C, x) with (x, β, γ) or (x, α, β), similar to the considerations when counting smooth or rough numbers in the literature. Alternatively, it seems also natural to fix x somehow in the interval ]B k , C k ] by introducing a parameter ξ by Now the parameters are (B, C, ξ) or (B, α, ξ). As a first observation, note that κ k B,C (x) is constantly 0 for x < B k and constantly (π(C) -π(B)) k for x ≥ C k and grows monotonically when Here π denotes the prime counting function. For 'middle' x-values the asymptotics can be derived from our main result: Corollary. Fix k ∈ N ≥2 , α > 0 and ε > 0. Then for large B and C = B 1+α we have uniformly for Later, we specify a piecewise smooth function κ k B,C which approximates κ k B,C with an error of order O x √ B where the hidden constants depend on ε, k and α only, see Theorem 3.2, Theorem 3.5, and subsequent considerations. The preceeding result is in contrast to the asymptotics at x = C k : This observation is explained as follows: Note that roughly half of the numbers up to x are in the interval [ 1 2 x, x] and similarly for primes. Thus the behaviour of those candidates largely rule κ k B,C (x) / x . For an x = B k-ξ C ξ with a fixed ξ ∈ ]0, k[, it is mostly determined by the requirement that the counted numbers are B-rough, and we thus observe a comparatively large fraction of ]B, C]-grained numbers. In the extreme case x = C k , most candidates are ruled out by the requirement to be C-smooth and thus we see a much smaller fraction of ]B, C]-grained numbers. The case that the intervals ]B k , C k ] are disjoint for considered values k is especially nice, as then the number of prime factors of a B-smooth and C-rough number n can be derived from the number n. So we assume in the entire paper that C < B s for some fixed s > 1. (Clearly, we cannot have a fixed s that grants disjointness for all intervals. But for the first few we can.) In the inspiring number field sieve application we have C < B 1.232 . 1 This ensures that the intervals B k , C k for k ≤ 5 are disjoint. A further application is related to RSA. Decker & Moree (2008) give estimates for the number of RSA-integers. They use one ad-hoc definition proposed by B. de Weger. However, it is not so clear which numbers we should call RSA-integers. A discussion and further calculations to adapt our results to the different shape are needed. We have treated these issues in Loebenberger & Nüsken (2011). As our basic field of interest is cryptography and there the largest occurring numbers are actually small in the number theorist's view, we assume the Riemann hypothesis throughout the entire paper. We use the following version of the prime number theorem: Prime number theorem 0.3 (Von Koch 1901, Schoenfeld 1976). If (and only if) the Riemann hypothesis holds then for x ≥ 2657 where li(x) = x 0 1 ln t dt. 1 Side remark: to indicate how a real number was rounded we append a special symbol. Examples: π = 3.14 = 3.142 = 3.1416 = 3.14159 . The height of the platform shows the size of the leftout part and the direction of the antenna indicates whether actual value is larger or smaller than displayed. We write, say, e = 2.72 = 2.71 as if the shorthand were exact. We have numerically verified this inequality for x ≤ 2 40 ≈ 1.1•10 12 based on Kulsha's tables (Kulsha 2008) and extensions built using the segmented siever implemented by Oliveira e Silva (2003). We are confident that we can extend this verification much further. In the inspiring application we have x < 2 37 and so for those x we can take this theorem for granted even if the Riemann hypothesis should not hold. We arrive at the following description of the desired count: Then for any (small) ε > 0 and B tending to infinity we have for Also without assuming the Riemann hypothesis we can achieve meaningful results provided we use a good unconditional version of the prime number theorem whose error estimate is at least in O x ln 3 x . The famous work by Rosser & Schoenfeld (1962,1975) is not sufficient. Yet, Dusart (1998) provides an explicit error bound of order O x ln 3 x , and Ford (2002a) provides explicit error bounds of order O x exp -A(ln x) 3/5 (ln ln x) 1/5 though this only applies for x beyond 10 171 or even much later depending on A and the O-constant, see Fact 6.1. The essential basis for the analysis of the counting functions κ k B,C is the following simple description. Lemma 1.1. For all k ∈ N >0 we have the recursion The case k = 0 is immediate from the definition. From the definition (0.2) or from Lemma 1.1, it is clear that This reveals that the case distinction in κ 0 B,C leaves its traces on higher κ k B,C . For further calculations it is vital that we make this precise. This will enable us later to do our estimates. For k ∈ N we distinguish k + 2 cases: where j ∈ N 0 → R ≥0 such that f and f are piecewise continuous, f + f is increasing, and where Moreover, g and g are piecewise continuous, and g + g is increasing. Proof. The assumption immediately implies . Shifting the second term to the correspondingly transformed error bound, we now have This bound always holds, yet we still have to estimate E(p) in it. Abbreviating h(p) = ( f + f )(x/p) we estimate the last integral: For the inequality we use that h(p) is decreasing in p. Collecting gives the claim. Finally, g and g being obviously piecewise continuous it remains to show that g + g is increasing. By the (defining) equation this follows from f + f and E being increasing. Based on Lemma 2.1 we recursively define approximation functions κ k B,C and error bounding functions κ k B,C . Recursive application of Lemma 2.1 will lead to a good estimate in Theorem 3.2. For the understanding we further need to determine the asymptotic order of the functions defined here, which we start in the remainder of this section. Theorem 3.5 relates the functions κ k B,C and κ k B,C to some easier manageable functions. These in turn are computed or estimated, respectively, in Section 4 and Section 5. Definition 3.1. For x ≥ 0 we define These functions now describe the behavior of κ k B,C nicely: Theorem 3.2. Given x ∈ R >0 and k ∈ N. Then the inequality holds. Proof. Using Lemma 2.1 the claim together with the fact that κ k B,C + κ k B,C is increasing follow simultaneously by induction on k based on κ 0 B,C = κ 0 B,C and κ 0 B,C = 0. In order to give a first impression we calculate κ 1 B,C and κ 1 B,C . Analogous to Lemma 1.2, we split the integration at x / B k-1-j C j so that the parts fall entirely into case (k -1, j -1) or into case (k -1, j): Also for κ k B,C this can be done, simply split the occurring integrals at x / B k-1-j C j . Now unfolding the recursive definition of κ 1 B,C gives: which corresponds exactly to the approximation of the prime counting function by the logarithmic integral Li. In case k = 2 we obtain This is now exactly the transformed version of (1.3), as announced there. You may ask where you can find a display of the error term corresponding to (3.3). Well, we have computed it. But the resulting terms are so complex that we didn't really learn much from it. Here is an expression for x ∈ B 2 , BC : Though this term is still handleable, it becomes apparent that things get more and more complicated with increasing k. We escape from this issue by loosening the bonds and weakening our bounds slightly. The first aim will be to obtain easily computable terms while retaining the asymptotic orders, the second aim will be to still retain meaningful bounds for the fixed values B = 1100 • 10 6 , C = 2 37 -1, k ∈ {2, 3, 4} from our inspiring application. Our next task is to describe the orders of κ k B,C and κ k B,C . The main problem in an exact calculation is that most of the time we cannot elementary integrate a function with a logarithm occurring in the denominator. But using B ≤ p ≤ C we can obtain a suitably good approximation instead by replacing 1 ln p with 1 ln B in the integrals. At this point we start using C ≤ B s and rewrite C = B 1+α where α is a new parameter (bounded by s -1). For the time being you can consider α as a constant, but actually we make no assumption on it. This leads to the following Definition 3.4. For x ≥ 0 we let and recursively for k > 0 We observe that ln B ≤ ln ln B dp for any positive integrable function f . By induction on k we obtain Theorem 3.5. Write C = B 1+α and fix k ∈ N >0 . Then for x ∈ R >0 we have In order to determine at least the asymptotic orders of κ k B,C and κ k B,C (along with precise estimates) it remains to solve the recursions for λ k and λ k , or at least to estimate these functions. As a first step, we rewrite all integrals in Definition 3.4 in terms of ̺ defined by p k = B 1+̺α . Definition 3.4 continued. Rewrite x = B k+ξα with a new parameter ξ and let ) for any family of functions f k .) Lemma 3.6. For k > 0 we now have the recursion To construct a useful description of the function λ k we make a small excursion and consider the following family of piecewise polynomial functions. Definition 4.1 (Polynomial hills). Initially, define the integral operator M by based on the rectangular function m 1 given by m 1 (ξ) = 1 for ξ ∈ [0, 1[ and m 1 (ξ) = 0 otherwise. Actually, m 1 = M m 0 if we let m 0 = δ be the 'left lopsided' Dirac delta distribution defined by its integral ξ -∞ δ(t) dt being 0 for ξ < 0 and 1 for ξ ≥ 0. Contrastingly, the standard Dirac delta function is balanced and has 0 -∞ δ(t) dt = 1 2 . However, we will stick to the lopsided variant throughout the entire paper. By D ξ we denote the differential operator with respect to ξ . (ix) The next derivate can only be correctly described as a linear combination of Dirac delta distributions: (x) For any 0 ≤ ℓ < k we obtain the following explicit description: Curiosity: The function √ k • m k can also be described as the volume of a slice through a (k -1)-dimensional unit hypercube of thickness 1 √ k orthogonal to a main diagonal. That's the same as saying it is Proof. From the definition (i) through (vii) follow directly. Further, note that (ix) follows similarly. To prove (x) consider As we do not need (v) and (xi), we leave these proofs to the interested reader. Most of the following is easier if we first renormalize λ k . So we let The recursion for λ k now turns into Theorem 4.4 (Approximation order). For any ξ ∈ R we have Proof. The definition for λ 0 turns into Now, since M commutes with this integration this immediately implies that which is the first stated equality noting that m k is zero outside [0, k]. By partial integration we obtain Using the description of D k ξ m k from Lemma 4.2(ix) the last integral turns into the claimed sum of the second stated equality. Expressing D k-ℓ-1 ξ m k (ξ -̺) using Lemma 4.2(x) and rearranging slightly yields the third equality. We are going to estimate the estimation of the error in the next section. To that aim we first need to estimate λ k norm . If k > 0 then, based on Theorem 4.4 and m k (ξ) ≤ 1, we obtain the upper bound For k = 0 we have λ 0 norm ξ = B -ξα for ξ ≥ 0 and so λ 0 norm ξ ≤ 1 will do for all ξ ∈ R. We first describe the qualitative behaviour of λ k norm . Actually its graph looks like a slighty biaswise hill. Lemma 4.5. The function and there is a position ξ k Proof. First, inspecting shows that for k = 1 all claims hold with ξ k 1 2 = 1. So in the remainder of this proof we assume k > 1. Next, compute λ k norm k inductively: Here we have substituted τ = ̺ -̺ k and collapsed the new integration interval Finally, compute the derivate of λ k norm differently Since this term starts at zero, it is positive for some time, begins to decrease at k 2 , traverses zero at some point ξ k 1 2 recalling that at k the value is negative, and stays negative til ξ = k since it continues to decrease. Thus the sign of , k[, and so λ k norm ξ is increasing till ξ k 1 2 and decreasing afterwards. k = 1,2,3,4,5,6,7 Lemma 4.7. Assume k ≥ 2, α ln B ≥ ln 16 k , and ξ ∈ [0, 1]. Then we have The assumptions are already true for C = 2B when k ≥ 4. Note that this is rather sharp as for ξ ∈ [0, 1] we have Proof. We use the integral representation from Theorem 4.4 and estimate the polynomial hill part m k (ξ -̺) of its kernel by a simple piecewise constant function as indicated in the picture. We obtain This holds for any ε ∈ [0, ξ] since 0 ≤ ξ ≤ k 2 ensures that m k is increasing. So we can optimize ε depending on ξ. We obtain a suitable value when setting ε by To make sure that now ε ≤ ξ we use the following simple fact. Fact 4.9. For any ϑ > 0 and τ = 1-exp(-ϑ) ϑ the map τ z z is bijective and increasing and for z ∈ In particular, ε ≤ ξ follows from kτ 1 α ln B = 1 ϑ 1 α ln B ≥ 1. By Fact 4.9 with ϑ = ln 2 we obtain τ = 1 2 ln 2 and 1 - (k-1)! and so Since ϑ 1 < α ln B we can define ϑ 2 by 1-exp(-ϑ 2 ) = ϑ 1 α ln B , and according to Fact 4.9 let and thus simplifies our above inequality to By our choices Though these inequalities get equalities with k → ∞, we need a precise estimate. Substituting k in -k ln 1 -1 k ≤ ln 4 with kα ln B ln 4 yields Though we know the value of λ k norm k , it is orders smaller than the above left lower bound. Thus let us consider Lemma 4.12. Proof. By definition we have To check the claimed inequality we substitute τ = (k -ξ)α ln B ∈ [0, α ln B] (eliminating ξ): We have to show that this is at least ) k in our situation, with equality for τ = α ln B or ξ = k -1. Thus it suffices to show for any τ > 0 0≤ℓ≤k-2 The remaining proof proceeds in four steps: • Brute-force: Prove (4.13) for k = 3. (Actually, for k ∈ {3, 4, 5, 6, 7, 8, 9}.) As Bernoulli's inequality is good for T close to 0 only, it is not surprising that this only gives a sufficient result for τ large enough. Low case: Assume is decreasing, this is always true on some interval [0, τ 0 ].) The condition on σ implies that e τ -1 σ + 1 τ k-1 (k-1)! is non-negative for all τ > 0: We use this to obtain: Notice that the value of σ only influences the set of values of τ that fall in this case. Covering: We choose σ := . Recall Stirling's formula: For any n > 0 there , see Robbins (1955). We thus have 1 σ +1 = 2π(k -1) < e k-1 (k-1)! (k-1) k-1 as required. Further, we define the value τ split where we split between the low and the high case: We start with the treatment of the cases k ≥ 8. We claim that for τ ≤ τ split we are in the low case, ie. Consider f 2 (τ ) = 1 -e -τ -ϑτ . It obviously vanishes at τ = 0. The derivative of f 2 shows that there is exactly one maximum at τ = -ln(ϑ), which is roughly ln k. To Well, now we have Checking that f 3 (e) = 0 and the derivative f ′ 3 (k) = (1 -e k ) 2 is positive for k > e shows that f 3 (k) > 0 for all k ≥ 3. Thus for τ ≤ τ split we are in the low case with the above choice of σ. It remains to check that for τ ≥ τ split we are in the high case. We use the Lagrange remainder estimate of the power series of the exponential function and again Stirling's formula to obtain 0≤ℓ≤k-2 As the left hand side is increasing in τ > 0 we consider the smallest τ in question: . Observe that the term (I) is obviously positive and increasing for k ≥ 8. The same is true for the term (II): its derivative 1 k-1 -1 k(ln k-1) is positive for k ≥ e 2 ≈ 7.39 and the value of term (II) at e 2 is positive. Using this we infer that f 6 is increasing and positive. Checking f 6 (10) > 1 (f 6 (10) ∈ ]1.19, 1.20[) now proves 0≤ℓ≤k-2 (2 ln k e ) ℓ ℓ! ≥ k for k ≥ 10. For k = 8 and k = 9 we just verify this inequality directly. It remains to consider 4 ≤ k ≤ 7. Here we use individual seperation positions as defined above: Just check that for this τ split the low and the high case conditions are both fulfilled. Summing up: the claim is proved for k ≥ 4. k < 10: For k = 3 we need an explicit check as the estimates done in the low and the high case are too sloppy. As the following actually is a general computational way to verify the inequality (4.13) we do describe it in general, show computational results for 3 ≤ k ≤ 10 and make the critical case k = 3 hand-checkable at the end. For the verification we use a small trick and brute force: First, we substitute occurrences of e -τ with a new variable T . The task turns into showing that the bivariate polynomial If we thus replaceln T with a lower bound and we can show that the resulting term is still non-negative then we are done. For T ∈ ]0, 1] we haveln T ≥ 1≤ℓ≤s (1-T ) ℓ ℓ . This lower bound even converges toln T , which actually ensures that we can always find some s that allows the following reduction. We consider the univariate polynomial By our reasoning, the claim follows if ∀T ∈ ]0, 1] : g k,s (T ) ≥ 0 for some s. This in turn is implied by The second statement can be checked using Sturm's theorem (Sturm 1835) by only evaluating certain rational polynomials at T = 0 and T = 1. However, this only works if the chosen s is large enough. We have determined the smallest s that make (4.15) true: Though we can always divide out (1 -T ) k from g k,s the degrees are in all cases quite high and the computations better done by a computer. The timings refer to our own (non-optimized) MuPad-program used to assert (4.15). As k = 3 is the only case that we do not cover otherwise we give g 3,4 here: With this description you can easily see that it is positive on [0, 1[. Finally, we put together the upper bound on λ k norm , Lemma 4.5, Lemma 4.7, and Lemma 4.12 in the following theorem. Theorem 4.16. For any k ≥ 1 we have Here we can choose is the maximum of λ k norm which in our experiments is always less then k-1 for k ≥ 3. However, proving that would result in a stronger version of Lemma 4.12, which even in the given form required quite some effort. However, we just want that to work for some small ε and that is always granted. Proof. The upper bound being proven we consider the lower bound. First, keeping in mind that α ln B ≥ ln 2 and α ln B ≥ ln 16 k , we consider the cases k ≥ 3. We know by Lemma 4.5 that λ k norm on [ε, k-ε] attains its minimal value at one of the boundaries since ξ k 1 2 ∈ [ε, k -ε] (by assumption). We thus only need to consider its values at ξ = ε and at ξ = k -ε. On the left hand side Lemma 4.7 gives using the conditions on α ln B. On the right hand side by Lemma 4.12 we find using again α ln B ≥ ln 2. This completes the proof for the cases k ≥ 3. We will now show corresponding lower bounds for the cases k = 1, 2. For k = 1 we have for ε ∈ ]0, 1] using α ln B ≥ ln 2: Using Fact 4.9 and ε ≤ 1, we have For k = 2 we again apply Lemma 4.7 for the left hand side as in the general case. For the right hand side we show that this proves the claim. Now, using Theorem 4.4 we write the left hand side minus the right hand side as is positive for ϑ > 0. Thus it suffices to show that f 5 (τ, ϑ) ≥ 0 for the smallest allowed ϑ, which is the larger of τ and ln 2. If τ ≥ ln 2 then we consider f 5 (τ, τ ) = exp(-τ ) + 1 2 ln 2 τ -1. This expression is increasing in this case (even for τ ≥ ln 2 + ln ln 2) and so it is greater than or equal to f 5 (ln 2, ln 2) = 0. If otherwise 0 ≤ τ ≤ ln 2 then we consider f 5 (τ, ln 2) = -3 4 exp(τ ) + 1 2 ln 2 τ + 1 which is decreasing even for τ ≥ 0 and so it is greater than or equal to f 5 (ln 2, ln 2) = 0. Corollary 4.17. For any ε ∈ ]0, 1] and any k ≥ 1 we have uniformly for ξ ∈ The recurrence Lemma 3.6 for λ k is more complex than the one for λ k , so instead of solving it we estimate it. We consider also here the normed version λ k norm ξ := To better understand how the error behaves we compute it for k = 1: From this we estimate λ 1 norm directly: We have also looked at precise expressions for larger k, yet they are huge and do not give rise to better bounds. Theorem 5.1. Define values c k recursively by Then for any k and ξ ∈ R we have we have for k ≥ 2 and large B the inequality For k = 1 the order of c 1 is necessarily slightly larger. More precisely, we have for large B that Instead of defining c 2 and c 1 we could have left that to the recursion. But the given values are smaller than the ones derived from the recursion based on c 0 only. For k = 1 the recursion would give 4+3 ln B 8πα √ B . For k ≥ 2 however the improvement due to these explicit settings is a factor of order ln B. Proof. We first show that the value c k is bounded as claimed. For k = 1 the claim follows directly from the definition, since we have for B > exp(1) that (2 + α) ln B ≤ 2(1 + α) ln B as α is positive. For k = 2 we have for ln 2 B ≤ √ B the inequality For k ≥ 2 we proceed inductively. We have Now we prove the remaining estimate by induction on k. The case k = 0 is true by definition of λ 0 (with equality). For k = 1 the inspection above proves the claim. The explicit calculation of λ 1 also shows that a bound of order O x / √ B is impossible. We defer the case k = 2 to the end of the proof as most of it will be as in the general case. So assume k ≥ 3. Using the definition of λ k from Lemma 3.6 we split λ k into three summands: The first summand of λ k norm is at most c k-1 by induction hypothesis. The second summand we estimate using (5.2) by The third summand is bounded by By (5.3) the third summand is at most This completes the proof of the case k ≥ 3: For k = 2 we have to improve the estimate of the first summand only, since the other terms anyways are of order at most O 1 / √ B . For ξ < 0 there is nothing to prove. For ξ ∈ [0, 1[ we find for this first summand For ξ ∈ [1, 2] we find As λ 1 norm decreases for ξ ≥ 1 this bound also holds for ξ ≥ 2. Putting everything together the above defined value c 2 bounds λ 2 norm ξ as claimed. It is tempting to guess that we can save more ln B factors for larger k. However, inspecting λ 2 norm shows that, say, If you do not want to assume the Riemann hypothesis then only weaker bounds E(x) on |π(x) -Li(x)| can be used. In Ford (2002a,b) we found the following explicit bounds, the first one he attributes to a paper by Y. Cheng which we could not find. Fact 6.1. • For x > 10 we have • There is a constant C and a frontier x 0 such that for x > x 0 we have Admittedly, these bounds only start to be meaningful at large values of x (eg. the first statement around 10 159 299 ). All those bounds are of the form: For all x > x 0 holds. Here, C > 0, x 0 > 0, c 0 ∈ R, c 1 > 0, c 2 > 0 and A > 0 are given parameters (which are not always known). Note that we have that is decreasing for large x. Actually, with the parameter sets from Fact 6.1 this is already true for x ≥ 5. Moreover, the quotient of the relative errors at x 1+α and at x is bounded (or even tends to zero) with x → ∞ for any α > 0. This follows from E(x) / x decreasing when α is constant, but you may also consider values for α that increase when x grows. Revisiting the proof of Theorem 5.1 shows that only (5.2), (5.3), and the initial values c 1 and c 2 depend on the specific bound E. We now use the following recursion for the bounds: for k ≥ 1 based on c 0 = 0, and possibly values for c 1 and c 2 . To bound u tightly the trickiest step is bounding the integral. As our interests lie elsewhere we take the easy way out. We integrate by parts and use that E(x) / x is decreasing for the following rough estimate Thus u is bounded by In the following we neglect the bounded term, as we can compensate its effect for example by a small additional factor. Since u is small for large B, we expect c k to be dominated by c 1 = u. Precisely, for k > 1 we have Theorem 6.2. Assume that E(x) bounds |π(x) -Li(x)| and E(x)/x is decreasing for x > x 0 and the relative error decreases fast, ie. is bounded under the chosen behavior of α. Then for any k ≥ 2 and B large we have This is close to optimal, we only loose a factor of order ln B in the relative error compared to the used error bound in the prime number theorem: The assumptions on E also hold for explicit error bounds with E(x) ∈ O x ln ℓ x . Due to the lost ln B the result is only meaningful if ℓ ≥ 3, so Rosser & Schoenfeld (1962) does not suffice. From Dusart (1998) we can use E(x) = 2.3854 x ln 3 x for x > 355 991, and obtain We have a look at the quality of Theorem 3.5 when applied to our inspiring application. There B = 1100 • 10 6 , C = 2 37 -1, α = ln(C)/ ln(B) -1 = 0.232 , and the largest k of interest is k = 4. For these parameters we find That is a great loss when we try to enclose the function of interest in a small interval. Actually, we can improve the theorem for the price of a slightly more complicated recursion. The present result was based on approximating κ(̺) = 1 ln p for p = B 1+̺α ∈ ]B, C] by λ(̺) = 1 ln B in the recursion of Definition 3.1: We get a better bound by using Definition 7.1. For x ≥ 0 we let ν 0 := κ 0 B,C , and recursively for k > 0 Theorem 7.2. Write C = B 1+α and fix k ∈ N >0 . Then for x ∈ R >0 we have In the light of the inspiring application we now find When we started to think about solving the recursion in Definition 7.1 our first trial was to reuse the polynomial hills m k . Yet, that didn't want to fit nicely. Instead we learned from our calculations that an exponential density instead of a linear one would be easier to connect to the polynomial hills. So we tried to approximate like this and got The exponent in η is chosen such that for ̺ = 0 and ̺ = 1 we have equality. It turns out that this approximation is even better than the one before and at the same time easier to handle. Thus we replace the functions ν k with another family η k : Definition 7.3. For ξ ∈ R we let η 0 := κ 0 B,C , and recursively for k > 0 B 1+̺α d̺ . Theorem 7.4. Write C = B 1+α and fix k ∈ N >0 . Then for x ∈ R >0 we have Proof. To prove this we have to relate the function κ : [0, 1] → R >0 , ̺ → 1/ ln B 1+̺α occurring in the definition of κ k B,C to the function η : [0, 1] → R >0 , ̺ → η(̺) replacing it in the definition of η k . Routine calculus shows that the function κ/η is at most 1, namely at ̺ = 0 and ̺ = 1, and assumes its minimum value ln(1+α) α (1 + α) Testing this with the parameters α = 0.232 and k = 4 from our inspiring application we obtain ln(1 To get a better impression we have plotted the lower interval boundary as a function of α for all three cases in Figure 7.1. We see that for small values of α we obtain good approximations of κ B,C and all our attempts give only weak results for large α, but the one with η is always best. If we now rewrite the recursion to one for we find that η k norm = M η k-1 norm . So we'll obtain the solution from the polynomial hills as in Theorem 4.4 for λ k : The only difference is that instead of α we have α -ln(1+α) ln B . With this replacement Theorem 4.4 becomes: Considering κ k B,C (x) we immediately observe that the ordering of the counted prime lists are not important and we can group together many such elements. To get a precise picture, we define the sorting of a tuple P = (p 1 , . . . , p k ) in the following way: Given any type T = (T 1 , . . . , T r ), there are exactly k T = k! T 1 !•••Tr! different sortings S of type T . This corresponds to possible reorderings of a specific vector P ∈ A S (x) for a sorting S of type T . The type of such a vector P is defined to be T (S). It is clear that the type of P is invariant under permutations, yet not its sorting. Lemma 8.1. Let T be a type for k elements. (i) There exists a sorting S(T ) of type T such that all vectors in A S(T ) (x) are increasing. (ii) If T (S) = T then there is a permutation σ of k elements such that for all x we have A S (x) = A S(T ) (x) σ . (iii) More precisely, for any sorting S of k elements the following are equivalent: Noting that # {S T (S) = T } = k T we have On the other hand we have . Actually, for large B (and C and x) we have π x). This stems from the following fact that #A S (x) is asymptotically much smaller than #A S(1,...,1) (x) for any sorting S of k elements of type different from (1, . . . , 1). Lemma 8.2. For any sorting S of k elements of type different from (1, . . . , 1) there is a sorting S ′ of k -1 elements such that we have # Proof. Take S as specified. Let t be a position which does not occur as a singleton in S. Further, say t ∈ S τ , and let r = #S τ ≥ 2. Let S -be the sorting with S τ removed, and S ′ the sorting with t removed. (Retaining the old indexing is easier, yet then indices run over {1, . . . , k} \ S τ , or {1, . . . , k} \ {t}, respectively.) Then Here, A ′ denotes a variant of A consisting of prime vectors where at the positions j which should have smaller primes than p t originally still satisfy that, and same for positions with primes larger than p t . For the inequality note that S -= (S ′ ) -. Since obviously #A S (z) ≤ z we are done. shows that there must be a large summand, which can be only #A S(1,...,1) (x). The number s(k) = T k T of sortings of k elements is called ordered Bell number. We can also recursively define them: Using Lemma 8.2 for a comparison yields the -now immediate -following Lemma 8.3. We have Compared to the error bound in We are going to combine the results of the last section with Theorem 4.16 and Theorem 5.1 to finally arrive at our main theorem. Analogously to Definition 3.1, we define an approximation function for the function π k B,C (x). Definition 9.1. For x ≥ 0 and k ≥ 0 we define Similarly to Definition 3.1 we can also recursively define π k B,C (x) by It is also possible to define π k B,C (x) similarly based on Definition 3.1. We can now describe the behavior of π k B,C nicely and give our main result. Theorem 9.2. Given x ∈ R >0 and k ∈ N. Then the inequality Proof. By Lemma 8.3 we have Thus using the triangle inequality and Theorem 3.2, we obtain which proves the claim. Theorem 9.3. Fix k ≥ 2. Then for any ε > 0 and B tending to infinity, there are for Proof. By Definition 9.1 we have Theorem 3.5 tells us that there is a value s ∈ implying that for the same value s we have Considering π k B,C (x) we have by Definition 9.1 that Applying Theorem 3.5 gives a value s ∈ This proves the theorem. Unrolling our results on λ k (x) and λ k (x), namely Theorem 4.16 and Theorem 5.1, gives a slightly weaker result. Theorem 9.4. Let B < C = B 1+α with α ≥ ln B √ B and fix k ≥ 2. Then for any (small) ε > 0 and B tending to infinity we have for Proof. By Theorem 9.3 we have values s, s ∈ 1 (1+α) k , 1 such that By Theorem 4.16 we have for ε small enough that and by Theorem 5.1 that This gives the claim. To discuss the quality of our results we consider again the example parameters B = 1100 • 10 6 , C = 2 37 -1, α = ln(C)/ ln(B) -1 = 0.232 , k = 4 from our inspiring application. For k = 2, k = 3 we can give similar pictures; of course, the errors are even smaller in these cases. At present we do not have efficient algorithms for computing κ k B,C itself. However, based on our estimates we can compute values for encapsulating intervals in three variants, listed in increasing quality: • λ estimate (Theorem 3.5): • η estimate (Theorem 7.4): • κ estimate (Theorem 3.2): The κ estimate was easiest to obtain and is of course the most accurate one, however, it is difficult to evaluate. The λ estimate was easy to obtain and compute. But it is of course the least accurate of the three. The η estimate was slightly more difficult to find, is as easy to evaluate as the prior, and it is much more accurate. As usual we write x = B k+ξα and use ξ as a running parameter. Figure 10.1 shows the absolute behavior of all estimates. We observe that the absolute errors at the right margin are huge. This is expected as also the error estimates in the prime number theorem only bound the relative error. However, the picture completely conceals information about the middle and the left part of the interval [0, k]. To see more we divide by x = B k+ξα and therefore obtain estimates for the ratio of ]B, C]-grained integers x in Figure 10.2. This reveals a lot about the quality of our estimates. The black area indicates the best that we could hope for, namely the estimate based merely on Prime number theorem 0.3. However, as this is difficult to evaluate we have to approximate once more. The λ estimate, shown in light gray, is clearly only of to get a rough idea. The η estimate, however, is rather close to the actual behavior and may well serve as a basis for stochastic fine tuning of algorithms like the general number field sieve. Last, Figure 10.3 illustrates the size of the various errors terms relative to κ k : λ error: 1 -1 (1+α) k λ k (x). λ error: λ k (x). Unconditional λ error: λ k (x). κ error: κ k B,C (x). Unconditional κ error: κ k B,C (x). Non-squarefree error bound: 2 k-1 k! -s(k) x B . Non-squarefree error: For the unconditional errors we use Dusart's unconditional bound on |π(x) -Li(x)| which is given by E(x) = 2.3854 x ln 3 x for x ≥ 355 991. The figure shows that all the error terms but the λ error are sufficiently small. Our best choice is the η estimate which is ruled by the η error and the λ error. Both are fairly less than 3% of the target value at least in the middle of the interval. The estimations are more difficult close to the boundaries. It is also positive that, at the parameters of our interest, most error terms are still comparative in size to the contributions of the κ error, which is induced by the prime number theorem. Loebenberger & Nüsken