We examine the problem of writing every sufficiently large even number as the
sum of two primes and at most $K$ powers of 2. We outline an approach that only
just falls short of improving the current bounds on $K$. Finally, we improve
the estimates in other Waring--Goldbach problems.Comment: Second version: 10 pages, another problem adde

Platt, Dave

Trudgian, Tim

English

arXiv

Platt, David J

Trudgian, Timothy

Explore Bristol Research

Linnik's approximation to Goldbach's conjecture, and other problems

We examine the problem of writing every sufficiently large even number as the sum of two primes and at most K powers of 2. We outline an approach that only just falls short of improving the current bounds on K. Finally, we improve the estimates in other Waring–Goldbach problems

Platt, D. J.

Trudgian, T. S.

The Australian National University

Linnik’s approximation to Goldbach’s conjecture, andother problemsD.J. PlattHeilbronn Institute for Mathematical ResearchUniversity of Bristol, Bristol, UKdave.platt@bris.ac.ukandT.S. Trudgian∗Mathematical Sciences InstituteThe Australian National University, ACT 0200, Australiatimothy.trudgian@anu.edu.auJanuary 20, 2015AbstractWe examine the problem of writing every sufficiently large even number as the sum oftwo primes and at most K powers of 2. We outline an approach that only just fallsshort of improving the current bounds on K. Finally, we improve the estimates inother Waring–Goldbach problems.1 IntroductionThe Goldbach conjecture is that every even n > 4 can be written as a sum of two primenumbers. Linnik proved that there exists a finite K such that, for all sufficiently large evenn, one may writen = p+ q + 2ν1 + 2ν2 + · · ·+ 2νr , (1)where p and q are primes, the νi are positive integers, and where r ≤ K. For a historicaldevelopment on bounding the size of K, see [4, §1].∗Supported by Australian Research Council DECRA Grant DE120100173.1Significant improvements on the size of K were made by Heath-Brown and Puchta [4] and,independently, by Pintz and Ruzsa [26]. Heath-Brown and Puchta showed that K ≤ 13, andon the Generalised Riemann Hypothesis (GRH) that K ≤ 7. Pintz and Ruzsa establishedthis latter result and announced that they could show unconditionally that K ≤ 8. Thispaper is yet to appear in print. Elsholtz, in an unpublished manuscript, showed that K ≤ 12;this was proved independently by Liu and Lü [21].The methods of Heath-Brown and Puchta and of Pintz and Ruzsa allow one to show thatK is admissible in (1) provided thatλK−2 <C3(C1 − 2)C2 + cC−10 log 2, (2)for certain constants C0, C1, C2, C3, λ and c. The inequality (2) follows from §6 in [4] andthe relations (8.20) and (8.21) in [26], which, although proved under GRH, give analogousresults unconditionally.We haveC0 =∏p>2(1− (p− 1)−2),where, according to Wrench [8],0.6601618158 < C0 < 0.6601618159. (3)Indeed, Wrench computed C0 to 45 digits; the truncated bound in (3) is certainly fit forpurpose.As for C2 we haveC2 =∞∑d=1k(2d− 1)ε(2d− 1),where k(n) is a multiplicative function defined byk(pe) ={0, p = 2 or e ≥ 2,(p− 2)−1, otherwise,and where ε(d) is the multiplicative order of 2 modulo d. Lemma 2′′ of [25] shows that1.2783521041 < C2C0 < 1.2784421041, which may be combined with (3) to give1.93642 < C2 < 1.93656.The constant C1 is that which appears in Chen’s work on the twin prime conjecture.Heath-Brown and Puchta take C1 = 7.8342 as in [2]; Liu and Lü take C1 = 7.8209 as in[30].1The constant C3 satisfies2∑d≤Dk(d)H(d;N,K)ε(d)−K ≥ C3,1It is claimed in [29] that one may take C1 = 7.81565; this proof appears to be incomplete.2in which D is a fixed number, andH(d;N,K) = #{(ν1, . . . , νK) : 1 ≤ νi ≤ ε(d), d|N −∑2νi}.Heath-Brown and Puchta used D = 5 to show that C3 ≥ 2.7895; Liu and Lü used D = 11to show that C3 ≥ 2.8096; it is remarked in [4] that Elsholtz used D = 21 to show thatC3 ≥ 2.96169.To define c and λ we first defineGL(x) =L−1∑j=0e(2jx), L = blogN/ log 2c, (4)where e(x) = exp(2πix). For c > 0 we seek to find the smallest positive value of λ such that∆ = meas(x ∈ [0, 1] : |GL(x)| > λL) < N−clog 2 . (5)Heath-Brown and Puchta showed that one may take c = 109/154 unconditionally; Pintz andRuzsa claim that one may take c = 3/5. Both sets of authors show that one may take c = 12on GRH. Heath-Brown and Puchta showed that λ ≤ 0.863665; Liu and Lü improved thisto λ ≤ 0.862327. On GRH, Heath-Brown and Puchta showed that λ ≤ 0.722428. Pintz andRuzsa used an improved method to show that λ ≤ 0.716344.Using(C0, C1, C2, C3, λ) = (0.6601618159, 7.8209, 1.93656, 2.96169, 0.862327),and the same quintuple with the last entry replaced by 0.716344, one may see that (2) issatisfied for K ≥ 11.4549 and K ≥ 6.1432. This proves only what is already known, viz.that one may take K = 12, and K = 7 on GRH.It is clear that there is little to be gained by pursuing improvements in C0 and in C2.It appears difficult to improve C1 at all substantially — see [30, Rem. 2, p. 253]. In §2 weimprove the value of C3; in §3 we investigate λ. Finally, in §4 we improve on estimates forsome related problems.2 Computing C3Heath-Brown and Puchta examined all those d ≤ D for which 2 is a primitive root modulod. They stated that, in this case,H(d;N,K) ={1d{(d− 1)K − (−1)K}, d - N,1d{(d− 1)K + (−1)K(d− 1)}, d | N.To improve on the value we could take for C3 we wished to consider more general d.Algorithm 1 describes the approach we adopted.3input: d > 1, an odd integer, and K > 1 an integerW ← a vector of length d indexed byW [0] . . .W [d− 1];Y ← a vector of length d indexed byY [0] . . . Y [d− 1];for i← 0 to d− 1 doW [i]← #{ν|1 ≤ ν ≤ ε(d), 2(K − 1) + 2ν ≡ i (mod d)}endfor k ← 2 to K dofor j ← 2 to ε(d)− 1 dor ← 2j − 2;for i← 0 to d− 1 doY [i+ r (mod d)]← W [i]endfor i← 0 to d− 1 doW [i]← W [i] + Y [i]endendendres← W [0];for i← 1 to d− 1 doif W [i] < res thenres← W [i]endendreturn res;Algorithm 1: Computing H(d;N,K).4Setting up the initial vector requires d+ ε(d) = O(d) steps. We then copy and add the dvector entries ε(d) times for each of the remaining K − 1 powers of 2 giving a total cost ofO(Kd2) for each d. Computing H for all the d less than D therefore has time complexity ofO(KD3) where each operation is on a number of size O(D log k) bits. The space requirementis O(D2 log k).Following a suggestion by Roger Heath-Brown, we introduce a slight variation of theabove argument. Let the worst residue class modulo d be that which contributes the least toH(d;N,K). Suppose, for example, that the worst residue class modulo 15 is N ≡ 0, and theworst residue class modulo 3 is N ≡ 1. Since there can be no values of N that belong to bothresidue classes, this ‘worst of the worst’ scenario does not arise. We therefore limit ourselvesto admissible values: that is, sets of residue classes for N that could be simultaneouslysatisfied. Modifying the above Algorithm 1 to take advantage of this is a trivial matter.2.1 Implementation and ResultsWe implemented the above algorithm in C++ using GMP [3] for the large integer arithmeticand MPFI [28] to handle floating point quantities as intervals, thereby avoiding any issueswith rounding. We summed over all d ≤ 40, 000 and considered admissibility modulo 255, 255to obtain C3 ≥ 3.011112 for K = 6 and C3 ≥ 3.02858417 for K = 11. The computationsrequired 58 hours in the case of K = 6 and 116 hours in the case of K = 11 on a single coreof a 1.8 GHz Intelr Xeonr E5-2603.By modern standards these run times are modest; it would be a simple matter to go higherin d. However, we expect the returns to be very small based on the following argument. Weshall ignore admissibility since the improvement we observed when introducing it to ouralgorithm was small. In this case, the best we can hope for is an even distribution of thecounts over all the residue classes modulo d. Thus we expect H(d;N,K) to be about ε(d)K/d,whence the contribution from each d to C3 will be no more than2k(d)d. We note that thistreatment removes the dependency on K. Further, we have trivially that k(d) ≤ (d − 2)−1for all d > 3. Therefore2∑d>40,000k(d)d< 2∞∫40,000dtt(t− 2)< 5.1× 10−5.It seems that the potential gains from further computation are limited.3 Computing λ and KTo estimate λ, we use the method given by Pintz and Ruzsa [26]. Essentially one wishes toapproximate GL(x) in (4) by estimating the error in G2h(x)−GL(x) for h large. It is to thispurpose that §§3-7 of [26] are dedicated. Heath-Brown and Puchta took h = 16; Liu andLü took h = 23. Using the method of Pintz and Ruzsa we are able to take h considerablylarger.5The authors are grateful to Alessandro Languasco who supplied his Pari [1] implemen-tation of this algorithm. We took h = 2138 with polynomials2 of degree 40 to obtainλ ≤ 0.8594000 unconditionally and λ ≤ 0.7163436 on GRH. The computations take about15 minutes using Pari, or about double that when implemented in C using the intervalarithmetic package ARB [7]. The latter approach confirms that the stated values for λ areaccurate to the precision given.With (C3, λ) = (3.02858417, 0.8594000) in the unconditional case and with (C3, λ) =(3.011112, 0.7163436) on GRH we haveK ≥ 11.0953 unconditionally, K ≥ 6.09353 on GRH. (6)This means we are just short of being able to take K = 11 and, on GRH, K = 6. Given thedifficulty in improving the values of C3 and λ with existing methods, it seems that a newidea is needed to improve the estimate on K.As a consolation prize, we applied the same code to some other problems. For theseproblems one has a possibly different value of c for which one wishes to calculate a smallvalue of λ. The results are summarised in Table 1.4 Other Waring–Goldbach problemsSuppose it is conjectured that for all sufficiently large N we have N = f1+ · · ·+fr for certainnumbers fi. Suppose that we can prove the following approximation of this conjecture, thatN = f1+ · · ·+fr+2ν1 + · · ·+2νr , where r ≤ K for some K. Just as in Linnik’s approximationto Goldbach’s conjecture, one seeks good bounds on K. Various approximations have beengiven to problems involving sums of powers of primes. We investigate ten of them below.For the following problems the value of c has been established in the literature. It maybe possible to improve this value and some of the other arguments that lead to the estimateson K in problems (A)-(J). We have not pursued this: we limited ourselves to improving thevalue of λ since this appears to be the most influential parameter.4.0.1 Even numbers as sums of four squares of primesN = p21 + p22 + p23 + p24 + 2ν1 + 2ν2 + · · ·+ 2νKA . (A)This was considered in [14, 13, 12, 10] and most recently by Zhao [31] who showed thatKA ≤ 46.4.0.2 Odd numbers as sums of a prime and two squares of primesN = p1 + p22 + p23 + 2ν1 + 2ν2 + · · ·+ 2νKB . (B)This was considered in [14, 15, 6, 11, 24] and most recently by Liu [19] who showed thatKB ≤ 35.2For details of these polynomials the reader is invited to examine §6 in [26].64.0.3 Even numbers as sums of eight cubes of primesN = p31 + · · ·+ p38 + 2ν1 + 2ν2 + · · ·+ 2νKC . (C)This was considered in [20] and most recently by Liu [16] who showed that KC ≤ 341.4.0.4 Odd numbers as sums of a prime and four cubes of primesN = p1 + p32 + · · ·+ p35 + 2ν1 + 2ν2 + · · ·+ 2νKD . (D)This considered by Liu and Lü [22] who showed that KD ≤ 106.4.0.5 Even numbers as sums of two squares of primes and four cubes of primesN = p21 + p22 + p33 · · ·+ p36 + 2ν1 + 2ν2 + · · ·+ 2νKE . (E)This was considered by Liu and Lü [22] who showed that KE ≤ 211.4.0.6 Even numbers as sums of a prime, a square of a prime and two cubes ofprimesN = p1 + p22 + p33 + p34 + 2ν1 + 2ν2 + · · ·+ 2νKF . (F)This was considered by Liu and Lü [23] who showed that KF ≤ 161.In the nextfour problems one asks when the equations are true simultaneously for positiveeven integers B1 and B2 with B1 > B2.4.0.7 Even numbers as sums of two primes, simultaneouslyB1 = p1 + p2 + 2ν1 + 2ν2 + · · ·+ 2νKGB2 = p3 + p4 + 2ν1 + 2ν2 + · · ·+ 2νKG .(G)This was considered by Kong [9] who showed that KG ≤ 63 and that, on GRH, KG ≤ 31.4.0.8 Even numbers as sums of four squares of primes, simultaneouslyB1 = p21 + p22 + p23 + p24 + 2ν1 + 2ν2 + · · ·+ 2νKHB2 = p25 + p26 + p27 + p28 + 2ν1 + 2ν2 + · · ·+ 2νKH .(H)This was considered in [18] and most recently by Hu and Liu [5] who showed that KH ≤ 142.34.0.9 Even numbers as sums of eight cubes of primes, simultaneouslyB1 = p31 + · · ·+ p38 + 2ν1 + 2ν2 + · · ·+ 2νKIB2 = p39 + · · ·+ p316 + 2ν1 + 2ν2 + · · ·+ 2νKI .(I)This was considered by Liu [18] who showed that KI ≤ 1432.43Though it seems that Hu and Liu’s proof actually gives KH ≤ 141.4Note that, on [18, p. 3347] Liu uses b = 268096, which comes from Ren [27]. This value has beenimproved in [16] to b = 147185.22. This gives at once that KI ≤ 1364.74.0.10 Odd numbers as sums of one prime and two squares of primes, simulta-neouslyB1 = p1 + p22 + p23 + 2ν1 + 2ν2 + · · ·+ 2νKJB2 = p4 + p25 + p26 + 2ν1 + 2ν2 + · · ·+ 2νKJ .(J)This was considered by Liu [17] who showed that KJ ≤ 332.Table 1: Improvements on (A)-(J)Required c Old λ New λ Old K New K(A) 3/4 0.887167 0.8844473 46 45(B) 3/4 0.887167 0.8844473 35 34(C) 19/21 0.965411 0.9642399 341 330(D) 113/126 0.961917 0.9606646 106 102(E) 53/63 0.935746 0.9339489 211 205(F) 109/126 0.947313 0.9457435 161 156(G) 109/154 0.862327 0.8594000 63 62(G) on GRH 1/2 0.716344 0.7163436 31 31(H) 3/4 0.887167 0.8844473 142 138(I) 19/21 0.965411 0.9642399 1432 1319(J) 3/4 0.887167 0.8844473 332 323References[1] C. Batut, K. Belabas, D. Bernardi, H. Cohen, and M. Olivier. User’s Guide to PARI-GP,2000.[2] J.-R. Chen. On the Goldbach’s problem and the sieve methods. Sci. 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Motivations for an arbitrary precision interval arithmeticand the MPFI library. Reliab. Comput., 11(4):275–290, 2005.[29] D.-H. Wu. An improvement of J. R. Chen’s theorem. Shanghai Keji Daxue Xuebao,(1):94–99, 1987.[30] J. Wu. Chen’s double sieve, Goldbach’s conjecture and the twin prime problem. ActaArith., 114(3):215–273, 2004.[31] L. Zhao. Four squares of primes and powers of 2. Acta Arith., 162(3):255–271, 2014.10

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D.J. Platt

T.S. Trudgian

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Linnik's approximation to Goldbach's conjecture, and other problems

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