CMST 20(2) 39-47 (2014) DOI:10.12921/cmst.2014.20.02.39-47

Some remarks on the Báez-Duarte criterion for the

Marek Wolf

Cardinal Stefan Wyszynski University Faculty of Mathematics and Natural Sciences. College of Sciences ul. Wóycickiego 1/3, Auditorium Maximum, (room 113) PL-01-938 Warsaw, Poland e-mail: [email protected]

Received: 4 March 2014; revised: 9 April 2014; accepted: 11 April 2014; published online: 25 April 2014

Abstract: In this paper we are going to describe the results of the computer experiment, which in principle can rule out validity of the Riemann Hypothesis (RH). We use the sequence ck appearing in the Báez-Duarte criterion for the RH and compare two formulas for these numbers. We describe the mechanism of possible violation of the Riemann Hypothesis. Next we calculate c100000 with a thousand digits of accuracy using two different formulas for ck with the aim to disprove the Riemann Hypothesis in the case these two numbers will differ. We found the discrepancy only on the 996th decimal place (accuracy of 10−996). The computer experiment reported herein can be of interest for developers of Mathematica and PARI/GP. Key words: Riemann Hypothesis, zeta function, Baez-Duarte criterion

I. INTRODUCTION to prove RH in Chapter 8. Presently the requirement that the 0 nontrivial zeros are simple ζ (ρl) 6= 0 is often added. With the advent of the computer era the computing ma- The first use of computers in connection with RH was by chines have been used to prove mathematical . Allan Turing checking whether the nontrivial zeros of ζ(s) 1 The most spectacular examples of such a use of computers have indeed real part 2 [5]. Turing suspected that the RH is were proofs of the four color [1, 2] and of the Ke- not true and the first is lying relatively low. pler about sphere packing in three-dimensional Let us quote the sentence from the first page of his paper: Euclidean space [3]. It seems to be not possible to use com- “The calculation were done in an optimistic hope that a zero puters for the proof of the Riemann Hypothesis (RH), but its would be found off the critical line”, but up to t = 1540 Tur- refutation by numerical calculations seems to be plausible. ing found that all zeros are on the critical line. The present The Riemann Hypothesis says that the series record belongs to Xavier Gourdon [6] who has checked that all 1013 first zeros of the Riemann ζ(s) lie on the critical ∞ X 1 line. checked that RH is true in different ζ(s) = , (s = σ + it, <(s) > 1) (1) 20 21 22 ns intervals around 10 [7], 10 [8], 10 [9], but his aim was n=1 not verifying the RH but rather provide evidence for con- analytically continued to the complex plane in addition to jectures that relate nontrivial zeros of ζ(s) to eigenvalues of trivial zeros ζ(−2n) = 0 has nontrivial zeros ζ(ρl) = 0 in random matrices. In fact, Odlyzko expressed the view that the critical strip 0 < <(s) < 1 only on the critical line: off critical line zeros could be encountered at least at t of the 1 1 1010000 <(ρl) = 2 i.e. ρl = 2 + iγl, see e.g. the modern guide to the order 10 , see [10]. Asked by Derbyshire “What do you RH [4]. In the same book there is a review of failed attempts think about this darn Hypothesis? Is it true, or not?” Odlyzko 40 M. Wolf replied: “Either it’s true, or else it isn’t”. Also other famous where mathematicians John E. Littlewood and Paul Turán did not 1  s  ξ(s) = s(s − 1)Γ ζ(s) believe RH is true. Aleksandar Ivic´ gave a few arguments 2 2 against the truth of the RH, see [4, ] and on arxiv [11]. fulfills: There were several attempts to use computers to disprove some related to RH in the past. Sometimes it λn ≥ 0 for n = 1, 2,... (5) was sufficient to find a counterexample to the given hypoth- The explicit expression has the form: esis, sometimes the disproof was not direct. For example, X n the Haselgrove [12] disproved the Pólya’s Conjecture stat- λn = (1 − (1 − 1/ρ) ). (6) ing that the function ρ X K. Maslanka´ [21-23] has performed extensive computer cal- L(x) := λ(n) (2) culations of these constants confirming (5) with large sur- n≤x plus. satisfies L(x) ≤ 0 for x ≥ 2, where λ(n) is Liouville’s Let us mention also the elementary Lagarias criterion function defined by [24]: to disprove the RH it suffices to find one n that has λ(n) = (−1)r(n) so many divisors, that: where r(n) is the number of, not necessarily distinct, prime X d > Hn + exp(Hn) ln(Hn), (7) r1 rn factors in n = p1 ··· pα(n), with multiple factors counted d|n according to their multiplicities: r = r +...+r . From the 1 n where H = Pn 1/k is the n-th harmonic number. truth of the Pólya Conjectures the RH follows, but the con- n k=1 The Lagarias criterion is not well suited for computer verifi- verse statement is not true. The Haselgrove proof was indi- cation and in [25] Keith Briggs has undertook instead the rect, and in 1960 Lehman [13] found on the computer ex- verification of the Robin [26] criterion for RH: plicit counter–example: L(906180359) = 1. X The next example is provided by the Mertens conjecture. RH ⇔ d < eγ n ln ln(n) for n > 5040 (8) Let M(x) denote the defined by d|n X M(x) = µ(n), (3) For appropriately chosen n Briggs obtained for the differ- n≤x ence between r.h.s. and l.h.s. of the above inequality value as −13 −6 where µ(n) is the Möbius function small as e ≈ 2.2 × 10 , hence again RH is very close to being violated.  1 n = 1  In this paper we are going to propose a method which in µ(n) = 0 p2|n principle can provide a refutation of the RH. The idea is to  (−1)r n = p p . . . p , p 1 2 r i distinct calculate a number with very high accuracy (one thousand From digits) in two ways: one without any knowledge on the zeros 1 |M(x)| < x 2 (4) of ζ(s) and second using the explicit formula involving all again the RH would follow. However, in 1985 A. Odlyzko ρl. Despite some estimation presented in Sect. 3 indicating and H. te Riele [14] disproved the Mertens conjecture, again that the discrepancy could be found merely at much higher not directly, but later it was shown by J. Pintz [15] that than a thousand decimal places we performed the calcula- the first counterexample appears below exp(3.21 × 1064). tions in an optimistic hope that we will find the discrepancy The upper bound has since been lowered to exp(1.59×1040) between these two numbers, paraphrasing the sentence of [16]. Turing. There is a lot of number theoretic functions often Especially interesting is the value of the de Bruijn- defined in an elementary way being expressed also by the Newman constant Λ, see e.g. §2.32 (pp. 203-205) in [17]. “explicit” formulas in terms of zeros of the ζ(s) function. Unconditionally Λ ≤ 1/2 and the Riemann Hypothesis is Let us mention here the Chebyshev function X equivalent to the inequality Λ ≤ 0. The fascinating run ψ(x) = Λ(n) (= ln(lcm(2, 3, · · · bxc))), Λ Λ > for the best lower bound on ended with the value n≤x −2.7 × 10−9 obtained by Odlyzko [18]. Such a narrow gap where the von Mangoldt function Λ(n) is defined as for values of Λ being compatible with RH allowed Odlyzko to make the remark: “the Riemann Hypothesis, if true, is just  ln p for n = pm Λ(n) = barely true”. However, Don Zagier [19] interpreted it as “the 0 in other cases RH, if false, is only slightly false”. The explicit formula reads, see eg. [27] (the term ln 1 − 1  In 1997 Xian-Jin Li proved [20], that Riemann Hypo- x2 comes from trivial zeros): thesis is true iff the sequence:   ρl n 1 1 X x 1 d n−1 ψ(x) = x−ln(2π)− ln 1 − 2 − (9) λn = n (s ln ξ(s))|s=1 2 x ρl (n − 1)! ds nontr. zeros ρl Some remarks on the Báez-Duarte criterion for the Riemann Hypothesis 41

Also the Mertens’s function has the explicit representa- where tion (the last term is comprising contribution from all trivial k   zeros) [27]: X k 1 c = (−1)j ≡ k j ζ(2j + 2) X X xρl j=0 µ(n) = 0 − 2 (14) ρlζ (ρl) k n≤x nontr. X k (4j + 4)! zeros ρl j+1 ≡ (−1) 2j+1 2j+2 . (10) j 2 B2j+2π ∞ 2n j=0 X 2π  1 − (−1)n x (2n)!nζ(2n + 1) n=1 The problem with these series is that they are extremely If additionally slow converging because the partial sums oscillate with  −3/4 amplitudes diminishing at very slow rates. For example ck = O k (15) ψ(1000001) = 999586.597 ..., while from (9) summing then all zeros of ζ(s) are simple. Báez-Duarte showed over 5,549,728 zeros gives 999587.15 ..., thus relative error unconditionally (regardless of validity of the RH) slower is 0.000055. decrease c = O(k−1/2). The plot of c for k = In the computer experiment reported here we were able k k 1, 2,..., 1000000 is presented in Fig. 1. We anticipate here to get discrepancy less than 10−996 between the quantity that the formula (14) contains contribution from all zeros of calculated from the generic formula and from an explicit ζ(s). one summed over only 2600 nontrivial zeros computed with 1000 significant digits. This paper can be regarded as a con- tinuation of our investigation reported in [28].

II. THE BÁEZ-DUARTE CRITERION FOR THE RIEMANN HYPOTHESIS

We begin by recalling the following representation of the ζ(s) function valid on the whole complex plane without s = 1 found by Krzysztof Maslanka´ [29, 30]:

∞ s  1 X Γ k + 1 − Ak ζ(s) = 2 (11) s − 1 Γ 1 − s  k! k=0 2 where Γ(z) is the Euler gamma function and

k X k A := (−1)j (2j + 1)ζ(2j + 2) ≡ 6 k j Fig. 1. The plot of the Báez-Duarte sequence ck for k ∈ (1, 10 ). j=0 The equation for the envelope was obtained from the explicit equa- (12) tion (27): for large k the oscillating part c˜k is dominant and for k   j+1 2j+1 2j+2 X k (−1) 2 B2j+2π k > 100000 ck fits well between the red lines, for details see [28] ≡ . j (4j + 4)! j=0

Above we have used the fact that ζ(2n) = In [32] Báez-Duarte was able to express ck as a sum over n−1 2n−1 2n (−1) 2 B2nπ /(2n)! where B2n are Bernoulli zeros of ζ(s). The explicit formula for the sequence ck can numbers. be written as a sum of two parts: quickly decreasing with k The expansion (11) provides an example of the analyt- trend c¯k arising from trivial zeros of ζ(s) and oscillations c˜k ical continuation of (1) to the whole complex plane except involving complex nontrivial zeros: s = 1. Since Ak tend to zero sufficiently fast as k → ∞ the expansion (11) converges uniformly on the whole com- ck =c ¯k +c ˜k (16) plex plane [31]. Based on the representation (11) Luis Báez- Because the derivatives ζ0(−2n) at trivial zeros are Duarte in [32] proved that RH is equivalent to the statement known analytically: that   (−1)nζ(2n + 1)(2n)! c = O k−3/4+ε , ∀ε > 0, (13) ζ0(−2n) = (17) k 22n+1π2n 42 M. Wolf

1 Maslanka´ in [33] was able to give the closed expression for Now we assume the RH: ρl = 2 + iγl. Then we get for −3/4 trend stemming from zeros ρn = −2n: c˜k the overall factor k – the dependence following from 1 RH, see (13) – multiplied by oscillating terms: c¯ = − k 2 iγ /2 3 i ! (2π) 1 X k l Γ( − γl) c˜ = < 4 2 . (27) ∞ m 2m (18) k 3/4 0 1 X Γ(k + 1)Γ(m) (−1) (2π) k ζ ( 2 + iγl) × . γl>0 Γ(k + m + 1)Γ(2m − 1) ζ(2m − 1) m=2 Using the formula (6.1.45) from [34]:

Báez-Duarte is skipping the trend c¯k remarking only that it 1 1 π|y| 1 −x lim √ |Γ(x + iy)|e 2 |y| 2 = 1 (28) is of the order o(1) (Remark 1.6 in [32]). It is an easy calcu- |y|→∞ 2π lation to show (see [28]) that for large k assuming RH we obtain for large γl > 0: 2   1 3 i √ γ  4 (2π) 1 −3 −πγl/4 l c¯k = − + O(k ), (19) Γ ∓ γl ≈ 2πe , (29) 2ζ(3) k2 4 2 2 −3/4 hence we get exponential decrease of summands in the sum thus the dependence ck = O(k ) in (13) is linked to the (27) over nontrivial zeros giving c˜k and (27) is very fast con- oscillating part c˜k. vergent. Because of that if RH is true the sum (27) will be For c˜k Báez-Duarte gives the formula [32]: dominated by first zero γ = 14.13472514 ..., which leads   1 1 X 1 to the approximate expression (for details see [28]): c˜k−1 = Res 0 ; s = ρl , (20) 2k ζ (s)Pk(s/2)   ρl A 1 c˜k = 3 sin φ − γ1 ln(k) where k 4 2 (30) −5 k A = 7.775062 ... × 10 , φ = 2.592433 .... Y  s P (s) := 1 − (21) For large k the above formula (30) gives a very fast method k r r=1 for calculating quite accurate values of ck, orders faster is the Pochhammer symbol. Assuming zeros of ζ(s) are sim- than (14). g ple we can write: In the following we will denote by ck the values of the Báez-Duarte sequence obtained from the generic for- 1 1 X mula (14) c˜k−1 = 0 . (22) 2k ζ (ρl)Pk(ρl/2) ρl k X k 1 cg = (−1)j An appropriate order of summation over nontrivial zeros is k j ζ(2j + 2) assumed in (20) and (22), see [32, Theorem 1.5]. Because j=0 and by ce the values obtained from explicit formula (16), i.e. (−1)kΓ(s) k Pk(s) = (23) in fact from (18) and (27) as no one zero off critical line is Γ(k + 1)Γ(s − k) known: ∞ collecting in pairs ρl and ρl we can convert (22) to the form: 1 X Γ(k + 1)Γ(m) ce = −   k (2π)2 Γ(k + m + 1)Γ(2m − 1) ρl m=2 k+1 X Γ(k + 1)Γ( 2 − k − 1) c˜k = (−1) < . !  ρl 0  m 2m kiγl/2Γ( 3 − i γ ) Γ( 2 )ζ (ρl) (−1) (2π) 1 X 4 2 l ρl, =(ρl)>0 × + < . 3/4 0 1 ζ(2m − 1) k ζ ( + iγl) (24) γl>0 2 We have found that the numerical calculation of Pk(ρl/2) in PARI/GP directly from the above product (21) is much slower than the use of the Γ(z) functions (23). III. THE SCENARIO OF VIOLATION Báez-Duarte proves in [32, Lemma 2.2] that OF THE RIEMANN HYPOTHESIS

s 1 3 lim Pk(s)k = (25) The condition (15) means that the combination k 4 ck k→∞ Γ(1 − s) should be contained between two paralel lines ±C for all thus for large k we can replace k +1 by k and transform (22) k, where C is the constant hidden in big-O in (15). The vio- in the following way: lation of the RH would manifest as an increase of the ampli- 3 4 ρl/2 tude of the combination k ck and for sufficiently large k (de- 1 X k 3 c˜ = = 4 k 0 ρ /2 pending on C) the product k ck will escape outside the strip 2k ζ (ρl)Pk(ρl/2)k l −5 ρl ±C (if RH is true we can take C = A = 7.7751 ...×10 ).   (26) ρl/2−1 ρl We will discuss below the possible mechanism of violation X k Γ(1 − 2 ) 3/4 of the inequality ck < C/k for the case of simple zeros =<  0  . ζ (ρl) 0 ρl, =(ρl)>0 of zeta function: ζ (s) 6= 0. Some remarks on the Báez-Duarte criterion for the Riemann Hypothesis 43

0 g e The derivatives ζ (ρl) in the denominator of (26) does discrepancy between ck and ck larger than assumed accuracy not pose any threat to RH. First of all it does not depend  sufficiently large value of k = K is needed. The point is 0 3/4 g on k, thus hypothetical extremely small values of ζ (ρl) will that k ck will escape outside the strip ±C for sufficiently only change the constant hidden in big-O in (15). Second, large k = K and the value of such K we can estimate ana- this derivative is taking moderate values for zeros used by lyzing the explicit formula for c˜k. 0 1 us: the smallest ζ ( 2 + γl) was 0.032050162 ... at γ1310 = We can estimate value of the index K from the re- 1771.212945 ... and the largest was 7.7852581838 ... at quirement that the term Kδl/2 (K−3/4 is present in front γ1773 = 2275.06866478 .... We remark that R.S. Lehman of the sum for c˜k) will defeat the smallness of the term (o) in [13] makes the conjecture: Γ(3/4−γl /2) and together their product will overcome the 1 first summand in (26) corresponding to γ1. In other words in = O(ρν ), where 0 < ν < 1. (o) 0 l (31) the series (26) all terms up to γ monotonically and fast ζ (ρl) l decrease but the terms corresponding to zeros off the criti- Some rigorous theorems about the possible large and small cal line can be made arbitrary large for sufficiently large k. values of ζ0(ρ ) proved under the assumption of RH can be l Instead of (29) now will have: found in [35]. We checked that for first 5549728 nontriv- ial zeros of ζ0(ρ ) the largest derivative was 9.38127677 at 3 δ  l Γ ∓ l c − iγ(o) ≈ γ5376610 and the smallest was 0.001028760514 at γ4161179. 4 2 l Let us suppose there are some zeros of ζ(s) off critical (o) √ −1/4±δl/2 −πγ(o)/4γl  line. Next let us assume that we can split the sum over ze- ≈ 2πe l . 2 ros ρl in (26) in two parts: one over zeros on the critical line and second over zeros off the critical line. This second The condition for such a K is roughly: sum should violate the overall term k−3/4 present in the first δ /2 −γ(o) 0 2γ(o)/δ (o) K l e l > C thus K > C e l l . (33) sum. Let ρl denote the zeros lying off the critical line (“o” (o) 1 (o) stands for “off”): ρl = 2 ± δl + iγl , 0 < δl < 1/2 Because δl can be arbitrarily close to zero and, as we ex- (o) (as it is well known the nontrivial zeros are symmetric with pect, γl is very large the value of K will be extremely respect to the critical line zero hence the combination ±δl huge – larger than the famous Skewes number and will look . .. plus there are appropriate complex conjugate zeros below something like 1010 . Thus it is practically impossible to ρl g real axis). In the factor |Γ(1 − 2 )| the off-critical line ze- disprove RH by comparing c and ce . Maslanka´ presented 1 1 k k 4 4 ∓δl/2 ros will lead only to the change of γl → γl in (29) in [33] a discussion of possible violation of (15) and he also and in view of exponential decrease present in (29) the pos- came to the pessimistic conclusion that disproving RH by g e sible violation of RH in (26) will manifest through the terms comparing ck and ck is “far beyond any numerical capabil- ρ /2−1 − 3 ±δ /2+iγ(o)/2 3 +δ /2 k l = k 4 l l . The combination 1/k 4 l ities”, see pp. 7-8. We wanted to find agreement between g e poses no problem as it leads to faster than required in (15) c100000 and c100000 within one thousand digits and to our ce decrease of some terms in the sum for c˜k. But the combi- surprise the first attempt to calculate 100000 resulted in the 3 −δ /2 th nation 1/k 4 l leads to violation of (15) and we want to difference already on the 87 place. We started to struggle elucidate how it can be detected. Say we want to compare cg with numerical problems to improve the accuracy and finally k g e we got 996 digits of c and ce the same. with ck with accuracy , where we are interested in values of 100000 100000 −10...0 g  of the order  = 10 . The expression for ck is a finite sum and we can in principle calculate its value in PARI with practically arbitrary exactness (however, for really large k it IV. THE COMPUTER EXPERIMENT g can take years of CPU time). Although ck contains informa- tion coming from all zeros, to see influence of the first off The idea of the experiment is to calculate to high preci- the critical line zero the value of sufficiently large k has to g e sion the values of c100000 and c100000 and try to find a dis- e g be examined. The sum for ck is infinite and we expect that to crepancy between them. We calculated one value c100000 get accuracy  we have to sum in (27) up to l = L given by from the generic formula (14) which contains contribu- 1/4 (as we skip |γl| we will skip also π/4 as our consideration tion from all zeros of ζ(s), even hypothetical zeros with are not rigorous in general): 1 <(ρl) 6= 2 . Because ζ(2n) very quickly tend 1 to get the −γL c  ≈ e , hence γL ≈ − ln(). (32) firm value of k it is necessary to perform calculations with many digits accuracy. An additional problem is fast growing Because values of the imaginary parts γ(o) of the hypothet- g l of binomial symbols. We performed a calculation of c100000 ical zeros off critical line should be extremely large, per- using the free package PARI/GP [36]. This package allows to 10000 haps even as large as 1010 , see [10], we suppose that perform very fast computations practically of arbitrary pre- (o) (o) g γL < γl . The contribution of γl is present in ck, but will cision. We set precision to 100000 decimals and below in e not be present in the explicit sum for ck cut at L. To detect Table I are the partial sums of (14) recorded after summation 44 M. Wolf of 10000, 20000, ... 100001 terms. In the middle of compu- The arguments given earlier in Sect. 3 suggested that no g 30100 g e tations the partial sums for c100000 were of the order 10 discrepancy should be found between ck and ck for k = to drop finally to 1.609757993 ... × 10−9 after summing up 100000 hence we paid attention to the numerical inaccura- 0 all 100001 terms. Separately we repeated these calculations cies in the computation of the derivative ζ (ρl) as a pos- g with precision set to 150000 places and we found the differ- sible explanation. We stress here that ck contains informa- ence only from the 69900th place. Because the sum (14) is tion on all zeros of ζ(s), even off the critical line, while e finite we got accurate say 50000 digits from this generic for- the sum for ck is truncated, thus ruling out numerical rea- g mula for c100000. This was an easy part which took approxi- sons or mistakes of the program should disprove the Rie- mately 14 hours for precision 100000 decimals and over 20 mann Hypothesis. We played with different options 20, 30, hours for precision 150000 digits on the AMD Opteron 2.6 40 terms and WorkingPrecision in ND[...], but fi- GHz 64 bits processor. nally we got only the moderate improvement: the difference e g e th Now we turn to the calculation of c100000 from the ex- between c100000 and c100000 shifted to 105 place: plicit formulas (18) and (27), which are infinite sums. It is g −105 c100000 −104 important not to make the replacement k + 1 → k even for 10 < − 1 < 10 . (35) ce k = 100000 if we want to get accuracy of the order of 1000 100000 digits. The series in (18) decreases very fast to zero and it Because in (27) gamma functions can be calculated in PARI with practically arbitrary digits of accuracy, the only is very easy to get arbitrary number of digits of c¯k using the e PARI/GP procedure sumalt. way to improve accuracy of calculation of c100000 is to find a reliable method for calculating ζ0(s) with certainty that say Next we want to calculate c˜k from (27) with 1000 digits −πγ /4 −1000 all 1000 digits are correct. From (1) it is easy to obtain the accuracy. From e l = 10 we get γl ≈ 2931.7 and a glance at the list of zeros of ζ(s) (e.g. [37]) gives l = 2402 modified expression for zeta: ∞ n−1 because γ2402 = 2931.0691 .... First 100 zeros of ζ(s) ac- 1 X (−1) ζ(s) = . (36) curate to over 1000 decimal places we took from [37]. Next 1 − 21−s ns 0 n=1 2500 zeros of ζ(s) and derivatives ζ (ρl) with precision 1000 digits we decided to calculate using the built in Mathematica By the Leibniz test for alternating series the above sum con- v.7 procedures ZetaZero[m] and numerical differentia- verges for <(s) = σ > 0. Then it uniformly converges on every compact subset lying interior to the half–plane of con- tion ND[...]. After a few days we got the values of γl and 0 vergence (see eg. [38, Th. 11.11 and Th. 11.12]) and thus can ζ (ρl). We checked using PARI/GP that these zeros were ac- curate within at least 996 places in the sense that in the worst be differentiated term by term: −996 1−s ∞ n−1 case |ζ(ρl)| < 10 , l = 1, 2,..., 2600. The formulas ln(2)2 X (−1) ζ0(s) = − (27) and (18) were implemented in PARI with precision set (1 − 21−s)2 ns to 1000 digits and to our surprise we found that the values of n=1 ∞ (37) cg ce th 1 X (−1)n ln(n) 100000 and 100000 coincide only up to 87 place: + . g 1 − 21−s ns c n=1 10−86 > 100000 − 1 > 10−87. e (34) PARI contains the numerical routine sumalt for sum- c100000 ming infinite alternating series in which an extremely effi- cient algorithm of Cohen, Villegas and Zagier [39] is imple- Tab. 1. The partial sums of (14) recorded after summation of 10000, 20000,... 100001 terms mented. As these authors point out on p. 6, their algorithm works even for series like (36) with s complex – (36) is al- n Pn (−1)j 100000 1 ternating only when s ∈ R. We used this routine sumalt j=0 j ζ(2j+2) outside scope of its applicability with success to calculate 14115 10000 5.65168726144550 × 10 0 ζ (ρl) from (37) with precision set to 1000 digits and then 21729 e 20000 4.00927204946289 × 10 calculated c100000 from (27) and (18). The result was aston- ishing: the difference between ce and cg appeared 30000 6.08771775660005 × 1026526 100000 100000 on the 625th place: 29225 40000 5.17938759373151 × 10 g −625 c100000 −624 50000 1.26030418446100 × 1030100 10 < e − 1 < 10 . (38) c100000 60000 3.45292506248767 × 1029225 Because we expected that a possible violation of RH should

26526 manifest at much larger k we were looking for a way to still 70000 2.60902189568574 × 10 0 improve the accuracy of ζ (ρl). We decided to make a frenzy 80000 1.00231801236572 × 1021729 0 think: we calculated again ζ (ρl) using sumalt with ze- 90000 6.27965251271723 × 1014114 ros having 1000 digits but this time with precision set to 0 2000 (however, values of ζ (ρl) were stored only with 1000 100000 1.60975799392038 × 10−9 digits). Thus the aim was to enlarge the number of terms Some remarks on the Báez-Duarte criterion for the Riemann Hypothesis 45 summed in (37), or in fact the number of iteration performed Because we got the precision (39) it is a posteriori proof 0 inside sumalt until the prescribed accuracy was attained. that 1000 digits of derivatives ζ (ρl) were calculated cor- After 18 hours on AMD Opteron 2.6 GHz we got the results. rectly from (37) by the PARI procedure sumalt with pre- g e And now bingo! The first 996 digits of c100000 and c100000 cision set to 2000 digits. were the same: In Fig. 2 we present a summary of these computer calcu- g lations. Since it is not possible to plot using standard plotting −996 c100000 −995 10 < − 1 < 10 . (39) −600 ce software as small values as 10 on the y-axis we present 100000 in Fig. 2 the following quantity measuring the distance from g c100000 to the partial sums over zeros γl in (27) and decreas- ing with a number of zeros included in the sum: y(n) = −1 n iγ /2 3 i ! !! 1 X k l Γ( − γl) = ln < 4 2 − cg k3/4 ζ0( 1 + iγ ) k l=1 2 l (40) where k = 100000 and the absolute value is necessary as the e differences between successive approximants to c100000 and g c100000 change sign erratically. The consecutive terms in the series (27) behave like e−πγl/4 hence we expect that y(n), by analogy with well known property of alternating series with decreasing terms, should behave like the first discarded term: 4 y(n) ∼ . (41) πγn+1 e Fig. 2. The plot of y(n) for c100000 obtained by three meth- ods. The plot (a) is the best result obtained with Mathemat- Fig. 2 confirms these considerations: initially y(n) for all ica with the procedure ND[...] with the option terms=30 curves follows the prediction (41) and starting with n for 0 and WorkingPrecision=1000. The accuracy was 105 dig- which the values of the ζ (ρn) are incorrect adding further its and the curve (a) departs from the red line at the zero terms does not improve accuracy. Because of the exponen- γ149 = 317.73480594237 ... for which exp(−πγ149/4) = tial decrease of Γ( 3 − γl ) the contribution of further terms 4.193107483×10−109. The plot (b) was obtained with PARI from 4 2 the formula (37) using the procedure sumalt with precision set to is suppressed and horizontal lines in Fig.2 are determined by 1000 digits. The y(n) reaches plateau 0.0006856926750 ... at zero the first γn corresponding to the bad value of the derivative 0 γ1412 = 1884.00577834967 ..., for which exp(−πγ1412/4) = ζ (ρn). 2.391868726 × 10−643. The curve (c) is the same as (b) 0 1 but derivatives ζ ( 2 + iγl) were calculated with sumalt and precision set to 2000 digits. Curve (c) reaches the plateau V. FINAL REMARKS 0.00043191361 ... at the zero γ2430 = 2960.033617812 ... for −1010 which exp(−πγ2430/4) = 2.298783954 × 10 Although we have reached the agreement between g e c100000 and c100000 up to 996 places:

g −9 c100000 = 1.60975799 ... ←− 980 digits −→ ... 291369630140 × 10

e −9 c100000 = 1.60975799 ... ←− 980 digits −→ ... 291369629833 × 10 it means nothing about the validity of the RH. Refutation of found a few striking examples of the approximate equalities the RH by computer methods seems to be as difficult as the correct to numerous digits which finally are not identities. analytical proof of its validity. We need to find an example of The most fraudulent is the following the quantity which can be expressed in two ways: one with- j √ k out the zeta zeros and the second formula containing contri- ∞ π 163/9 X ne . butions from all zeta zeros so that there is strong sensitivity = 1280632, (42) 2n on the location of high zeros. Then perhaps it can be possible n=1 to overthrow RH using a computer. As a possible precaution let us mention the paper “Strange Series and High Precision which is valid up to accuracy at least 10−450,000,000 but is Fraud” written by Borwein brothers [40]. In this paper we not an identity. 46 M. Wolf

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Marek Wolf, obtained PhD degree in physics (1982) and habilitation (1993) at the Wroclaw University, where he was employed till 2011. Since 1984 he has been doing computer experiments in physics and mathematics. In 1991 and 1993 he was a research fellow at the Center for Polymer Studies at the Boston University. Since 2011 he is a member of the Faculty of Mathematics and Natural Sciences, Col- lege of Sciences at the Cardinal Stefan Wyszynski University in Warsaw. His hobby is photography.

CMST 20(2) 39-47 (2014) DOI:10.12921/cmst.2014.20.02.39-47