Arias de Reyna, J.

Lune, J. (Jan) van de

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A NOTE ON ALMOST FLAT NUMBERSJUAN ARIAS DE REYNA AND JAN VAN DE LUNEDedicated to Herman J. J. te Riele on the occasion of his retirement from theCWI in January 2012In this note we present a solution of the followingProblem. For n ∈ N let β(n) be the product of all prime divisors of n( not counting multiplicities ).So, in the notation of Hardy & Wright, if n =∏pe then β(n) =∏pwith β(1) = 1.A positive integer n is called flat iff n = β(n) ( or, equivalently, iff|µ(n)| = 1, where µ(n) is the Mo¨bius function ).A positive integer is called almost flat iff n/β(n) is prime.(A) Show that the sequence of almost flat numbers has a positivenatural density ( denoted by d1 ), and indicate how this density can becomputed to any degree of accuracy. ( This is the case k = 1 in thenext, more general problem (B). )(B) Let ω(n) denote the number of different prime divisors of n ∈ N( not counting multiplicities ), and let k be any ( fixed ) positive integer.Show that the sequence of all n ∈ N such that q := q(n) := n/β(n) isflat with ω(q) = k, has positive natural density ( denoted by dk ), andindicate how this density can be computed to any degree of accuracy.(B1) Solve this problem for k = 2.(B2) How to proceed for k ≥ 3? Compute d3, d4 and d5.Solution of (A), the case k = 1.The generating Dirichlet series of the almost flat numbers is obtainedby expanding(1)∑q prime1q2s∏p prime(1 + 1ps)1 + 1qs=ζ(s)ζ(2s)∑q prime1qs(qs + 1),(s = σ + it, σ > 1).Invoking the well-known Wiener-Ikehara Tauberian theorem ( whichapplies indeed see [3, p. 259–266] ), we find that the required natural51A NOTE ON ALMOST FLAT NUMBERSdensity d1 exists and equals(2) d1 =6pi2∑q prime1q(q + 1).The sum of the last series may be approximated by observing that(3) σ1 :=∑q prime1q(q + 1)=∑q prime1q211 + 1q=∑q prime1q2∞∑n=0(−1)nqn==∞∑n=0(−1)n∑q prime1q2+n=∞∑n=0(−1)nP (n+ 2)where, for s > 1,(4) P (s) :=∑p prime1ps=∞∑r=1µ(r)rlog ζ(rs)( see Titchmarsh [4, p. 12, formula (1.6.1)] ) and that, also for s > 1∣∣∣µ(r)rlog ζ(rs)∣∣∣ ≤ 1rlog ζ(rs) <1rlog(1 +12rs+∫ +∞2dxxrs)<<1rlog(1 +32rs)<32rsso that, for n ≥ 2,(5) P (n) <∞∑r=11rlog ζ(rs) <∞∑r=132rs<32n − 1 ≤42n.A combination of these ingredients is sufficient for a high precisioncomputation of∑∞n=2(−1)nP (n).For n ≥ 2 we may approximate P (n) by evaluating∑Rr=1 µ(r)r log ζ(rn)for a sufficiently large R.If we want an accuracy of ε then it suffices to take R such that∑∞r=R+11rlog ζ(rn) < ε. It is easily seen that(6) R =⌊ 1nlog(4ε)log 2⌋suffices. Using Mathematica we findσ1 ≈ 0.3302299262 6420324101 5094588086 7447606442 5941947407 . . .so that ( recall that d1 =6pi2σ1 )d1 ≈ 0.2007557220 1926598699 6250723114 4047658535 3555535256 . . .52A NOTE ON ALMOST FLAT NUMBERSSolution of (B1), the case k = 2.Similarly as in (A) the required density d2 equals(7) d2 =6pi2∑p<q1p(p+ 1)q(q + 1)( p and q denoting primes ).The last series may also be written asσ2 :=∑p<q1p(p+ 1)q(q + 1)=12(∑p<q1p(p+ 1)q(q + 1)+∑q<p1q(q + 1)p(p+ 1))==12∑p 6=q1p(p+ 1)q(q + 1)=12(∑p,q1p(p+ 1)q(q + 1)−∑p=q1p(p+ 1)q(q + 1))==12[(∑p1p(p+ 1))2−∑p1p2(p+ 1)2]=σ212− 12∑p1p4(1 + 1p)2( with σ1 =pi26d1 ) and this in its turn may be reduced to a form “only”containing P (n)’s, so that we can compute σ2 ( and hence d2 ) to anydegree of accuracy. Indeed, it is easily verified that(8) σ2 =12σ21 −12∞∑n=1(−1)n+1nP (n+ 3).Using Mathematica we find thatσ2 ≈ 0.0727869325 3120878610 0250493970 5431864431 8060075841 . . .so thatd2 ≈ 0.0221245744 7327116398 0012002355 9483175788 6781598850 . . .Solution of (B2), the case k ≥ 3.For k ≥ 3 we make use of the well-known (Girard-) Newton formulas.We briefly recall some pertinent details :We consider ( formally ) the equation f(x) =∑n(−1)nσnxn = 0with roots 1αk. In our application we will have αn =1pn(pn+1)where pnis the n-th prime. Then we have∑n(−1)nσnxn =∏n(1−αnx) = 0, with σn =∑j1<j2<···<jnαj1αj2 · · ·αjn( we assume here that σ0 = 1 ). We defineSk :=∑nαkn.53A NOTE ON ALMOST FLAT NUMBERSAlthough the (Girard-) Newton formulas usually express the sums Snin terms of the coefficients σn, we will turn things around and expressthe ( elementary symmetric functions ) σn in terms of the ( exponentialsums ) Sn. In order to do so we compute f′(x) in two different ways :f ′(x) =∑n(−1)nnσnxn−1andf ′(x) = −f(x)∑nαn1− αnx = −f(x)∑n∞∑k=0αk+1n xk = −f(x)∞∑k=0Sk+1xk.Comparing coefficients we find that ( see [1, Chap. 8, p. 166], [2, p. 140]or [5, p. 99] )(9) (−1)mmσm = −m−1∑n=0(−1)nσnSm−nwhich leads directly to the recurrence ( with σ0 = 1 )σm =(−1)m+1m(Sm−σ1Sm−1+σ2Sm−2−· · ·+(−1)m−1σm−1S1), (m ≥ 1).In this way we easily obtain, for example,σ1 = S1σ2 =12(S21 − S2)σ3 =16(S31 − 3S1S2 + 2S3)σ4 =124(S41 − 6S21S2 + 3S22 + 8S1S3 − 6S4)σ5 =1120(S51 − 10S31S2 + 15S1S22 + 20S21S3 − 20S2S3 − 30S1S4 + 24S5).For k = 3 we have to deal with the sumσ3 :=∑p1p(p+ 1)∑p<q1q(q + 1)∑q<r1r(r + 1), (p, q, r primes).In this case we have :(10) αn =1pn(pn + 1), σ3 =∞∑n=1αn∑n<mαm∑m<rαr ==∑n1pn(pn + 1)∑n<m1pm(pm + 1)∑m<r1pr(pr + 1)54A NOTE ON ALMOST FLAT NUMBERSandσ3 =16(S31 − 3S1S2 + 2S3).Note that we can compute the Sn =∑p1pn(p+1)nby the formula(11)Sn =∑p1pn(1 + p)n=∑p1p2n(1 +1p)−n=∑p1p2n∞∑k=0(−nk)1pk==∞∑k=0(−nk)∑p1p2n+k=∞∑k=0(−1)k(n+ k − 1k)P (2n+ k).Proceeding similarly for k > 3 we find ( using Mathematica )d1 = 0.2007557220 1926598699 6250723114 4047658535 3555535256 . . .d2 = 0.0221245744 7327116398 0012002355 9483175788 6781598850 . . .d3 = 0.0010728279 2166161493 7597184179 0511299854 7080207983 . . .d4 = 0.0000267593 5151889275 7741972284 4743787780 5157715943 . . .d5 = 0.0000003834 9005273872 2348794555 0178910921 5013442743 . . .d6 = 0.0000000034 4999551430 8580387444 6993630085 9120389312 . . .d7 = 0.0000000000 2082589566 1766505646 3194316856 4945749335 . . .d8 = 0.0000000000 0008875408 1001607125 3428410234 4454925913 . . .d9 = 0.0000000000 0000027791 2994465580 9631134694 5089946028 . . .d10 = 0.0000000000 0000000066 0331441112 0947527899 3022631397 . . .References[1] W. S. Burnside & A. W. Panton, The Theory of Equations, Vol. 1, Dover,1960.[2] N. Jacobson, Basic Algebra, Vol. 1, 2 ed., W. H. Freeman and Company, 1996.[3] H. L. Montgomery & R. C. Vaughan, Multiplicative Number Theory I,Classical Theory, Cambridge University Press, Cambridge, 2006.[4] E. C. Titchmarsh, The Theory of the Riemann Zeta-function, Second ed.revised by D. R. Heath-Brown, Oxford University Press, 1986.[5] B. L. van der Waerden, Algebra, Vol. 1, Springer, 1991.Facultad de Matema´ticas, Universidad de Sevilla,Apdo. 1160, 41080-Sevilla, SpainE-mail address: arias@us.esLangebuorren 49, 9074 CH Hallum, The Neterlands(Formerly at CWI, Amsterdam )E-mail address: j.vandelune@hccnet.nl55

A note on almost flat numbers

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