The vanishing properties of Fourier coefficients of integral powers of the Dedekind eta function correspond to the existence of integral roots of integer-valued polynomials Pn(x) introduced by M. Newman. In this paper we study the derivatives of these polynomials. We obtain non-vanishing results at integral points. As an application we prove that integral roots are simple if the index n of the polynomial is equal to a prime power pm or to pm + 1. We obtain a formula for the derivative of Pn(x) involving the polynomials of lower degree

Heim, B.

Neuhauser, M.

English

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#A97 INTEGERS 18 (2018)POLYNOMIALS RELATED TO POWERS OF THE DEDEKINDETA FUNCTIONBernhard HeimGerman University of Technology in Oman, Athaibah, Sultanate of Omanbernhard.heim@gutech.edu.omMarkus NeuhauserGerman University of Technology in Oman, Athaibah, Sultanate of Omanmarkus.neuhauser@gutech.edu.omReceived: 12/30/17, Revised: 07/20/18, Accepted: 11/21/18, Published: 12/10/18AbstractThe vanishing properties of Fourier coe cients of integral powers of the Dedekindeta function correspond to the existence of integral roots of integer-valued poly-nomials Pn(x) introduced by M. Newman. In this paper we study the derivativesof these polynomials. We obtain non-vanishing results at integral points. As anapplication we prove that integral roots are simple if the index n of the polynomialis equal to a prime power pm or to pm + 1. We obtain a formula for the derivativeof Pn(x) involving the polynomials of lower degree.1. IntroductionIn 1955, Newman [7] studied a family of polynomials Pn(x) with remarkable prop-erties. These polynomials are recursively defined by P0(x) := 1 andPn(x) :=xnnXk=1 (k)Pn k(x). (1)Here  (m) :=Pd|m d is the divisor sum. The polynomials Pn have degree n andare integer-valued functions.Let ⌘(⌧) be the Dedekind eta function ([1], [4]):⌘(⌧) := q1241Yn=1(1  qn) , (2)INTEGERS: 18 (2018) 2where ⌧ 2 H := {⌧ 2 C | Im(⌧) > 0}. One puts q := e2⇡i ⌧ . Then1Xn=0Pn(z) qn =1Yn=1(1  qn) z . (3)Let r be an integer. Then Pn( r) is essentially equal to the n-th Fourier coe cientof ⌘r (see also [12], introduction). The number of partitions [10] of n is given byP1(1) = 1, P2(1) = 2, P3(1) = 3, P4(1) = 5, . . . , P200(1) = 3972999029388, . . .Euler already studied the values of the Fourier coe cients for r =  1, which arerelated to pentagonal numbers [8]. The first few are given byP1( 1) =  1, P2( 1) =  1, P3( 1) = 0, P4( 1) = 0, P5( 1) = 1.In contrast to partition numbers, where r is negative, the Fourier coe cients for rpositive have sign changes and can vanish. The study of these properties, especiallythe vanishing property is of special interest (see for example [13], [2] (page 94), [3]).Let ⌧(n) be the Ramanujan tau-function [6, 11], the coe cients of the discrimi-nant function   (see Section 2 for more details). Then the values of Pn(x) at  24are given byP0( 24) = ⌧(1) = 1,P1( 24) = ⌧(2) =  24,P2( 24) = ⌧(3) = 252.The roots of Pn(x) dictate the vanishing of the Fourier coe cients of the corre-sponding powers of the Dedekind eta function. For exampleP19(x) =x19!(x+ 1)(x+ 3)(x+ 4)(x+ 6)(x+ 8)(x+ 14)Q(x), (4)where Q(x) is an irreducible polynomial over Q. Hence only the 19-th Fouriercoe cient of the zeroth, first, third, sixth, eighth, and fourteenth integral powerof ⌘(⌧) vanishes. This implies that ⌧(20) 6= 0. Actually the Lehmer conjecture [6]predicts that ⌧(n) never vanishes. It is known that the sum ⌃n and product ⇧n ofthe roots of n!x Pn(x) are given by⌃n =3n(n  1)2, ⇧n = (n  1)!  (n). (5)Note that several roots as  1, 2, 3, 4, 6, 8, 10, 14, 26 occur frequently(see [3]) and precise criteria for the related Fourier coe cients exist [13]. Neverthe-less up to the moment only  5, 7 and  15 have been identified as further integralroots. To make use of the explicit formula for ⌃n,⇧n to study the distribution ofINTEGERS: 18 (2018) 3the roots, especially the integral roots, it is essential to know if all non-trivial rootshave negative real part and their multiplicity. Note that the converse is not true.For example(X + 3) (X   1  3i) (X   1 + 3i) = X3 +X2 + 4X + 30.Numerical calculations show that the polynomials n!x Pn(x) for n  1000 are Hur-witz polynomials, which supports the assumption that all roots have negative realpart.In this paper we study the multiplicities of possible integral roots of Pn(x). Ourmain ingredient is a new formula for the derivatives of Pn(x). Here we obtain resultsfor specific indices n and general x. Additionally we study the casesx =  1, 2, . . . , 10 and x =  24.For example we have P 0n( 1) 62 Z for 2  n  107.2. ResultsThe Dedekind eta function ⌘(⌧) is a modular form of weight 1/2 for SL2(Z). Dueto Euler, the Fourier coe cients are easy to calculate. The same is true for ⌘(⌧)3due to Jacobi (see also [9], [5]) and1Yn=1(1  qn) =1Xn= 1( 1)n q 3n2+n2 , (6)1Yn=1(1  qn)3 =1Xn=0( 1)n (2n+ 1) q n2+n2 . (7)Here (3n2   n)/2 are the pentagonal numbers. Note that ⌘(⌧) and ⌘(⌧)3 are theonly superlacunary odd integral ⌘ powers.For general powers the situation is much more complicated. Let  (⌧) be thediscriminant function with Fourier coe cients ⌧(n): (⌧) := ⌘(⌧)24 =1Xn=1⌧(n) qn. (8)This is the unique newform of weight 12 for SL2(Z). The ⌧(n) have many remark-able properties. They are multiplicative with an additional law for ⌧(pm) (p primeand m 2 N) since   is a Hecke eigenform. Further   satisfies the Ramanujanconjecture proven by Deligne, which implies a precise estimation on the growth ofthe numbers ⌧(n). It is also known that the Fourier coe cients of   have infinitelyINTEGERS: 18 (2018) 4many sign changes. But it is still an open problem (Lehmer conjecture) if any coef-ficient vanishes. Lehmer proved that the smallest n, for which ⌧(n) is zero, has tobe a prime [6].Let n 2 N. We note thatn!xPn(x) =n 1Xk=0an,k xk 2 Z[x]is a normalized polynomial of degree n   1 and that all its coe cients an,k arepositive. It is easy to show that an,0 = (n   1)!  (n) and an,n 2 = 3n(n 1)2 . Seealso [7]. These simple formulas are misleading, since there is no explicit closedformula known for an,k in general.Our first result on the derivatives of Pn(x) is the following one.Theorem 1. Let n = pm, where p is a prime and m 2 N. ThenP 0n(x0) 2 Q \Z (9)for any integer x0.In the proof we use the property that for these specific n we have that  (n) andn are coprime. We hope to extend our theorem to all n with this property. On theother hand, our theorem is true for all integers x0, which has not been expected.Let n be a prime power that is an element of the set{2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, . . .} , (10)then P 0n(x0) is not an integer. This implies:Corollary 1. Suppose n = pm, where p is a prime and m 2 N. Then all integralroots of Pn(x) are simple.In the course of the proof we obtain the following useful formula, expressing thederivatives of Pn(x) in terms of linear combinations of Pm(x) for m  n  1:P 0n(x) =nXk=1 (k)kPn k(x) (n 2 N). (11)Inverting this formula leads to:Corollary 2. There is a sequence of bk, k    1, so that we can express Pn(x) asa linear combination of derivatives P 0m(x), wherePn(x) =n 1Xk= 1bkP0n k(x).INTEGERS: 18 (2018) 5The coe cients are obtained recursively from b 1 = 1. For k   0:bk =  k 1Xm= 1bm (k + 1 m)k + 1 m . (12)If we further let n = pm+1, then we do not get the full result of Theorem 1, butstill an integral result for the derivative of the related polynomialFn(x) :=nxPn(x)at integral points x0. This leads to the simplicity of integral roots at n = pm + 1independent of x0.Theorem 2. Let n = pm + 1, where p is a prime and m 2 N. ThenF 0n(x0) 2 Q \Z (13)for any integer x0.This shows immediately:Corollary 3. Suppose n = pm+1, where p is a prime and m 2 N. Then all integralroots of Pn(x) are simple.This adds the following n to the above list (10):6, 10, 12, 14, 18, 20, 24, 26, 28, 30, 33, . . . , (14)Hence the first index not covered by these two corollaries is n = 15.3. Proofs of Theorem 1 and Theorem 2Before we prove the two theorems, we list the derivatives of Pn(x) for n  10. Fora list of the polynomials for n  20 we refer the reader to [3].n P 0n (x)0 01 12 (2x+ 3) /23 3x2 + 18x+ 8 /3!4 4x3 + 54x2 + 118x+ 42 /4!5 5x4 + 120x3 + 645x2 + 900x+ 144 /5!6 6x5 + 225x4 + 2260x3 + 7425x2 + 6788x+ 1440 /6!7 7x6 + 378x5 + 6125x4 + 37380x3 + 84882x2 + 61824x+ 5760 /7!8 (8x7 + 588x6 + 14028x5 + 138600x4 + 591556x3 + 1020348x2 + 586584x+ 75600)/8!9 (9x8 + 864x7 + 28518x6 + 417312x5 + 2896845x4 + 9365328x3+ 13006788x2 + 6064416x+ 524160)/9!10 (10x9 + 1215x8 + 53040x7 + 1080450x6 + 11145078x5 + 58723875x4+ 152199680x3 + 173321100x2 + 72581472x+ 6531840)/10!INTEGERS: 18 (2018) 6It is well-known and not di cult to show that the derivative of a Hurwitz polynomialis again a Hurwitz polynomial. Hence we expect that all roots of P 0n(x) have negativereal part. Similar to the Pn(x), the P 0n(x) can also have non-real roots. For examplefor n = 21.Proof of Theorem 1. We deduce from [7] that1Xn=0Pn(z) qn =1Yn=1(1  qn) z (z 2 C). (15)See also Section 2, where some further remarks are given. The proof is based onexamining the logarithmic derivative of (15), certain grouping of terms and usingthe uniqueness of the coe cients of the involved power series in q. We have analyticfunctions on the upper half space, since |q| < 1. It is well known that1Yn=1(1  qn) = exp  1Xn=1 (n)qnn!.Note the logarithmic derivative of the Dedekind eta function is essentially theholomorphic Eisenstein series of weight 2. Putting both sides of our equation to thepower of  z leads to1Yn=1(1  qn) z = exp z1Xn=1 (n)qnn!.Hence we obtain the useful identity1Xn=0Pn(z) qn = exp z1Xn=1 (n)qnn!. (16)Now we di↵erentiate both sides of (16) with respect to z. The right side of theformula reproduces the power series multiplied with1Xn=1 (n)qnn.After comparing the Fourier coe cients with respect qn of both sides we obtainthe following identity:P 0n(x) =nXk=1 (k)kPn k(x) (n 2 N). (17)It expresses the derivative of the polynomial Pn(x) in terms of Pn k(x) and  (k)/kinstead of  (k) in the recursion formula of Pn(x). To examine the possible denom-inator of the right side, we write down the sum explicitly:P 0n(x) = Pn 1(x) + (2)2Pn 2(x) + . . .+ (n  1)n  1 x+ (n)n. (18)INTEGERS: 18 (2018) 7Recall that the polynomial Pn(x) 2 Q[x] is an integer-valued function, hence pos-sible denominators can only be obtained by the quotients  (k)/k. Let n = pm.Then the greatest common divisor of  (n) and n is equal to 1 (gcd( (n), n) = 1).Further we note that n = pm is exactly the highest power of p appearing in thedenominator of P 0n(x0), where x0 2 Z. Hence P 0n(x0) is a rational number, but notan integer.The same argument is expected to work if we assume that  (n) and n are coprime.But if n splits into di↵erent coprime numbers, the denominator could be eliminatedby other terms in the sum (18). We put this as an open problem.Proof of Theorem 2. For all natural numbers n we have:Fn(x) =nXk=1 (k) Pn k(x). (19)Taking the derivative with respect to x leads toF 0n(x) =n 1Xk=1 (k) P 0n k(x)= P 0n 1(x) +  (2)P0n 2(x) + . . .+  (n  1)P 01(x).Note that P 00(x) = 0 and P 01(x) = 1. Using the formula (17) leads toF 0n(x) =n 1Xk=1n kXl=1 (k) (l)lPn k l(x). (20)Then the claim follows from taking special values for k and l. We observe that fork = 1 and l = n 1, the summand  (n 1)/ (n 1) has the denominator pm, whichdominates the denominator of F 0n(x0) for x0 2 Z. Here it is essential that Pn(x) isan integer-valued function.4. On the Derivatives of Pn(x) at x =  1We have proven that Pn(x) does not have multiple zeros for n or n   1 equal toprime powers at integral points. It would definitely be interesting to understandwhat happens for general n. Fixing the argument x 2 Z and varying the n givessome new insight.In this section we study Pn(x0) at x0 =  1. Our Theorems already show thatP 0n( 1) 6= 0 for n  14, if n is a prime power. Hence  1 is a single root for n  14.A calculation shows thatP 015( 1) =1256310920=17 · 73923 · 3 · 5 · 7 · 13 .INTEGERS: 18 (2018) 8Our interest is the multiplicity of ( 1) as a root of Pn(x). The n’s label in (6)the Fourier coe cients of1Ym=1(1  qm) = 1 +1Xm=1( 1)mq(3m+1)m/2 =1Xn=0Pn( 1) qn= 1  q   q2 + q5 + q7   q12   q15 + q22 + . . . .Let Pn( 1) = 1 then Pn(x) is forced to have a root in ( 1, 0). Further investigationsin the values of the derivatives of Pn(x) at  1 lead to:Theorem 3. If n  N = 107, then P 0n( 1) 6= 0. Refined calculations show thatP 0n( 1) 62 Z for 2  n  N .If x0 2 { 2, 3, . . . , 10}, then P 0n(x0) 62 Z for 2  n  104. We also checkedthe special case x0 =  24. We have that P 0n( 24) is not an integer for 2  n  105.For further studies and for the convenience of the reader we record the explicitfactorization of the values P 0n( 1) for n  20. They reveal sign changes. Furthernote that 2 and 3 do not always appear in the denominator.n 1 2 3 4 5P 0n ( 1) 1 12   72·3   1322·3   11322·3·5P 0n+5 ( 1) 122·5   3·132·5·7 17923·3·7 5·13223·32·77323·32P 0n+10 ( 1) 29·10922·32·5·11 2·10313·5·7·11 181123·3·11·13   9972·32·7·13 17·73923·3·5·7·13P 0n+15 ( 1)  83·106324·32·5·7·11  85324·3·17  23281124·3·11·13·17  23·120122·3·7·17·19  5·7·25723·3·13·19Acknowledgement. The main part of the paper was written when both authorshad been invited guests at the Max-Planck-Institute for Mathematics in Bonn inAugust 2017. A revision of the results had been made in 2018 when both authorshad a research stay at the RWTH Aachen. Both authors thank the MPIM and theRWTH for their hospitality and excellent research conditions. We thank the Ger-man University in Oman for supporting our research. Finally we thank the Integersjournal for the excellent reviewing process and support in finalizing the paper.References[1] T. Apostol, Modular Functions and Dirichlet Series in Number Theory, Second Edition,Springer, Berlin–Heidelberg–New York, 1990.[2] J. H. Bruinier, Borcherds Products on O(2, l) and Chern Classes of Heegner Divisors, LectureNotes in Math. 1780, Springer Verlag, 2002.INTEGERS: 18 (2018) 9[3] B. Heim, M. Neuhauser, and F. Rupp, Imaginary powers of the Dedekind eta function,Experimental Mathematics (2018), DOI: 10.1080/10586458.2018.1468288.[4] M. Koecher and A. Krieg, Elliptische Funktionen und Modulformen, Springer, Berlin–Heidelberg–New York, 2007.[5] G. Ko¨hler, Eta Products and Theta Series Identities, Springer Monographs in Mathematics,Springer, Berlin–Heidelberg–New York, 2011.[6] D. Lehmer, The vanishing of Ramanujan’s ⌧(n), Duke Math. J. 14 (1947), 429–433.[7] M. Newman, An identity for the coe cients of certain modular forms, J. London Math. Soc.30 (1955), 488–493.[8] E. Neher, Jacobis Tripleprodukt-Identita¨t und ⌘-Identita¨ten in der Theorie der a nen Lie-Algebren, Jahresbericht d. Dt. Math.-Vereinigung 87 (1985), 164–181.[9] K. Ono and S. Robins, Superlacunary cusp forms, Proceedings of the AMS 123 (1995),1021–1029.[10] K. Ono, The Web of Modularity: Arithmetic of the Coe cients of Modular Forms andq-series, Conference Board of Mathematical Sciences 102, 2003.[11] K. Ono, Lehmer’s conjecture on Ramanujan’s tau-function, Journal of the Indian Math.Society Special Centenary Issue (2008), 149–163.[12] J. Rouse, Bounds for the coe cients of powers of the  -function, Bull. London Math. Soc.40 (2008), 1081–1090.[13] J. Serre, Sur la lacunarite´ des puissances de ⌘, Glasgow Math. J. 27 (1985), 203–221.

Polynominals related to powers of the Dedekind eta function

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