SOME REMARKS ON THE RIEMANN HYPOTHESIS

M. Ram Murty^{∗}

August 3, 2000

1. Pólya and Turán conjectures. The Liouville function λ(n) is defined as (−1)^{Ω(n)} where Ω(n) is the total number of prime factors of n counted with multiplicity. It is a completely multiplicative function and it is easy to see that

∑_{n=1}^{∞} \frac{λ(n)}{n^{s}} = \frac{ζ(2s)}{ζ(s)}  (1)

for Re(s) > 1. If we define

S(x) := ∑_{n≤x} λ(n)  (2)

then, by partial summation, we have

∑_{n=1}^{∞} \frac{λ(n)}{n^{s}} = s ∫_{1}^{∞} \frac{S(t)}{t^{s+1}}dt.  (3)

Based on numerical data, Pólya [Po] conjectured that

S(x) ≤ 0

for all x ≥ 2. It should be noted that Pólya’s conjecture implies the Riemann hypothesis. Indeed, by a well-known theorem of Landau, the integral expression in (3) converges to the right of Re(s) > σ_{0} where σ_{0} is the first real singularity of ζ(2s)/ζ(s). For Landau’s theorem, see for example, [EM, Theorem 10.4.2, p. 132], where the proof is given for Dirichlet series with non-negative coefficients. However, the proof also works, mutatis mutandis, for Dirichlet integrals of the form

∫_{1}^{∞} \frac{S(t)}{t^{s+1}}dt,

where S(t) is of fixed sign for t sufficiently large. In the case under discussion, ζ(s) has no real zeros in 1/2 ≤ s ≤ 1, and so the first real singularity is at s = 1/2 coming from the pole of ζ(2s) on the numerator. Thus, ζ(2s)/ζ(s) is regular for Re(s) > 1/2 which implies that there are no zeros of ζ(s) in Re(s) > 1/2 since ζ(2s) is regular and non-vanishing in that region.

∗Research partially supported by NSERC and a Killam Research Fellowship.

Even if we have S(x) ≤ 0 for x sufficiently large, a similar argument allows us to deduce the Riemann hypothesis. Unfortunately, Haselgrove [Ha] has shown that S(x) changes sign infinitely often and so the Pólya conjecture is false. The smallest counterexample is x = 906, 150, 257 for which S(x) = 1. It is to be noted that the estimate

S(x) = O(x^{1/2+ϵ})  (4)

for any ϵ > 0 (where the implied constant may depend on ϵ would also allow us to deduce the Riemann hypothesis. Indeed, (4) implies that the integral expression in (3) is regular for Re(s) > 1/2. Thus, ζ(2s)/ζ(s) is regular in that half-plane and by the same reasoning, we deduce the Riemann hypothesis. In fact, it is not hard to show that (4) is equivalent to the Riemann hypothesis.

Our goal in this paper is to formulate automorphic generalizations of the Pólya conjecture and (4) and then investigate when we can expect them to be true.

There is a related conjecture of Turán [T], namely that the sum

∑_{n≤x} \frac{λ(n)}{n} ≥ 0

for x sufficiently large. This too has been disproved by Haselgrove [H]. Below, we shall also investigate modular analogues of the Turán conjecture. In an appendix by Nathan Ng, we present some numerical evidence related to the modular versions of the Pólya and Turán conjectures.

Acknowledgements. I would like to thank Michael Rosen for his comments on preliminary version of this paper. I also thank Nathan Ng for doing the computations recorded in the Appendices.

2. Modular analogues of Pólya’s conjecture.

Let f be a normalized eigenform of weight k and level N and trivial nebentypus. Let us write

f(z) = ∑_{n=1}^{∞} a_{f}(n)n^{\frac{k−1}{2}} e(nz)

where e(z) = e^{2πiz}, as usual. Then,

a_{f}(m)a_{f}(n) = ∑_{d|m,n} a_{f}(mn/d^{2}).

It is easy to prove the following:

Lemma 1. Let

F (m, n) = ∑_{d|m,n} G(m/d, n/d).

Then

G(m, n) = ∑_{d|m,n} μ(d)F (m/d, n/d)

and conversely.

We can apply the lemma to deduce that

a_{f}(mn) = ∑_{d|m,n} μ(d)a_{f}(m/d)a_{f}(n/d).  (5)

Now, let us observe that from (1),

∑_{d|n} λ(d) = \{ \begin{matrix}1 & if n is a square \\ 0 & otherwise. \end{matrix}  (6)

Then,

∑_{n=1}^{∞} a_{f}(n^{2})/n^{2s} = ∑_{n=1}^{∞} \frac{a_{f}(n)}{n^{s}}( ∑_{d|n} λ(d))

by (6). Interchanging summations, using (5) and observing that λ is completely multiplicative, we find that

∑_{n=1}^{∞} a_{f}(n^{2})/n^{2s} = \frac{1}{ζ(2s)}L(s, f)L(s, fλ)  (7)

where L(s, f) = ∑_{n=1}^{∞} a_{f}(n)/n^{s} and L(s, fλ) = ∑_{n=1}^{∞} a_{f}(n)λ(n)/n^{s}. Since

L(s, f)L(s, fλ) = L(2s, Sym^{2}(f))/ζ(2s),  (8)

as is easily seen by examin Euler factors, we deduce the identity

ζ^{2}(s) ∑_{n=1}^{∞} a_{f}(n^{2})/n^{s} = L(s, Sym^{2}(f)),  (9)

which is of independent interest. Thus, from the previous equation, we have

L(s, fλ) = \frac{L(2s, Sym^{2}(f))}{ζ(2s)L(s, f)}.  (10)

Now suppose that a_{f}(n) are real and consider the hypothesis

∑_{n≤x} a_{f}(n)λ(n) ≥ 0.  (11)

Then, writing the left hand side of (10) as an integral via partial summation, we find that the right hand side of (10) converges for Re(s) > σ_{0} where σ_{0} is the first real singularity of L(2s, Sym^{2}(f))/ζ(2s)L(s, f). Since L(s, f) has infinitely many zeros on Re(s) = 1/2, and because L(2s, Sym^{2}(f))/ζ(2s) doesn’t vanish in the half-plane Re(s) > 1/2, we deduce that this singularity must occur in the half-plane Re(s) ≥ 1/2. This leads to:

Theorem 2. Suppose that L(s, f) ≠ 0 for 1/2 < s ≤ 1 and that ord_{s=1/2} L(s, f) ≤ 1. Then,

S_{f}(x) := ∑_{n≤x} a_{f}(n)λ(n)

changes sign infinitely often.

Proof. Let us first consider the case L(1/2, f) ≠ 0. If S_{f}(x) is of constant sign for x sufficiently large, then

L(2s, Sym^{2}(f))/ζ(2s)L(s, f)

is regular for Re(s) > α where α is the first real singularity of the right hand side of (10). By hypothesis, L(s, f) does not vanish for any real s between 1/2 and 1. Also, ζ(2s) has no real zeros between 1/4 and 1 and the numerator is regular by a celebrated theorem of Shimura [Sh]. Thus, the right hand side of (10) is regular for Re(s) > α with α < 1/2. We also know that L(2s, Sym^{2}(f)) does not vanish on Re(s) = 1/2. Thus L(s, f) has no zeros for Re(s) ≥ 1/2 which is a contradiction. This deals with the case L(1/2, f) ≠ 0. If now, L(1/2, f) = 0, and s = 1/2 is a simple zero, then ζ(2s)L(s, f) is non-zero at s = 1/2. Thus, L(2s, Sym^{2}(f))/ζ(2s)L(s, f) is regular for Re(s) ≥ 1/2. But this is a contradiction since L(s, f) has infinitely many zeros on Re(s) = 1/2.

It is easy to give examples of f which satisfy the hypothesis of Theorem 2.

Thus, the modular analogue of Pólya’s conjecture is false in general. A necessary condition for it to be true is that L(1/2, f) = 0 for then the right hand side of (10) will have a singularity at s = 1/2.

It is quite possible that if E is an elliptic curve with large Mordell-Weil rank, then

S_{E}(x) = ∑_{n≤x} a(n)λ(n)/\sqrt{n} ≥ 0

for all x sufficiently large.

Gonek [Go] and Hejhal [He] have independently conjectured that for Riemann zeta function, we should have

∑_{|Im(ρ)|≤T} \frac{1}{|ζ′(ρ)|^{2}} ≪ T  (12)

where the summation is over zeros of the zeta function. If we suppose that all the zeros of L(s, f) are simple (apart from the zero at s = 1/2), then the analogue of the above is

∑_{0<|Im(ρ)|≤T} \frac{1}{|L′(ρ)|^{2}} ≪ T  (13)

Murty and Perelli [MP] have shown that almost all zeros of L(s, f) are simple if we assume the Riemann hypothesis for L(s, f) and the pair correlation conjecture for it. For the discussion below, we do not need an estimate as strong as the above estimate. If r is the order of the zero at s = 1/2, what is actually needed is that the order of every zero on the critical line have order ≤ r − 1 and one would need a similar estimate for

∑_{0<|Im(ρ)|<T} | Res_{s=ρ} \frac{1}{L(s, f)}|^{2} ≪ T.  (14)

In fact, one can prove the following.

Theorem 3. Assume the Riemann hypothesis for L(s, f) and suppose that L(s, f) has a zero at s = 1/2 of order r. Suppose further that all zeros of L(s, f) on Re(s) = 1/2 are of order ≤ r − 1 apart from s = 1/2 and that (14) is satisfied. Then,

∑_{n≤x} a_{f}(n)λ(n) = x^{1/2}p_{r−2}(log x) + O(x^{1/2}(log x)^{3/2})

where p_{r−1} is a polynomial of degree r − 2.

Here is an indication of the proof. For the sake of simplicity we shall suppose all zeros of L(s, f) apart from s = 1/2 are simple. The sum

∑_{n≤x} a_{f}(n)λ(n)

can be written for c > 1,

\frac{1}{2πi} ∫_{c−iT}^{c+iT} \frac{L(2s, Sym^{2}(f))x^{s}ds}{sζ(2s)L(s, f)} + O(x^{c}/T)

by Perron’s formula. We will choose T = T_{j} with T_{j}→∞ along an appropriate sequence that doesn’t coincide with any ordinate of a zero of L(s, f). Moving the line of integration to the left and picking up the residues arising from the zeros of L(s, f), we obtain

S_{f}(x) = x^{1/2}p_{r−2}(log x) + ∑_{|Im(ρ)|<T} \frac{L(2ρ, Sym^{2}(f))}{ρζ(2ρ)L′(f, ρ)} + \frac{1}{2πi} ∫_{C} \frac{L(2s, Sym^{2}(f))x^{s}}{sζ(2s)L(s, f)}

where C denotes the semi-rectangular path beginning at c+iT_{j} to a+iT_{j} and then to a−iT_{j} ending at c−iT_{j}. The horizontal and vertical integrals are easily estimated by the functional equation. For the sum over zeros one can use

∑_{0<|Im(ρ)|<T} \frac{1}{|L′(ρ, f)|^{2}} ≪ T

or the more general (14), which is a modular analogue of a conjecture of Gonek [Go]. Breaking up the sum over the zeros into dyadic intervals of type [U, 2U] we obtain an error term of

O(x^{1/2}(log x)^{3/2}).

3. Modular analogues of the Turán conjecture.

If we expect that

S_{f}(x) = ∑_{n≤x} a_{f}(n)λ(n) ∼ cx^{1/2}(log x)^{r−2}

for r ≥ 4, then by partial summation we deduce that

∑_{n≤x} \frac{a_{f}(n)λ(n)}{\sqrt{n}} = ∫_{1}^{x} \frac{S_{f}(t)dt}{t^{3/2}} ∼ c(log x)^{r−1}

as x → ∞, for some constant c > 0, so that the sums

T_{f}(x) = ∑_{n≤x} \frac{a_{f}(n)λ(n)}{\sqrt{n}} ≥ 0

for sufficiently large x. Unlike the Turán case, these sums are not partial sums of the corresponding series at the edge of the critical strip. They have the disadvantage of being the partial sums of the series at the center of the critical strip. It is not difficult to show that these series actually converge at the center of the critical strip (see for example, [KM, p. ]).

Thus, we see that if the modular analogue of the Pólya conjecture is true, then so is the modular analogue of the Turán conjecture.

4. Automorphic analogues.

Let L(s, π) be an automorphic L-function on GL(n). If π is self-dual, then it is reasonable to ask if

S_{π}(x) = ∑_{n≤x} a_{n}(π)λ(n) ≥ 0.

Certainly the Riemann hypothesis for L(s, π) follows from

S_{π}(x) = O(x^{1/2+ϵ})

since an easy calculation shows that

L(s, π)L(s, π ⊗ λ) = ∏_{p} ∏_{i=1}^{d}(1 − α_{p,i}^{2}p^{−2s})^{−1}

The above reasoning suggests that if there is a high-order zero at s = 1/2, then the analogue of the Pólya conjecture should be true for a function which is “primitive” in the sense of Selberg. It would be interesting to test the conjecture for automorphic forms of higher dimension.

5. Certain sums of Fourier coefficients.

In this section and the next, we indicate an approach to proving a quasi-Riemann hypothesis. To this end, we will need some estimates on averages of Fourier coefficients of modular forms. We use the notation m ∼ M to mean M ≤ m ≤ 2M. We will need to consider sums of the form

∑_{m∼M} a_{f}(mj)

for j fixed. We will prove that

Theorem 4. We have

∑_{m∼M} a_{f}(mj) = O(M^{1/3}j^{ϵ}),

where the implied constant is independent of M.

Proof. We have

∑_{m∼M} a_{f}(mj) = ∑_{m∼M} ∑_{d|m,j} μ(d)a_{f}(m/d)a_{f}(j/d)  (15)
= ∑_{d|j} μ(d)a_{f}(j/d) ∑_{t∼M/d} a_{f}(t)  (16)

and the inner sum is by an estimate of Rankin [Ra], O((M/d)^{1/3}) from which we easily deduce the stated estimate.

The interest in knowing the asymptotics of such sums is due to the following:

Theorem 5. Suppose that

∑_{k<X} ( ∑_{d|k,d≤V} μ(d))a_{f}(mk) = O(X^{1/2}m^{ϵ}V^{ϵ})

then L(s, f) has no zeros for Re(s) > 3/4.

Remark. We say a few words about the hypothesis in Theorem 5. Firstly, if V = 1, then the hypothesis holds by Theorem 4. If V is bounded then the same is true. If V = X, then the sum is just a_{f}(m) which is clearly m^{ϵ}. If we write k = dt in the inner sum and interchange the sums, we can estimate the inner sum by Theorem 4 to get an upper bound of O(m^{ϵ}X^{1/3}V^{2/3}). This means that the hypothesis is satisfied for V ≤ X^{1/4}. In fact, if even we can replace the above upper bound by O(m^{ϵ}X^{1/3}V^{2/3−δ}) for some small δ > 0, then we will be able to deduce some quasi-Riemann hypothesis for L(s, f). Thus, the hypothesised estimate (which can be viewed as a generalization of Theorem 4) seems to lie deeper. We make some further remarks about it in the final section.

6. Proof of Theorem 5.

We will apply the method of Vaughan to study sums of the form

∑_{n≤x} a(n)λ(n)

where a(n) = a_{f}(n). Vaughan’s identity can be stated in the following way. It is based on the formal identity:

A/B = (1 − BG)(A/B) + AG  (17)
= (F + (A/B − F))(1 − BG) + AG  (18)
= F + AG − BF G + (A/B − F)(1 − BG)  (19)

Suppose now we are given two Dirichlet series

A(s) = ∑_{n=1}^{∞} \frac{a(n)}{n^{s}}, B(s) = ∑_{n=1}^{∞} \frac{b(n)}{n^{s}},

and write

\frac{A(s)}{B(s)} = ∑_{n=1}^{∞} \frac{c(n)}{n^{s}}.

Set

F (s) = ∑_{n≤U} c(n)/n^{s}, G(s) = ∑_{n≤V} b̃(n)/n^{s}

Then, we have

c(n) = a_{1}(n) + a_{2}(n) + a_{3}(n) + a_{4}(n)

where

a_{1}(n) = c(n) for n ≤ U  (20)
= 0 otherwise  (21)
a_{2}(n) = ∑_{d|n,d≤V} a(n/d)b̃(d)  (22)
a_{3}(n) = − ∑_{et=n,e≤U} c(e)( ∑_{df=t,f≤V} b(d)b̃(f))  (23)
a_{4}(n) = − ∑_{de=n,d>U,e>V} c(d)( ∑_{rs=e,s≤V} b(r)b̃(s))  (24)

which is the essence of Vaughan’s identity. In the case of interest, A(s) = ζ(2s) and B(s) = ζ(s) so that

λ(n) = a_{1}(n) + a_{2}(n) + a_{3}(n) + a_{4}(n),

where

a_{1}(n) = λ(n) if n ≤ U  (25)
= 0 otherwise  (26)
a_{2}(n) = ∑_{{h^{2}d=n \atop d≤V}} μ(d),  (27)

a_{3}(n) = − ∑_{{mdr=n \atop m≤U,d≤V}} λ(m)μ(d),  (28)
a_{4}(n) = − ∑_{{mk=n \atop m>U, k≥V}} λ(m)( ∑_{d|k,d≤V} μ(d)).  (29)

Thus, we can write

∑_{n≤x} a(n)λ(n)

as S_{1} + S_{2} + S_{3} + S_{4} with appropriate notation. We now suppose that the a(n) are the coefficients (normalized) of our eigenform f. By Cauchy-Schwarz and Rankin-Selberg, we easily deduce that S_{1} ≪ U. We can write S_{2} as

∑_{n≤x} ( ∑_{{h^{2}d=n \atop d≤V}} μ(d))a(n) = ∑_{d≤V} μ(d) ∑_{h≤(x/d)^{1/2}} a(h^{2}d).

The inner sum can be estimated trivially by O((x/d)^{1/2}). This gives S_{2} ≪ x^{1/2+ϵ}V^{1/2}. For S_{3}, we have

S_{3} = − ∑_{t≤UV} ( ∑_{md=t,m≤U,d≤V} μ(d)λ(m)) ∑_{r≤x/t} a(rt).

By Theorem 4, the inner sum is O((x/t)^{1/3}t^{ϵ}), so we get easily S_{3} ≪ x^{1/3}(UV)^{2/3+ϵ}. Finally, for S_{4}, we have

∑_{V ≤k≤x/U} ( ∑_{d|k,d≤V} μ(d)) ∑_{U<m<x/k} λ(m)a(mk).

this can be re-written as

∑_{U<m<x/V} λ(m) ∑_{V <k<x/m} ( ∑_{d|k,d≤V} μ(d))a(mk).

By hypothesis, the inner sum is ≪ (x/m)^{1/2}m^{ϵ} so that we get S_{4} ≪ x^{1+ϵ}/\sqrt{V} . We choose V = x^{1/2} and U = X^{ϵ} to get a final estimate of x^{3/4+ϵ}. Thus, L(s, f) has no zeros for Re(s) > 3/4.

7. Concluding Remarks.

It is clear that the obstacle in proving a quasi-Riemann hypothesis is really the estimation of the sum S_{4}. It is interesting to note that if the sum

∑_{m<x} λ(m)a(mk)

are positive, then one can get the following estimate for S_{4}:

(x/U)^{ϵ} ∑_{V <k≤x/U} ∑_{U<m<x/k} λ(m)a(mk)

which is

≪ (x/U)^{ϵ} ∑_{U<m<x/V} λ(m) ∑_{V <k<x/m} a(mk)

which by Theorem 4 gives a final estimate of x^{1+ϵ}/V^{2/3} which would give a quasi Riemann hypothesis.

Appendices: by Nathan Ng

Modular Analogues of Polya’s Conjecture

Let E be an elliptic curve. The coefficients of its L-series will be denoted a(n). The normalized coefficients will be denoted a_{E}(n) where a_{E}(n) = a(n)/n^{\frac{1}{2}} . The Liouville function is denoted λ(n) where λ(n) = (−1)^{Ω(n)} and Ω(n) is the total number of prime factors of n (counted with multiplicity). Let S_{E}(x) = ∑_{n≤x} a_{E}(n)λ(n) be the generalized Polya sum. Note: In the tables, only the integer part for S_{E} is given.

1 E1 : y^{2} = x^{3} + x^{2} − 7x + 36

rank(E1) = 4

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 201404 & 2 & 322163 & 3 & 422250 & 4 & 511622 & 5 & 592659 \\ 6 & 669422 & 7 & 740673 & 8 & 807658 & 9 & 873727 & 10 & 935762 \\ 11 & 998750 & 12 & 1055369 & 13 & 1111007 & 14 & 1164917 & 15 & 1218562 \\ 16 & 1271467 & 17 & 1324716 & 18 & 1373508 & 19 & 1421993 & 20 & 1468089 \\ 21 & 1516194 & 22 & 1564940 & 23 & 1609313 & 24 & 1653517 & 25 & 1697040 \\ 26 & 1742414 & 27 & 1788221 & 28 & 1829214 & 29 & 1873512 & 30 & 1912127 \\ 31 & 1951990 & 32 & 1994299 & 33 & 2034881 & 34 & 2075782 & 35 & 2113478 \\ 36 & 2152129 & 37 & 2191081 & 38 & 2224929 & 39 & 2262398 & 40 & 2298416 \\ 41 & 2335326 & 42 & 2368912 & 43 & 2407780 & 44 & 2442943 & 45 & 2477384 \\ 46 & 2511918 & 47 & 2546599 & 48 & 2583300 & 49 & 2618861 & 50 & 2652814 \end{matrix}

2 E2 : y^{2} − 21y = x^{3} + 67x^{2} − 10x + 30

rank(E2) = 5

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 217561 & 2 & 353203 & 3 & 467854 & 4 & 570499 & 5 & 664760 \\ 6 & 752802 & 7 & 836816 & 8 & 916978 & 9 & 993251 & 10 & 1066276 \\ 11 & 1136854 & 12 & 1205474 & 13 & 1273073 & 14 & 1339060 & 15 & 1402266 \\ 16 & 1465722 & 17 & 1526688 & 18 & 1586506 & 19 & 1645289 & 20 & 1702981 \\ 21 & 1758113 & 22 & 1814534 & 23 & 1869888 & 24 & 1923348 & 25 & 1976276 \\ 26 & 2028424 & 27 & 2081935 & 28 & 2133258 & 29 & 2184795 & 30 & 2233014 \\ 31 & 2283240 & 32 & 2331103 & 33 & 2380388 & 34 & 2429313 & 35 & 2475573 \\ 36 & 2522469 & 37 & 2569446 & 38 & 2614393 & 39 & 2660464 & 40 & 2706789 \\ 41 & 2750564 & 42 & 2795057 & 43 & 2841453 & 44 & 2885226 & 45 & 2928576 \\ 46 & 2970948 & 47 & 3014348 & 48 & 3056984 & 49 & 3098133 & 50 & 3138632 \end{matrix}

3 E3 : y^{2} − 63y = x^{3} + 351x^{2} + 56x + 22

rank(E3) = 6

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 386697 & 2 & 645957 & 3 & 869445 & 4 & 1072938 & 5 & 1261476 \\ 6 & 1439449 & 7 & 1608641 & 8 & 1771245 & 9 & 1926524 & 10 & 2078573 \\ 11 & 2224311 & 12 & 2369104 & 13 & 2506776 & 14 & 2643033 & 15 & 2777310 \\ 16 & 2908091 & 17 & 3035366 & 18 & 3160920 & 19 & 3283870 & 20 & 3407035 \\ 21 & 3526513 & 22 & 3642749 & 23 & 3760472 & 24 & 3877013 & 25 & 3989843 \\ 26 & 4101297 & 27 & 4211884 & 28 & 4322482 & 29 & 4432330 & 30 & 4539339 \\ 31 & 4646646 & 32 & 4749538 & 33 & 4853587 & 34 & 4957684 & 35 & 5059171 \\ 36 & 5161085 & 37 & 5261785 & 38 & 5358391 & 39 & 5458689 & 40 & 5556704 \\ 41 & 5653294 & 42 & 5751511 & 43 & 5845392 & 44 & 5941619 & 45 & 6034557 \\ 46 & 6128691 & 47 & 6224164 & 48 & 6315399 & 49 & 6409947 & 50 & 6499323 \end{matrix}

4 E4 : y^{2} − 168y = x^{3} + 1641x^{2} + 161x − 8

rank(E4) = 7

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 594145 & 2 & 1015656 & 3 & 1385905 & 4 & 1725542 & 5 & 2043874 \\ 6 & 2346273 & 7 & 2634736 & 8 & 2914172 & 9 & 3183595 & 10 & 3445294 \\ 11 & 3699511 & 12 & 3948636 & 13 & 4191263 & 14 & 4430532 & 15 & 4663520 \\ 16 & 4893186 & 17 & 5118437 & 18 & 5341917 & 19 & 5560982 & 20 & 5776124 \\ 21 & 5989072 & 22 & 6197620 & 23 & 6406369 & 24 & 6612722 & 25 & 6814634 \\ 26 & 7014126 & 27 & 7213935 & 28 & 7410973 & 29 & 7604352 & 30 & 7796756 \\ 31 & 7987525 & 32 & 8177016 & 33 & 8362978 & 34 & 8549392 & 35 & 8733795 \\ 36 & 8918625 & 37 & 9099551 & 38 & 9279117 & 39 & 9458557 & 40 & 9636586 \\ 41 & 9813116 & 42 & 9989408 & 43 & 10161495 & 44 & 10332620 & 45 & 10503675 \\ 46 & 10675408 & 47 & 10847600 & 48 & 11016174 & 49 & 11182080 & 50 & 11350545 \end{matrix}

5 E5 : y^{2} − 2xy + 737y = x^{3} + 531x^{2} + 1262x − 110

rank(E5) = 8

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 746346 & 2 & 1295215 & 3 & 1782625 & 4 & 2234026 & 5 & 2658572 \\ 6 & 3063518 & 7 & 3453141 & 8 & 3830537 & 9 & 4194361 & 10 & 4549210 \\ 11 & 4896000 & 12 & 5234904 & 13 & 5568477 & 14 & 5892719 & 15 & 6213424 \\ 16 & 6529903 & 17 & 6837707 & 18 & 7142781 & 19 & 7444932 & 20 & 7740555 \\ 21 & 8035595 & 22 & 8326564 & 23 & 8611872 & 24 & 8896498 & 25 & 9176337 \\ 26 & 9456621 & 27 & 9731143 & 28 & 10004300 & 29 & 10276113 & 30 & 10542562 \\ 31 & 10810469 & 32 & 11073349 & 33 & 11331322 & 34 & 11591076 & 35 & 11847572 \\ 36 & 12104436 & 37 & 12360929 & 38 & 12611653 & 39 & 12861357 & 40 & 13109258 \\ 41 & 13357360 & 42 & 13602367 & 43 & 13847412 & 44 & 14090376 & 45 & 14332387 \\ 46 & 14571373 & 47 & 14810372 & 48 & 15048835 & 49 & 15282605 & 50 & 15515199 \end{matrix}

6 E6 : y^{2} + 3576y = x^{3} + 9767x^{2} + 425x − 2412

rank(E6) = 9

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 628669 & 2 & 1090005 & 3 & 1498764 & 4 & 1878154 & 5 & 2232601 \\ 6 & 2572880 & 7 & 2898629 & 8 & 3215406 & 9 & 3521342 & 10 & 3820162 \\ 11 & 4108589 & 12 & 4394015 & 13 & 4671069 & 14 & 4946030 & 15 & 5213297 \\ 16 & 5477051 & 17 & 5738393 & 18 & 5994435 & 19 & 6248832 & 20 & 6499274 \\ 21 & 6742563 & 22 & 6985878 & 23 & 7225992 & 24 & 7467909 & 25 & 7702909 \\ 26 & 7934087 & 27 & 8166383 & 28 & 8396313 & 29 & 8621645 & 30 & 8847970 \\ 31 & 9068998 & 32 & 9289189 & 33 & 9509889 & 34 & 9725722 & 35 & 9941257 \\ 36 & 10156603 & 37 & 10369435 & 38 & 10582542 & 39 & 10791065 & 40 & 11003125 \\ 41 & 11209192 & 42 & 11415744 & 43 & 11619274 & 44 & 11824137 & 45 & 12026375 \\ 46 & 12226343 & 47 & 12427274 & 48 & 12629308 & 49 & 12827095 & 50 & 13024838 \end{matrix}

7 E7 : y^{2} − 15336y = x^{3} + 1461695x^{2} − 1414x − 80334

rank(E7) = 10

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 863765 & 2 & 1518178 & 3 & 2103843 & 4 & 2650750 & 5 & 3167285 \\ 6 & 3661074 & 7 & 4138930 & 8 & 4601567 & 9 & 5049942 & 10 & 5490045 \\ 11 & 5918105 & 12 & 6337736 & 13 & 6750994 & 14 & 7154920 & 15 & 7552743 \\ 16 & 7945953 & 17 & 8332984 & 18 & 8714975 & 19 & 9092725 & 20 & 9463593 \\ 21 & 9832013 & 22 & 10197337 & 23 & 10556331 & 24 & 10913126 & 25 & 11265934 \\ 26 & 11616719 & 27 & 11961429 & 28 & 12304890 & 29 & 12645915 & 30 & 12983303 \\ 31 & 13318006 & 32 & 13653228 & 33 & 13983816 & 34 & 14311650 & 35 & 14638627 \\ 36 & 14963131 & 37 & 15283241 & 38 & 15604378 & 39 & 15923548 & 40 & 16237957 \\ 41 & 16551140 & 42 & 16863976 & 43 & 17174866 & 44 & 17485161 & 45 & 17789400 \\ 46 & 18095174 & 47 & 18400360 & 48 & 18702538 & 49 & 19001829 & 50 & 19300679 \end{matrix}

Modular Analogues of Turan’s Conjecture

Let E be an elliptic curve. The coefficients of its L-series will be denoted a(n). The normalized coefficients will be denoted a_{E}(n) where a_{E}(n) = a(n)/n^{\frac{1}{2}} . The Liouville function is denoted λ(n) where λ(n) = (−1)^{Ω(n)} and Ω(n) is the total number of prime factors of n (counted with multiplicity). Let T_{E}(x) = ∑_{n≤x} a_{E}(n)λ(n)/n^{\frac{1}{2}} be the generalized Turan sum. Note: In the tables, only the integer part for T_{E} is given.

8 E1 : y^{2} = x^{3} + x^{2} − 7x + 36

rank(E1) = 4

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 347 & 2 & 608 & 3 & 709 & 4 & 773 & 5 & 821 \\ 6 & 859 & 7 & 920 & 8 & 945 & 9 & 967 & 10 & 987 \\ 11 & 1007 & 12 & 1024 & 13 & 1039 & 14 & 1054 & 15 & 1068 \\ 16 & 1082 & 17 & 1095 & 18 & 1106 & 19 & 1118 & 20 & 1128 \\ 21 & 1139 & 22 & 1149 & 23 & 1159 & 24 & 1168 & 25 & 1176 \\ 26 & 1185 & 27 & 1194 & 28 & 1202 & 29 & 1210 & 30 & 1218 \end{matrix}

9 E2 : y^{2} − 21y = x^{3} + 67x^{2} − 10x + 30

rank(E2) = 5

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 630 & 2 & 743 & 3 & 816 & 4 & 871 & 5 & 916 \\ 6 & 953 & 7 & 986 & 8 & 1016 & 9 & 1042 & 10 & 1066 \\ 11 & 1087 & 12 & 1108 & 13 & 1127 & 14 & 1145 & 15 & 1161 \\ 16 & 1177 & 17 & 1192 & 18 & 1207 & 19 & 1220 & 20 & 1233 \\ 21 & 1246 & 22 & 1258 & 23 & 1269 & 24 & 1280 & 25 & 1291 \\ 26 & 1302 & 27 & 1312 & 28 & 1322 & 29 & 1331 & 30 & 1340 \end{matrix}

10 E3 : y^{2} − 63y = x^{3} + 351x^{2} + 56x + 22

rank(E3) = 6

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 1034 & 2 & 1250 & 3 & 1392 & 4 & 1501 & 5 & 1591 \\ 6 & 1667 & 7 & 1733 & 8 & 1792 & 9 & 1846 & 10 & 1895 \\ 11 & 1940 & 12 & 1983 & 13 & 2022 & 14 & 2059 & 15 & 2094 \\ 16 & 2127 & 17 & 2159 & 18 & 2189 & 19 & 2217 & 20 & 2245 \\ 21 & 2272 & 22 & 2297 & 23 & 2321 & 24 & 2345 & 25 & 2368 \\ 26 & 2390 & 27 & 2412 & 28 & 2433 & 29 & 2453 & 30 & 2474 \end{matrix}

11 E4 : y^{2} − 168y = x^{3} + 1641x^{2} + 161x − 8

rank(E4) = 7

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 1498 & 2 & 1848 & 3 & 2084 & 4 & 2266 & 5 & 2417 \\ 6 & 2546 & 7 & 2659 & 8 & 2761 & 9 & 2854 & 10 & 2939 \\ 11 & 3017 & 12 & 3091 & 13 & 3159 & 14 & 3224 & 15 & 3286 \\ 16 & 3344 & 17 & 3399 & 18 & 3453 & 19 & 3504 & 20 & 3553 \\ 21 & 3600 & 22 & 3645 & 23 & 3689 & 24 & 3731 & 25 & 3772 \\ 26 & 3811 & 27 & 3850 & 28 & 3888 & 29 & 3924 & 30 & 3960 \end{matrix}

12 E5 : y^{2} − 2xy + 737y = x^{3} + 531x^{2} + 1262x − 110

rank(E5) = 8

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 1821 & 2 & 2278 & 3 & 2589 & 4 & 2831 & 5 & 3031 \\ 6 & 3204 & 7 & 3357 & 8 & 3495 & 9 & 3620 & 10 & 3735 \\ 11 & 3842 & 12 & 3942 & 13 & 4037 & 14 & 4125 & 15 & 4209 \\ 16 & 4289 & 17 & 4365 & 18 & 4438 & 19 & 4508 & 20 & 4575 \\ 21 & 4641 & 22 & 4703 & 23 & 4763 & 24 & 4822 & 25 & 4879 \\ 26 & 4934 & 27 & 4988 & 28 & 5040 & 29 & 5091 & 30 & 5140 \end{matrix}

13 E6 : y^{2} + 3576y = x^{3} + 9767x^{2} + 425x − 2412

rank(E6) = 9

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 1548 & 2 & 1932 & 3 & 2192 & 4 & 2396 & 5 & 2563 \\ 6 & 2708 & 7 & 2836 & 8 & 2952 & 9 & 3057 & 10 & 3154 \\ 11 & 3243 & 12 & 3327 & 13 & 3405 & 14 & 3480 & 15 & 3550 \\ 16 & 3617 & 17 & 3682 & 18 & 3743 & 19 & 3802 & 20 & 3859 \\ 21 & 3913 & 22 & 3965 & 23 & 4016 & 24 & 4066 & 25 & 4113 \\ 26 & 4159 & 27 & 4204 & 28 & 4248 & 29 & 4290 & 30 & 4332 \end{matrix}

14 E7 : y^{2} − 15336y = x^{3} + 1461695x^{2} − 1414x − 80334

rank(E7) = 10

\begin{matrix}n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] & n & [S_{E}(n · 10^{6})] \\ 1 & 2060 & 2 & 2604 & 3 & 2977 & 4 & 3270 & 5 & 3514 \\ 6 & 3725 & 7 & 3913 & 8 & 4082 & 9 & 4236 & 10 & 4379 \\ 11 & 4511 & 12 & 4635 & 13 & 4751 & 14 & 4861 & 15 & 4966 \\ 16 & 5066 & 17 & 5161 & 18 & 5252 & 19 & 5340 & 20 & 5424 \\ 21 & 5506 & 22 & 5584 & 23 & 5660 & 24 & 5734 & 25 & 5805 \\ 26 & 5874 & 27 & 5941 & 28 & 6007 & 29 & 6071 & 30 & 6133 \end{matrix}

REFERENCES

[EM] J. Esmonde and M. Ram Murty, Problems in Algebraic Number Theory, Graduate Texts in Mathematics, 190, Springer-Verlag, 1999.

[He] D. Hejhal, On the distribution of log |ζ′(1/2 + it)|, in Number Theory, Trace Formulas, and Discrete Groups, (Oslo, 1987), pp. 343-370, Academic Press, MA, 1989.

[Go] S. Gonek, On negative moments of the Riemann zeta functions, Mathematika, 36 (1989), no. 1, pp. 71-88.

[Po] G. Pólya, Verschiedene Bemerkungen zur Zahlentheorie, Jahresbericht der deutschen Math.-Vereinigung, 28 (1919) pp. 31-40.

[Ha] C.B. Haselgrove, A disproof of a conjecture of Pólya, Mathematika, 5 (1958), pp. 141-145.

[KM] V. Kumar Murty, Modular Elliptic Curves, in Seminar on Fermat’s Last Theorem, ed. K. Murty CMS Conference Proceedings, Volume 17, amer. Math. Soc., Providence, RI, 1995.

[MP] M. Ram Murty and A. Perelli, The pair correlation of zeros of functions in the Selberg class, Inter. Math. Res. Notices, 10 (1999) pp. 531-546.

[T] P. Turán, On some approximative Dirichlet polynomials in the theory of the zeta function of Riemann, Danske Vid. Selsk. Mat.-Fys. Medd., 24 (1948), no. 17, 36 pp. (see also his Collected Papers.)