#  Riemann Hypothesis and Random Walks: the Zeta case

André<EMAIL_ADDRESS>Cornell University, Physics
Department, Ithaca, NY 14850

###### Abstract

In previous work it was shown that if certain series based on sums over primes
of non-principal Dirichlet characters have a conjectured random walk behavior,
then the Euler product formula for its $L$-function is valid to the right of
the critical line $\Re(s)>\tfrac{1}{2}$, and the Riemann Hypothesis for this
class of $L$-functions follows. Building on this work, here we propose how to
extend this line of reasoning to the Riemann zeta function and other principal
Dirichlet $L$-functions. We apply these results to the study of the argument
of the zeta function. In another application, we define and study a 1-point
correlation function of the Riemann zeros, which leads to the construction of
a probabilistic model for them. Based on these results we describe a new
algorithm for computing very high Riemann zeros, and we calculate the googol-
th zero, namely $10^{100}$-th zero to over 100 digits, far beyond what is
currently known.

##  I Introduction

There are many generalizations of Riemann’s zeta function to other Dirichlet
series, which are also believed to satisfy a Riemann Hypothesis. A common
opinion, based largely on counterexamples, is that the $L$-functions for which
the Riemann Hypothesis is true enjoy both an Euler product formula and a
functional equation. However a direct connection between these properties and
the Riemann Hypothesis has not been formulated in a precise manner. In EPF1 ;
EPF2  a concrete proposal making such a connection was presented for Dirichlet
$L$-functions, and those based on cusp forms, due to the validity of the Euler
product formula to the right of the critical line. In contrast to the non-
principal case, in this approach the case of principal Dirichlet
$L$-functions, of which Riemann zeta is the simplest, turned out to be more
delicate, and consequently it was more difficult to state precise results. In
the present work we address further this special case.

Let $\chi(n)$ be a Dirichlet character modulo $k$ and $L(s,\chi)$ its
$L$-function with $s=\sigma+it$. It satisfies the Euler product formula

$L(s,\chi)=\sum_{n=1}^{\infty}\dfrac{\chi(n)}{n^{s}}=\prod_{n=1}^{\infty}\left(1-\dfrac{\chi(p_{n})}{p_{n}^{s}}\right)^{-1}$
(1)

where $p_{n}$ is the $n$-th prime. The above formula is valid for $\Re(s)>1$
since both sides converge absolutely. The important distinction between
principal verses non-principal characters is the following. For non-principal
characters the $L$-function has no pole at $s=1$, thus there exists the
possibility that the Euler product is valid partway inside the strip, i.e. has
abscissa of convergence $\sigma_{c}<1$. It was proposed in EPF1 ; EPF2  that
$\sigma_{c}=\tfrac{1}{2}$ for this case. In contrast, now consider
$L$-functions based on principal characters. The latter character is defined
as $\chi(n)=1$ if $n$ is coprime to $k$ and zero otherwise. The Riemann zeta
function is the trivial principal character of modulus $k=1$ with all
$\chi(n)=1$. $L$-functions based on principal characters do have a pole at
$s=1$, and therefore have abscissa of convergence $\sigma_{c}=1$, which
implies the Euler product in the form given above cannot be valid inside the
critical strip $0<\sigma<1$. Nevertheless, in this paper we will show how a
truncated version of the Euler product formula is valid for
$\sigma>\tfrac{1}{2}$.

The primary aim of the work EPF1 ; EPF2  was to determine what specific
properties of the prime numbers would imply that the Riemann Hypothesis is
true. This is the opposite of the more well-studied question of what the
validity of the Riemann Hypothesis implies for the fluctuations in the
distribution of primes. The answer proposed was simply based on the
multiplicative independence of the primes, which to a large extent underlies
their pseudo-random behavior. To be more specific, let
$\chi(n)=e^{i\theta_{n}}$ for $\chi(n)\neq 0$. In EPF1 ; EPF2  it was proven
that if the series

$B_{N}(t,\chi)=\sum_{n=1}^{N}\cos\left(t\log p_{n}+\theta_{p_{n}}\right)$ (2)

is $O(\sqrt{N})$, then the Euler product converges for $\sigma>\tfrac{1}{2}$
and the formula (1) is valid to the right of the critical line. In fact, we
only need $B_{N}=O(\sqrt{N})$ up to logs (see Remark 1); when we write write
$O(\sqrt{N})$, it is implicit that this can be relaxed with logarithmic
factors. For non-principal characters the allowed angles $\theta_{n}$ are
equally spaced on the unit circle, and it was conjectured in EPF2  that the
above series with $t=0$ behaves like a random walk due to the multiplicative
independence of the primes, and this is the origin of the $O(\sqrt{N})$
growth. Furthermore, this result extends to all $t$ since domains of
convergence of Dirichlet series are always half-planes. Taking the logarithm
of (1), one sees that $\log L$ is never infinite to the right of the critical
line and thus has no zeros there. This, combined with the functional equation
that relates $L(s)$ to $L(1-s)$, implies there are also no zeros to the left
of the critical line, so that all zeros are on the line. The same reasoning
applies to cusp forms if one also uses a non-trivial result of Deligne EPF2 .

In this article we reconsider the principal Dirichlet case, specializing to
Riemann zeta itself since identical arguments apply to all other principal
cases with $k>1$. Here all angles $\theta_{n}=0$, so one needs to consider the
series

$B_{N}(t)=\sum_{n=1}^{N}\cos(t\log p_{n})$ (3)

which now strongly depends on $t$. On the one hand, whereas the case of
principal Dirichlet $L$-functions is complicated by the existence of the pole,
and, as we will see, one consequently needs to truncate the Euler product to
make sense of it, on the other hand $B_{N}$ can be estimated using the prime
number theorem since it does not involve sums over non-trivial characters
$\chi$, and this aids the analysis. This is in contrast to the non-principal
case, where, however well-motivated, we had to conjecture the random walk
behavior alluded to above, so in this respect the principal case is
potentially simpler. To this end, a theorem of Kac (Theorem 1 below) nearly
does the job: $B_{N}(t)=O(\sqrt{N})$ in the limit $t\to\infty$, which is also
a consequence of the multiplicative independence of the primes. This suggests
that one can also make sense of the Euler product formula in the limit
$t\to\infty$. However this is not enough for our main purpose, which is to
have a similar result for finite $t$ which we will develop.

This article is mainly based on our previous work EPF1 ; EPF2  but provides a
more detailed analysis and extends it in several ways. It was suggested in
EPF1  that one should truncate the series at an $N$ that depends on $t$.
First, in the next section we explain how a simple group structure underlies a
finite Euler product which relates it to a generalized Dirichlet series which
is a subseries of the Riemann zeta function. Subsequently we estimate the
error under truncation, which shows explicitly how this error is related to
the pole at $s=1$, as expected. The remainder of the paper, sections IV-VI,
presents various applications of these ideas. We use them to study the
argument of the zeta function. We present an algorithm to calculate very high
zeros, far beyond what is currently known. We also study the statistical
fluctuations of individual zeros, in other words, a 1-point correlation
function.

In many respects, our work is related to the work of Gonek et. al. Gonek1 ;
Gonek2 , which also considers a truncated Euler product. The important
difference is that the starting point in Gonek1  is a hybrid version of the
Euler product which involves both primes and zeros of zeta. Only after
assuming the Riemann Hypothesis can one explain in that approach why the
truncated product over primes is a good approximation to zeta. In contrast,
here we do not assume anything about the zeros of zeta, since the goal is to
actually understand their location.

We are unable to provide fully rigorous proofs of some of the statements
below, however we do provide supporting calculations and numerical work. In
order to be clear on this, below “Proposal” signifies the most important
claims that we could not rigorously prove.

##  II Algebraic structure of finite Euler products

The aim of this section is to define properly the objects we will be dealing
with. In particular we will place finite Euler products on the same footing as
other generalized Dirichlet series. The results are straightforward and are
mainly definitions.

######  Definition 1.

Fix a positive integer $N$ and let $\\{p_{1},p_{2},\ldots p_{N}\\}$ denote the
first $N$ primes where $p_{1}=2$. From this set one can generate an abelian
group $\mathbb{Q}_{N}$ of rank $N$ with elements

$\mathbb{Q}_{N}=\Bigl{\\{}p_{1}^{n_{1}}p_{2}^{n_{2}}\cdots
p_{N}^{n_{N}},~{}~{}n_{i}\in\mathbb{Z}~{}\forall_{i}\Bigr{\\}}$ (4)

where the group operation is ordinary multiplication. Clearly
$\mathbb{Q}_{N}\subset\mathbb{Q}^{+}$ where $\mathbb{Q}^{+}$ are the positive
rational numbers. There are an infinite number of integers in $\mathbb{Q}_{N}$
which form a subset of the natural numbers $\mathbb{N}=\\{1,2,\ldots\\}$. We
will denote this set as $\mathbb{N}_{N}\subset\mathbb{N}$, and elements of
this set simply as $\mathfrak{n}$.

######  Definition 2.

Fix a positive integer $N$. For every integer $n\in\mathbb{N}$ we can define
the character $c(n)$:

$\displaystyle c(n)$ $\displaystyle=$ $\displaystyle 1~{}~{}~{}~{}~{}{\rm
if}~{}n\in\mathbb{N}_{N}\subset\mathbb{Q}_{N}$ (5) $\displaystyle=$
$\displaystyle 0~{}~{}~{}~{}~{}{\rm otherwise}$

Clearly, for a prime $p$, $c(p)=0$ if $p>p_{N}$.

######  Definition 3.

Fix a positive integer $N$ and let $s$ be a complex number. Based on
$\mathbb{Q}_{N}$ we can define the infinite series

$\zeta_{N}(s)=\sum_{n=1}^{\infty}\dfrac{c(n)}{n^{s}}=\sum_{\mathfrak{n}\
\in\mathbb{N}_{N}}\dfrac{1}{\mathfrak{n}^{s}}$ (6)

which is a generalized Dirichlet series. There are an infinite number of terms
in the above series since $\mathbb{N}_{N}$ is infinite dimensional.

######  Example 1.

For instance

$\zeta_{2}(s)=1+\dfrac{1}{2^{s}}+\dfrac{1}{3^{s}}+\dfrac{1}{4^{s}}+\dfrac{1}{6^{s}}+\dfrac{1}{8^{s}}+\dfrac{1}{9^{s}}+\dfrac{1}{12^{s}}+\ldots$

Because of the group structure of $\mathbb{Q}_{N}$, $\zeta_{N}$ satisfies a
finite Euler product formula:

######  Proposition 1.

Let $\sigma_{c}$ be the abscissa of convergence of the series $\zeta_{N}(s)$
where $s=\sigma+it$, namely $\zeta_{N}(s)$ converges for $\Re(s)>\sigma_{c}$.
Then in this region of convergence, $\zeta_{N}$ satisfies a finite Euler
product formula:

$\zeta_{N}(s)=\prod_{n=1}^{N}\left(1-\dfrac{1}{p_{n}^{s}}\right)^{-1}$ (7)

###### Proof.

Based on the completely multiplicative property of the characters,

$c(nm)=c(n)c(m)$ (8)

one has

$\zeta_{N}(s)=\prod_{n=1}^{\infty}\left(1-\dfrac{c(p_{n})}{p_{n}^{s}}\right)^{-1}$

The result follows then from the fact that $c(p_{n})=0$ if $n>N$. ∎

######  Example 2.

Let $N=1$, so that $\\{\mathfrak{n}\\}=\\{1,2,2^{2},2^{3}\ldots\\}$. Then the
above Euler product formula (7) is simply the standard formula for the sum of
a geometric series:

$\zeta_{1}(s)=\sum_{n=0}^{\infty}\dfrac{1}{2^{ns}}=\dfrac{1}{1-2^{-s}}$ (9)

Here the abscissa of convergence is $\sigma_{c}=0$.

The series $\zeta_{N}(s)$ defined in (6) has some interesting properties:

(i) For finite $N$ the product is finite for $s\neq 0$, thus the infinite
series $\zeta_{N}(s)$ converges for $\Re(s)>0$ for any finite $N$.

(ii) Since the logarithm of the product is finite, for finite $N$,
$\zeta_{N}(s)$ has no zeros nor poles for $\Re(s)>0$. Thus the Riemann zeros
and the pole at $s=1$ arise from the primes at infinity $p_{\infty}$, i.e. in
the limit $N\to\infty$. In this limit all integers are included in the sum (6)
that defines $\zeta_{N}$ since $\mathbb{N}_{\infty}=\mathbb{N}$. This is in
accordance with the fact that the pole is a consequence of there being an
infinite number of primes.

The property (ii) implies that, in some sense, the Riemann zeros condense out
of the primes at infinity $p_{\infty}$. Formally one has

$\lim_{N\to\infty}\zeta_{N}(s)=\zeta(s)$ (10)

However since $N$ is going to infinity, the above is true only where the
series formally converges, which, as discussed in the Introduction, is
$\Re(s)>1$. Nevertheless, for very large but finite $N$, the function
$\zeta_{N}$ can still be a good approximation to $\zeta(s)$ inside the
critical strip since for $N$ finite there is convergence of $\zeta_{N}(s)$ for
$\Re(s)>0$. This is the subject of the next section, where we show that a
finite Euler product formula is valid for $\Re(s)>\tfrac{1}{2}$ in a manner
that we will specify.

##  III Finite Euler product formula at large $N$ to the right of the critical
liine.

In this section we propose that the Euler product formula can be a very good
approximation to $\zeta(s)$ for $\Re(s)>\tfrac{1}{2}$ and large $t$ if $N$ is
chosen to depend on $t$ in a specific way which was already proposed in EPF1 ;
EPF2 . The new result presented here is an estimate of the error due to the
truncation.

The random walk property we will build upon is based on a central limit
theorem of Kac Kac , which largely follows from the multiplicative
independence of the primes:

######  Theorem 1.

(Kac) Let $u$ be a random variable uniformly distributed on the interval
$u\in[T,2T]$, and define the series

$B_{N}(u)=\sum_{n=1}^{N}\cos(u\log p_{n})$ (11)

Then in the limit $N\to\infty$ and $T\to\infty$, $B_{N}/\sqrt{N}$ approaches
the normal distribution ${\cal N}(0,1)$, namely

$\lim_{N\to\infty}\lim_{T\to\infty}P\Biggl{\\{}\dfrac{x_{1}}{\sqrt{2}}<\dfrac{B_{N}(u)}{\sqrt{N}}<\dfrac{x_{2}}{\sqrt{2}}\Biggr{\\}}=\dfrac{1}{\sqrt{2\pi}}\int_{x_{1}}^{x_{2}}e^{-x^{2}/2}dx$
(12)

where $P$ denotes the probability for the set.

We wish to use the above theorem to conclude something about $B_{N}(t)$ for a
fixed, non-random $t$. Based on Theorem 1, we first conclude the following for
non-random, but large $t$:

######  Corollary 1.

For any $\epsilon>0$,

$\lim_{t\to\infty}B_{N}(t)=O(N^{1/2+\epsilon})$ (13)

###### Proof.

This is straightforward: as $T\to\infty$, even though $u$ is random, all $u$
in the range $[T,2T]$ are tending to $\infty$. One then uses the normal
distribution in Theorem 1. ∎

######  Remark 1.

The proof of convergence of the Euler product in EPF2  is not spoiled if the
bound on $B_{N}$ is relaxed up to logs. For instance, if in the limit
$t\to\infty$, $B_{N}=O(\sqrt{N\log\log N})$, as suggested by the law of
iterated logarithms relevant to central limit theorems, this is fine, as is
$B_{N}=O(\sqrt{N}\log^{a}N)$ for any positive power $a$.

A consequence of Theorem 1 and the Corollary 1 is that the Euler product
formula is valid to the right of the critical line in the limit $t\to\infty$,
at least formally. Namely for $\sigma>\tfrac{1}{2}$,

$\lim_{t\to\infty}\zeta(\sigma+it)=\lim_{N\to\infty}\lim_{t\to\infty}\prod_{n=1}^{N}\left(1-\dfrac{1}{p_{n}^{\sigma+it}}\right)^{-1}$
(14)

As shown in EPF1 ; EPF2  and discussed in the Introduction, this formally
follows from the $\sqrt{N}$ growth of $B_{N}$. The problem with the above
formula is that due to the double limit on the RHS, it is not rigorously
defined. For instance, it could depend on the order of limits. It is thus
desirable to have a version of (14) where $N$ and $t$ are taken to infinity
simultaneously. Namely, we wish to truncate the product at an $N(t)$ that
depends on $t$ with the property that $\lim_{t\to\infty}N(t)=\infty$. One can
then replace the double limit on the RHS of (14) with one limit $t\to\infty$,
or equivalently $N(t)\to\infty$.

There is no unique choice for $N(t)$, but there is an optimal upper limit,
$N(t)<N_{\rm max}(t)\equiv[t^{2}]$, with $[t^{2}]$ its integer part, which we
now describe. We can use the prime number theorem to estimate $B_{N}(t)$:

$\displaystyle B_{N}(t)\approx\int_{2}^{p_{N}}\dfrac{dx}{\log x}\cos(t\log x)$
$\displaystyle=$ $\displaystyle\Re\left({\rm Ei}\left((1+it)\log
p_{N}\right)\right)$ $\displaystyle\approx$ $\displaystyle\dfrac{p_{N}}{\log
p_{N}}\left(\dfrac{t}{1+t^{2}}\right)\sin\left(t\log p_{N}\right)$

where ${\rm Ei}$ is the usual exponential-integral function, and we have used

${\rm Ei}(z)=\dfrac{e^{z}}{z}\left(1+O\left(\dfrac{1}{z}\right)\right)$ (16)

The prime number theorem implies $p_{N}\approx N\log N$. Using this in (III)
and imposing $B_{N}(t)<\sqrt{N}$ leads to $N<[t^{2}]$.

Based on the above, henceforth we will always assume the following properties
of $N(t)$:

$N(t)\leq N_{\rm max}(t)\equiv[t^{2}]~{}~{}~{}~{}{\rm
with}~{}~{}~{}~{}~{}\lim_{t\to\infty}N(t)=\infty,$ (17)

and will not always display the $t$ dependence of $N$. Equation (14) now
formally becomes

$\lim_{t\to\infty}\zeta(s)=\lim_{t\to\infty}\prod_{n=1}^{N(t)}\left(1-\dfrac{1}{p_{n}^{s}}\right)^{-1},~{}~{}~{}~{}~{}{\rm
for}~{}\Re(s)>\tfrac{1}{2}$ (18)

Extensive and compelling numerical evidence supporting the above formula was
already presented in EPF1 .

Based on the above results we are now in a position to study the following
important question. If we fix a finite but large $t$, and truncate the Euler
product at $N(t)$, which is finite, what is the error in the approximation to
$\zeta$ to the right of the critical line? We estimate this error as follows:

######  Proposal 1.

Let $N=N(t)$ satisfy (17). Then for $\Re(s)>\tfrac{1}{2}$ and large $t$,

$\zeta(s)=\prod_{n=1}^{N(t)}\left(1-\dfrac{1}{p_{n}^{s}}\right)^{-1}\exp\left(R_{N}(s)\right)$
(19)

where $\zeta(s)$ is the actual $\zeta$ function defined by analytic
continuation and

$R_{N}(s)=\dfrac{1}{(s-1)}\,O\left(\dfrac{N^{1-s}}{\log^{s}N}\right).$ (20)

$R_{N}$ is finite (except at the pole $s=1$) and satisfies

$\lim_{t\to\infty}R_{N(t)}(s)=0,$ (21)

namely the error goes to zero as $t\to\infty$.

We provide the following supporting argument, although not a rigorous proof,
for this Proposal. From (18), one concludes that (19) must hold in the limit
of large $t$ with $R_{N}$ satisfying (21). The logarithm of (19) reads

$\log\zeta(s)=-\sum_{n=1}^{N}\log\left(1-\dfrac{1}{p_{n}^{s}}\right)+R_{N}(s)$
(22)

First assume $\Re(s)>1$. Then in the limit of large $t$, the error upon
truncation is the part that is neglected in (18):

$R_{N}(s)=-\sum_{n=N+1}^{\infty}\log\left(1-\dfrac{1}{p_{n}^{s}}\right)$ (23)

Expanding out the logarithm, one has

$\displaystyle R_{N}(s)$ $\displaystyle\approx$
$\displaystyle\sum_{n=N}^{\infty}\dfrac{1}{p_{n}^{s}}$ (24)
$\displaystyle\approx$ $\displaystyle\int_{p_{N}}^{\infty}\dfrac{dx}{\log
x}\dfrac{1}{x^{s}}\approx\dfrac{1}{(s-1)}\dfrac{p_{N}^{1-s}}{\log p_{N}}$

where in the second line we again used the prime number theorem to approximate
the sum over primes. Next using $p_{N}\approx N\log N$, one obtains (20).
Finally, the above expression can be continued into the strip
$\sigma>\tfrac{1}{2}$ if $N(t)<[t^{2}]$ since $N(t)^{1-s}/t<N^{1/2-s}$ which
goes to zero as $N\to\infty$ if $\Re(s)>\tfrac{1}{2}$. The latter also implies
(21).

Proposal 1 makes it clear that the need for a cut-off $N<N_{\rm max}$
originates from the pole at $s=1$, since as long as $s\neq 1$, the error
$R_{N}(s)$ in (20) is finite. The error becomes smaller and smaller the
further one is from the pole, i.e. as $t\to\infty$. In Figure 1 we numerically
illustrate Proposal 1 inside the critical strip.

######  Remark 2.

For estimating errors at large $t$ the following formula is useful:

$|R_{N(t)}(s)|\sim\dfrac{N(t)^{1-\sigma}}{t}\sim\dfrac{1}{t^{2\sigma-1}}$ (25)

Figure 1: The error term $|R_{N}(s)|$ with $N(t)=N_{\rm max}(t)=[t^{2}]$ for
$\Re(s)=3/4$ inside the critical strip as a function of $t$. The fluctuating
(blue) curve is $|R_{N}|$ computed directly from the definition (19) with
$\zeta(s)$ the usual analytic continuation into the strip. The smooth (yellow)
curve is the approximation
$R_{N}(s)=\dfrac{1}{(s-1)}\,\dfrac{N^{1-s}}{\log^{s}N}$ based on (20).

######  Theorem 2.

Assuming Proposal 1, all non-trivial zeros of $\zeta(s)$ are on the critical
line.

###### Proof.

Taking the logarithm of the truncated Euler product, one obtains (22). If
there were a zero $\rho$ with $\Re(\rho)>\tfrac{1}{2}$, then
$\log\zeta(\rho)=-\infty$. However the right hand side of (22) is always
finite, thus there are no zeros to the right of the critical line. The
functional equation relating $\zeta(s)$ to $\zeta(1-s)$ shows there are also
no zeros to the left of the critical line. ∎

######  Remark 3.

Interestingly, Proposal 1 and Theorem 2 imply that proving the validity of the
Riemann Hypothesis is under better control the higher one moves up the
critical line. For instance, it is known that all zeros are on the line up to
$t\sim 10^{13}$, and beyond this, the error $R_{N}$ is too small to spoil the
validity of the Riemann Hypothesis. Henceforth, we assume the RH.

##  IV The argument of the $\zeta$-function near the critical line

In our work Trans , $a(t)$ was defined as follows:

$a(t)=\lim_{\delta\to
0^{+}}\dfrac{1}{\pi}\arg\zeta\left(\tfrac{1}{2}+\delta+it\right)$ (26)

It is important in the above definition that $\delta$ is not allowed to be
strictly zero. It will also be important that the $\delta\to 0$ limit
approaches the critical line from the right  because this is the region where
the (truncated) Euler product formula is valid in the sense described above.
The above definition for $a(t)$ is not identical to that of the conventional
$S(t)=\arg\zeta(\tfrac{1}{2}+it)/\pi$, and one should not assume they are the
same. For instance, it is well known that $S(t)$ is not defined at the
ordinate of a zero, whereas $a(t)$ is. (More generally, the argument of any
analytic function at a zero is well-defined once the contour by which it is
approached is specified.) The behavior of $a(t)$ would be completely different
if it were defined as a limit from the left.

######  Proposal 2.

The function $a(t)$ is well defined and finite for all $t$, i.e. $a(t)=O(1)$.

For the remainder of the section we provide arguments supporting this
Proposal. Based on the Euler product formula (Proposal 1) one has

$a(t)=\dfrac{1}{\pi}\Im\log\zeta\left(\tfrac{1}{2}+\delta+it\right)=-\dfrac{1}{\pi}\Im\sum_{n=1}^{N(t)}\log\left(1-\dfrac{1}{p_{n}^{1/2+\delta+it}}\right)+\dfrac{1}{\pi}\Im\,R_{N}(\tfrac{1}{2}+\delta+it)$
(27)

where the limit $\delta\to 0^{+}$ is implicit. Recall that as $t\to\infty$,
$R_{N}$ actually goes to zero. One can check numerically that the above
formula works rather well with $R_{N}$ disregarded; see Figure 2. From this
figure one clearly sees that the above formula knows about all the Riemann
zeros even if one neglects the $R_{N}$ error term, since it jumps by one at
each zero.

Figure 2: The exact $a(t)$ (blue line) verses $a(t)$ calculated from the
primes, i.e. the Euler product formula (27) (yellow line). Here we took
$\delta=0.01$ and $N=10^{5}<N_{\rm max}(t)$.

It is clear that based on (27), $a(t)=O(1)$ because it is finite for all $t$.
Let us try to be more specific based on our results thus far. Under the
assumption of Proposal 1, which implies the Euler product formula (27) for
$a(t)$, then $a(t)$ is well-defined for all $t$. Let us fix $N=N(t)$
satisfying (17). Expanding the logarithm, one has

$a(t)=\lim_{\delta\to
0^{+}}\dfrac{1}{\pi}\Im\sum_{n=1}^{N(t)}\dfrac{1}{p_{n}^{1/2+\delta+it}}+O(1)$
(28)

We neglected the $R_{N}$ error since it is also $O(1)$ by (25). The first term
is finite, thus $a(t)$ is finite.

As for other functions defined by sums over primes, such as the prime number
counting function $\pi(x)$, there is a leading smooth part which is determined
by the prime number theorem, and a sub-leading fluctuating part that depends
on the exact locations of the primes. We can therefore write

$a(t)=a_{{\rm pnt}}(t)+\Delta a(t)$ (29)

where $a_{{\rm pnt}}(t)$ is the smooth part coming from the prime number
theorem, and $\Delta a(t)$ are the fluctuating corrections. Consider first the
smooth part:

$\displaystyle a_{{\rm pnt}}(t)$ $\displaystyle=$
$\displaystyle\dfrac{1}{\pi}\Im\int_{2}^{p_{N}}\dfrac{dx}{\log
x}\dfrac{e^{-it\log x}}{\sqrt{x}}$ $\displaystyle=$
$\displaystyle\dfrac{1}{\pi}\Im\left({\rm Ei}\left[(\tfrac{1}{2}-it)\log
p_{N}\right]-{\rm Ei}\left[(\tfrac{1}{2}-it)\log 2\right]\right)$

For $y>0$:

$\Im({\rm Ei}(-iy))=-\pi+\dfrac{\cos y}{y}+O\left(\dfrac{1}{y^{2}}\right)$
(31)

Thus $\lim_{y\to\infty}\Im\left({\rm Ei}(-iy)\right)=-\pi$. Now, as
$t\to\infty$, in (IV) one can replace $\tfrac{1}{2}-it$ with $-it$, and the
two terms cancel:

$\lim_{t\to\infty}a_{{\rm pnt}}(t)=0$ (32)

Let us now turn to the fluctuating term $\Delta a(t)$ which actually knows
about the locations of the zeros since at each zero it jumps by its
multiplicity. Since the leading contribution $a_{{\rm pnt}}$ goes to zero,
$\Delta a(t)$ has no growth and consists only of these jumps, all occurring
around $a=0$, which is consistent with the average of $a(t)$ being zero.

If one assumes all zeros of $\zeta$ are simple, as Theorem 3 below would
imply, then one can further argue that $a(t)$ is nearly always on the
principal branch: $-1\lesssim a(t)\lesssim 1$. If all zeros are simple, then
$a(t)$ jumps by only $1$ at each zero. Thus the largest value of $|\Delta
a(t)|$ is approximately $1$ corresponding to a jump beginning at $t\approx 0$.
In other words, $a(t)$ is never very far from zero so that most of the jumps
pass through $a=0$ as seen in Figure 2.

Figure 2 provides numerical evidence for the above statements. Simply stated,
the Proposal 2 says that there is no change in behavior of $a(t)$ as $t$
increases to infinity, such that the pattern in Figure 2 persists. We checked
its validity all the way up to $t=10^{12}$. Only rarely is $|a(t)|$ slightly
above $1$. Over this whole range we found $|a(t)|<1.2$.

##  V $1$-point correlation function of the Riemann zeros

Montgomery conjectured that the pair correlation function of ordinates of the
Riemann zeros on the critical line satisfy GUE statistics Montgomery . Being a
2-point correlation function, it is a reasonably complicated statistic. In
this section we propose a simpler 1-point correlation function that captures
the statistical fluctuations of individual zeros.

Let $t_{n}$ be the exact ordinate of the $n$-th zero on the critical line,
with $t_{1}=14.1347...$ and so forth. The single equation $\zeta(\rho)=0$ is
known to have an infinite number of non-trivial solutions
$\rho=\tfrac{1}{2}+it_{n}$. In Trans , by placing the zeros in one-to-one
correspondence with the zeros of a cosine function, the single equation
$\zeta(\rho)=0$ was replaced by an infinite number of equations, one for each
$t_{n}$ that depends only on $n$:

$\vartheta(t_{n})+\lim_{\delta\to
0^{+}}\arg\zeta(\tfrac{1}{2}+\delta+it_{n})=(n-\tfrac{3}{2})\pi$ (33)

where $\vartheta$ is the Riemann-Siegel function:

$\vartheta(t)=\Im\log\Gamma(\tfrac{1}{4}+\tfrac{it}{2})-t\log\sqrt{\pi}$ (34)

The $\arg\zeta$ term equals $\pi a(t)$ discussed in the last section. It is
important that the $\delta\to 0^{+}$ approaches the critical line from the
right, since this is where the Euler product formula is valid in the sense
described above. This equation was used to calculate zeros very accurately in
Trans , up to thousands of digits. There is no need for a cut-off $N_{\rm
max}$ in the above equation since the $\arg\zeta$ term is defined for
arbitrarily high $t$ by standard analytic continuation. One aspect of this
equation is the following theorem:

######  Theorem 3.

(França-LeClair) If there is a unique solution to the equation (33) for every
positive integer $n$, then the Riemann Hypothesis is true, and furthermore,
all zeros are simple.

######  Remark 4.

Details of the proof are in Trans . The main idea is that if there is a unique
solution, then the zeros are enumerated by the integer $n$ and can be counted
along the critical line, and the resulting counting formula coincides with a
well known result due to Backlund for the number of zeros in the entire
critical strip. The zeros are simple because the zeros of the cosine are
simple. The above theorem is another approach towards proving the Riemann
Hypothesis, however it is not entirely independent of the above approach based
on the Euler product formula, in particular Theorem 2. In Trans , we were
unable to prove there is a unique solution because we did not have sufficient
control over the relevant properties of the function $a(t)$. The previous
section helps close this gap in our understanding of $a(t)$ by showing that
$a(t)$ is indeed well defined at the zeros and consequently there should be a
unique solution to (33) for all $n$.

If the $\arg\zeta$ term is ignored, then there is indeed a unique solution for
all $n$ since $\vartheta(t)$ is a monotonically increasing function of $t$.
Using its asymptotic expansion for large $t$, equation (40) below, and
dropping the $O(1/t)$ term, then the solution is

$\widetilde{t}_{n}=\dfrac{2\pi(n-\tfrac{11}{8})}{W\left((n-\tfrac{11}{8})/e\right)}$
(35)

where $W$ is the Lambert $W$-function. The only way there would fail to be a
solution is if $a(t)$ is not well defined for all $t$. However, as discussed
above, this would appear to contradict the analysis of the last section, in
particular Proposal 2.

The fluctuations in the zeros come from $a(t)$ since $\widetilde{t}_{n}$ is a
smooth function of $n$. These small fluctuations are shown in Figure 3. Let us
define $\delta t_{n}=t_{n}-\widetilde{t}_{n}$. One needs to properly normalize
$\delta t_{n}$, taking into account that the spacing between zeros decreases
as $2\pi/\log n$. To this end we expand the equation (33) around
$\widetilde{t}_{n}$. Using
$\vartheta(\widetilde{t}_{n})\approx(n-\tfrac{3}{2})\pi$, one obtains $\delta
t_{n}\approx-\pi a(t_{n})/\vartheta^{\prime}(\widetilde{t}_{n})$ where
$\vartheta^{\prime}(t)$ is the derivative with respect to $t$. Using
$\vartheta^{\prime}(t)\approx\tfrac{1}{2}\log(t/2\pi e)$, this leads us to
define

$\delta_{n}\equiv\dfrac{(t_{n}-\widetilde{t}_{n})}{2\pi}\log\left(\dfrac{\widetilde{t}_{n}}{2\pi
e}\right)\approx-a(t_{n})$ (36)

The probability distribution of the set

$\Delta_{M}\equiv\Bigl{\\{}\delta_{1},\delta_{2},\ldots,\delta_{M}\Bigr{\\}}$
(37)

for large $M$ is then an interesting property to study. Here “probability” is
defined as frequency of occurrence. The equation (36) together with (27) makes
it clear that the origin of the statistical fluctuations of $\Delta_{M}$ is
the fluctuations in the primes.

Figure 3:  The first 30 Riemann zeros $t_{n}$. The smooth curve is the
approximation $\widetilde{t}_{n}$ in (35), whereas the dots are the actual
zeros $t_{n}$.

In Figure 4 we plot the distribution of $\Delta_{M}$ for $M=10^{5}$. It
closely resembles a normal distribution, however as we will argue, we believe
it is not exactly normal. Let us first suppose $\Delta_{M}$ does satisfy a
normal distribution ${\cal N}(\mu,\sigma_{1})$. Using the properties of
$a(t_{n})$ described in the last section, together with the equation (36), we
can propose then the following. First, one expects that the average of
$\delta_{n}$ is zero since it is known that the average of $a(t)$ is zero,
thus $\mu=0$. Secondly, if $a(t)$ is nearly always on the principal branch, as
argued in the last section, then at each jump by $1$ at $t_{n}$, on average
$a(t_{n})$ passes through zero. This implies that the average
$\overline{|a(t_{n})|}\approx 1/4$. For a normal distribution
$\overline{|a(t_{n})|}=\sqrt{\tfrac{2}{\pi}}\,\sigma_{1}$. Thus one expects
the standard deviation $\sigma_{1}$ of $\Delta_{M}$ to be
$\sigma_{1}\approx\sqrt{\pi/32}=0.313..$. In Figure 4 we present results for
the first $10^{5}$-th known exact zeros. The distribution function fits a
normal distribution with $\sigma_{1}=\sqrt{\pi/32}$ rather well. Performing a
fit, one finds $\sigma_{1}\approx 0.27$. For higher values of $M$ around
$10^{6}$, a fit gives $\sigma_{1}\approx 0.3$, which is closer to the
predicted value.

Figure 4: The probability distribution for the set $\Delta_{M}$ defined in
(37) for $M=10^{5}$. The smooth curve is the normal distribution ${\cal
N}(0,\sigma_{1})$ with $\sigma_{1}=0.274$.

The p-values for the fit to the normal distribution in Figure 4 are quite low.
For instance $p-{\rm value}<0.01$ for the Pearson $\chi^{2}$ test. Other tests
have even worse p-values. This leads us to believe that the distribution is
not precisely normal. If the distribution were exactly normal, then some
$\delta_{n}$ would be arbitrarily large. Equation (36) would then imply that
$a(t_{n})$ could also be arbitrarily large, which contradicts Proposal 2.
These issues played an important role in our proposal that the normalized gaps
between zeros have an upper bound ZeroGaps . For these reasons, we believe
that $\Delta_{M}$ has a distribution that is only approximately normal.

If we approximate the distribution of $\Delta_{M}$ as normal, then we can
construct a simple probabilistic model of the Riemann zeros:

######  Definition 4.

A probabilistic model of the Riemann zeros. Let $\mathfrak{r}$ be a random
variable with normal distribution ${\cal N}(0,\sigma_{1})$. Then a
probabilistic model of the zeros $t_{n}$ can be defined as the set
$\\{\widehat{t}_{n}\\}$, where

$\widehat{t}_{n}\equiv\widetilde{t}_{n}+\dfrac{2\pi\,\mathfrak{r}}{\log(\widetilde{t}_{n}/2\pi
e)}$ (38)

and $\widetilde{t}_{n}$ is defined in (35). In the above formula
$\mathfrak{r}$ is chosen at random independently for each $n$.

The statistical model (38) is rather simplistic since it is just based on a
normal distribution for $\mathfrak{r}$ and $\widetilde{t}_{n}$ is smooth and
completely deterministic. A natural question then arises. Does the pair
correlation function of $\\{\widehat{t}_{n}\\}$ satisfy GUE statistics as does
the actual zeros $\\{t_{n}\\}$? We expect the answer is no, since the only
correlation between pairs of $\widehat{t}_{n}$’s is the smooth, predictable
part $\widetilde{t}_{n}$. Nevertheless, it is interesting to study the 2-point
correlation function of $\\{\widehat{t}_{n}\\}$. Montgomery’s pair correlation
conjecture can be stated as follows. Let ${\cal N}(T)$ denote the number of
zeros up to height $T$, where ${\cal
N}(T)\approx\tfrac{T}{2\pi}\log\left(\tfrac{T}{2\pi e}\right)$. Let
$t,t^{\prime}$ denote zeros in the range $[0,T]$. Then in the limit of large
$T$:

$\dfrac{1}{{\cal
N}(T)}\sum_{\alpha<d(t,t^{\prime})<\beta}1\sim\int_{\alpha}^{\beta}du\left(1-\dfrac{\sin^{2}(\pi
u)}{\pi^{2}u^{2}}\right)$ (39)

where $d(t,t^{\prime})$ is a normalized distance between zeros
$d(t,t^{\prime})=\tfrac{1}{2\pi}\log\left(\tfrac{T}{2\pi
e}\right)(t-t^{\prime})$.

In Figure 5 we plot the pair correlation function for the first $10^{5}$-th
$\widehat{t}_{n}$’s. We chose $\sigma_{1}=0.274$ since in this range of $n$
this gives a better fit to the normal distribution of the 1-point function.
The results are reasonably close to the GUE prediction (39), especially
considering that for just the first $10^{5}$ true zeros the fit to the GUE
prediction is not perfect; for much higher zeros it is significantly better
OdlyzkoPair . We interpret the deviation from the GUE prediction to be
additional evidence that the distribution of the set $\Delta_{M}$ is not
exactly normal.

Figure 5: The pair correction function of $\\{\widehat{t}_{n}\\}$ defined in
(38) for $n$ up to $10^{5}$ where the standard deviation of $\mathfrak{r}$ was
taken to be $\sigma_{1}=0.274$. The solid curve is the GUE prediction. The
parameters in (39) are $\beta=\alpha+0.05$ with
$\alpha=(0,0.05,0.10,\ldots,3)$ and the $x$-axis is given by
$x=(\alpha+\beta)/2$.

##  VI Computing very high zeros from the primes

This section can be viewed as providing additional numerical evidence for some
of the previous results. Since we will be calculating $a(t)$ from the primes
using (27), which requires $\Re(s)\to\tfrac{1}{2}^{+}$, this is pushing the
limit of the validity of the Euler product formula, nevertheless we will
obtain reasonable results. We emphasize that this method has nothing to do
with the random model for the zeros in Definition 4, but rather relies on the
Euler product formula to calculate $a(t)$.

Many very high zeros of $\zeta$ have been computed numerically, beginning with
the work of Odlyzko. All zeros up to the $10^{13}$-th have been computed and
are all on the critical line Gourdon . Beyond this the computation of zeros
remains a challenging open problem. However some zeros around the $10^{21}$-st
and $10^{22}$-nd are known Odlyzko . In this section we describe a new and
simple algorithm for computing very high zeros based on the results of Section
IV. It will allow us to go much higher than the known zeros since it does not
require numerical implementation of the $\zeta$ function itself, but rather
only requires knowledge of some of the lower primes.

Let us first discuss the numerical challenges involved in computing high zeros
from the equation (33) based on the standard Mathematica package. The main
difficulty is that one needs to implement the $\arg\zeta$ term. Mathematica
computes ${\rm Arg}\,\zeta$, i.e. on the principal branch, however near a zero
this is likely to be valid based on the discussion in section IV. The main
problem is that Mathematica can only compute $\zeta$ for $t$ below some
maximum value around $t=10^{10}$. This was sufficient to calculate up to the
$n=10^{9}$-th zero from (33) in Trans . The $\log\Gamma$ term must also be
implemented to very high $t$, which is also limited in Mathematica.

We deal with these difficulties first by computing $\arg\zeta$ from the
formula (27) involving a finite sum over primes. We will neglect the $R_{N}$
term at first in (27) since it vanishes in the limit $t\to\infty$; however we
will return to it when we will estimate the error in computing zeros this way.
Then, the $\log\Gamma$ term can be accurately computed using corrections to
Stirling’s formula:

$\vartheta(t)=\dfrac{t}{2}\log\left(\dfrac{t}{2\pi
e}\right)-\dfrac{\pi}{8}+\dfrac{1}{48\,t}+O(1/t^{3})$ (40)

Let ${t_{n;N}}$ denote the ordinate of the $n$-th zero computed using the
first $N$ primes based on (33). For high zeros, it is approximately the
solution to the following equation

$\dfrac{{t_{n;N}}}{2}\log\left(\dfrac{{t_{n;N}}}{2\pi
e}\right)-\dfrac{\pi}{8}-\lim_{\delta\to
0^{+}}\Im\sum_{k=1}^{N}\log\left(1-\dfrac{1}{p_{k}^{1/2+\delta+i{t_{n;N}}}}\right)=(n-\tfrac{3}{2})\pi$
(41)

where it is implicit that $N<N_{\rm max}(t)=[t^{2}]$. The important property
of this equation is that it no longer makes any reference to $\zeta$ itself.
It is straightforward to solve the above equation with standard root-finder
software, such as FindRoot in Mathematica.

One can view the computation of $t_{n}$ as a kind of Markov process. If one
includes no primes, i.e. $N=0$, and drops the next to leading $1/t$
corrections, then the solution is unique and explicitly given by
$t_{n;0}=\widetilde{t}_{n}$ in terms of the Lambert $W$-function in (35). One
then goes from $t_{n;0}$ to $t_{n;1}$ by finding the root to the equation for
$t_{n;1}$ in the vicinity of $t_{n;0}$, then similarly $t_{n;2}$ is calculated
based on $t_{n;1}$ and so forth. At each step in the process one includes one
additional prime, and this slowly approaches $t_{n}$, so long as $N(t)<N_{\rm
max}(t)$. In practice we did not follow this iterative procedure, but rather
fixed $N$ and simply solved (41) in the vicinity of $\widetilde{t}_{n}$.

We can estimate the error in computing the zero $t_{n}$ from the primes using
equation (41) as follows. As in Section V, we expand the equation (33) now
around $t_{n;N}$ rather than $\widetilde{t}_{n}$. One obtains

$t_{n}-{t_{n;N}}=-\pi\,da_{N}/\vartheta^{\prime}({t_{n;N}})$

where $da_{N}$ is the error in computing $a(t)$ from the primes (27). Using
(24), we have

$da_{N}=\tfrac{1}{\pi}\Im
R_{N}(s=\tfrac{1}{2}+it)\approx\dfrac{\sqrt{p_{N}}}{\pi t\log p_{N}}\cos(t\log
p_{N}).$

Now from the prime number theorem, $p_{N}\approx N\log N$. Recall $N$ is cut
off at $N_{\rm max}=[t^{2}]$, which cancels the $1/t$ in the previous formula.
Finally it is meaningful to normalize the error by the mean spacing $2\pi/\log
n$. The result is

$\dfrac{t_{n}-{t_{n;N}}}{\scriptstyle{2\pi/\log
n}}\approx\dfrac{1}{\pi\sqrt{\log N}}\cos\left(t_{n}\log p_{N}\right)$ (42)

where we have used ${t_{n;N}}\approx\widetilde{t}_{n}\approx 2\pi n/\log n$.
The left hand side represents the ratio of the error to the mean spacing
between zeros at that height. Again, it is implicit that $N<[t_{n}^{2}]$. The
interesting aspect of the above formula is that the relative error decreases
with $N$, although rather slowly. The cosine factor also implies there are
large scale oscillations around the actual $t_{n}$.

For very high $t$, $N_{\rm max}(t)=[t^{2}]$ is extremely large and it is not
possible in practice to work with such a large number of primes. This is the
primary limitation to the accuracy we can obtain. We will limit ourselves to
the relatively small $N=5\times 10^{6}$ primes. Let us verify the method by
comparing with some known zeros around $n=10^{21}$ and $10^{22}$. The results
are shown in Table 1. Equation (42) predicts $t_{n}-t_{n;N}\approx 0.01$ for
these $n$ and $N$, and inspection of the table shows this is a good estimate.
Odlyzko was of course able to calculate more digits; our accuracy can be
improved by increasing $N$ in principle. We also checked some zeros around the
$n=10^{33}$-rd computed by Hiary Hiary , again with favorable results.

$n$ |  ${t_{n;N}}$ |  $t_{n}$ (Odlyzko)
---|---|---
$10^{21}-1$ | $144176897509546973538.205$ | $\sim.225$
$10^{21}$ | $144176897509546973538.301$ | $\sim.291$
$10^{21}+1$ | $144176897509546973538.505$ | $\sim.498$
$|$ | $|$ | $|$
$10^{22}-1$ | $1370919909931995308226.498$ | $\sim.490$
$10^{22}$ | $1370919909931995308226.614$ | $\sim.627$
$10^{22}+1$ | $1370919909931995308226.692$ | $\sim.680$
Table 1: Zeros around the $n=10^{21}$-st and $10^{22}$-nd computed from (41)
with $N=5\times 10^{6}$ primes. We fixed $\delta=10^{-6}$. Above, $\sim$
denotes the integer part of the second column.

Having made this check, let us now go far beyond this and compute the
$n=10^{100}$-th zero by the same method. Again using only $N=5\times 10^{6}$
primes, we found the following $t_{n}$:

$\displaystyle n=10^{100}{\rm-th~{}zero}:$
$\displaystyle{t_{n}=280690383842894069903195445838256400084548030162846}\atop{~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}045192360059224930922349073043060335653109252473.244....}$

Obtaining this number took only a few minutes on a laptop using Mathematica.
We are confident that the last $3$ digits $\sim.244$ are correct since we
checked that they didn’t change between $N=10^{6}$ and $5\times 10^{6}$.
Furthermore, $3$ digits is consistent with (42), which predicts that for these
$n$ and $N$, $t_{n}-{t_{n;N}}\approx 0.002$. We calculated the next zero to be
$\sim.273$.

We were able to extend this calculation to the $10^{1000}$-th zero without
much difficulty. As equation (42) shows, the relative error only decreases as
one increases $t$. It is also straightforward to extend this method to all
primitive Dirichlet $L$-functions and those based on cusp forms using the
transcendental equations in Trans  and the results in EPF2 .

## Acknowledgments

We wish to thank Denis Bernard, Guilherme França, Ghaith Hiary, Giuseppe
Mussardo, and German Sierra for discussions. We also thank the Isaac Newton
Institute for Mathematical Sciences for their hospitality in the final stages
of this work (January 2016).

## References

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https://people.math.osu.edu/hiary.1/outd3/out.88837796029624663862630219091085.zeros