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\bibitem{Note01}
{\bf How not to factor the numerator of a Bernoulli number:} \par As mentioned
  in Chapter~\ref {ch:envision}, the coefficient $B_k$ of the linear term of
  the polynomial $$ S_k(n)= 1^k+2^k+3^k+\dots +(n-1)^k $$ is (up to sign) the
  $k$-th {\bf Bernoulli number}. These numbers are rational numbers and,
  putting them in lowest terms, their numerators play a role in certain
  problems, and their denominators in others. (This is an amazing story, which
  we can't go into here!) \par One of us (Barry Mazur) in the recent article
  {\em How can we construct abelian Galois extensions of basic number fields?}
  (see \url
  {http://www.ams.org/journals/bull/2011-48-02/S0273-0979-2011-01326-X/}) found
  himself dealing (for various reasons) with the fraction $-B_{200}/400$, where
  $B_{200}$ is the two-hundredth Bernoulli number. The numerator of this
  fraction is quite large: it is---hold your breath---\ $$389\cdot 691\cdot
  5370056528687 \ \ \ \ \text {times this 204-digit number:} $$\vskip 10pt
  $N:=3452690329392158031464109281736967404068448156842396721012$\newline
  $\mbox {}\,\,\,\qquad
  9920642145194459192569415445652760676623601087497272415557$\newline $\mbox
  {}\,\,\,\qquad
  0842527652727868776362959519620872735612200601036506871681$\newline $\mbox
  {}\,\,\,\qquad 124610986596878180738901486527$ \vskip 10pt and he {\it
  incorrectly asserted} that it was prime. Happily, Bartosz Naskr\c {e}cki
  spotted this error: our $204$-digit $N$ is {\it not} prime. \par How did he
  know this? By using the most basic test in the repertoire of tests that we
  have available to check to see whether a number is prime: we'll call it the
  ``{\bf Fermat $2$-test.}'' We'll first give a general explanation of this
  type of test before we show how $N$ {\it fails the Fermat $2$-test.} \par
  \par The starting idea behind this test is the famous result known as {\em
  Fermat's Little Theorem} where the ``little'' is meant to alliteratively
  distinguish it from you-know-what. \begin {theorem}[Fermat's Little Theorem]
  If $p$ is a prime number, and $a$ is any number relatively prime to $p$ then
  $a^{p-1} - 1$ is divisible by $p$. \end {theorem} A good exercise is to try
  to prove this, and a one-word hint that might lead you to one of the many
  proofs of it is {\em induction}.\footnote {Here's the proof: $$(N+1)^p \equiv
  N^p +1 \equiv (N+1)\ \mod p,$$ where the first equality is the binomial
  theorem and the second equality is induction.} \par Now we are going to use
  this as a criterion, by---in effect---restating it in what logicians would
  call its {\it contrapositive}: \begin {theorem}[The Fermat $a$-test] If $M$
  is a positive integer, and $a$ is any number relatively prime to $M$ such
  that $a^{M-1} - 1$ is {\it not} divisible by $M$, then $M$ is {\it not} a
  prime. \end {theorem} \par Well, Naskr\c {e}cki computed $2^{N-1}-1$ (for the
  $204$-digit $N$ above) and saw that it is {\it not} divisible\footnote {The
  number $2^{N-1}-1$ has a residue after division by $N$ of\\
  $3334581100595953025153969739282790317394606677381970645616725285996925$\newline
  $
  6610000568292727335792620957159782739813115005451450864072425835484898$\newline
  $ 565112763692970799269335402819507605691622173717318335512037457$.} by $N$.
  Ergo, our $N$ fails the Fermat $2$-test so is {\it not} prime. \par But then,
  given that it is so ``easy'' to see that $N$ is not prime, a natural question
  to ask is: what, in fact, is its prime factorization? This---it turns
  out---isn't so easy to determine; Naskr\c {e}cki devoted $24$ hours of
  computer time setting standard factorization algorithms on the task, and that
  was not sufficient time to resolve the issue. The factorization of the
  numerators of general Bernoulli numbers is the subject of a very interesting
  website run by Samuel Wagstaff (\url
  {http://homes.cerias.purdue.edu/~ssw/bernoulli}). Linked to this web page one
  finds (\url {http://homes.cerias.purdue.edu/~ssw/bernoulli/composite}) which
  gives a list of composite numbers whose factorizations have resisted all
  attempts to date. The two-hundredth Bernoulli number is 12th on the list.
  \par The page \url
  {http://en.wikipedia.org/wiki/Integer_factorization_records} lists record
  challenge factorization, and one challenge that was completed in 2009
  involves a difficult-to-factor number with 232 digits; its factorization was
  completed by a large team of researchers and took around 2000 years of CPU
  time. This convinced us that with sufficient motivation it would be possible
  to factor $N$, and so we asked some leaders in the field to try. They
  succeeded! \begin {quote}\sf Factorisation of B200 \par by Bill Hart on 4 Aug
  05, 2012 at 07:24pm \par We are happy to announce the factorization of the
  numerator of the 200th Bernoulli number: \par \begin {align*} N &= 389 \cdot
  691 \cdot 5370056528687 \cdot c_{204}\\ c_{204} &= p_{90} \cdot p_{115}\\
  p_{90} &= 149474329044343594528784250333645983079497454292\\ &=
  838248852612270757617561057674257880592603\\ p_{115} &=
  230988849487852221315416645031371036732923661613\\ &=
  619208811597595398791184043153272314198502348476\\ &= 2629703896050377709
  \end {align*} \par The factorization of the 204-digit composite was made
  possible with the help of many people: \begin {itemize} \item William Stein
  and Barry Mazur challenged us to factor this number. \item Sam Wagstaff
  maintains a table of factorizations of numerators of Bernoulli numbers at
  \url {http://homes.cerias.purdue.edu/~ssw/bernoulli/bnum}. According to this
  table, the 200th Bernoulli number is the 2nd smallest index with unfactored
  numerator (the first being the 188th Bernoulli number). \item Cyril Bouvier
  tried to factor the c204 by ECM up to 60-digit level, using the TALC cluster
  at Inria Nancy - Grand Est. \item {\sf yoyo@home} tried to factor the c204 by
  ECM up to 65-digit level, using the help of many volunteers of the
  distributed computing platform \url {http://www.rechenkraft.net/yoyo/}. After
  ECM was unsuccessful, we decided to factor the c204 by GNFS. \item Many
  people at the Mersenne forum helped for the polynomial selection. The best
  polynomial was found by Shi Bai, using his implementation of Kleinjung's
  algorithm in CADO-NFS: \url
  {http://www.mersenneforum.org/showthread.php?p=298264##post298264}. Sieving
  was performed by many volunteers using {\sf NFS@home}, thanks to Greg
  Childers. See \url {http://escatter11.fullerton.edu/nfs} for more details
  about {\sf NFS@home}. This factorization showed that such a distributed
  effort might be feasible for a new record GNFS factorization, in particular
  for the polynomial selection. This was the largest GNFS factorization
  performed by {\sf NFS@home} to date, the second largest being $2^{1040}+1$ at
  183.7 digits. \item Two independent runs of the filtering and linear algebra
  were done: one by Greg Childers with msieve (\url
  {http://www.boo.net/~jasonp/qs.html}) using a 48-core cluster made available
  by Bill Hart, and one by Emmanuel Thom\'{e} and Paul Zimmermann with CADO-NFS
  (\url {http://cado-nfs.gforge.inria.fr/}), using the Grid 5000 platform.
  \item The first linear algebra run to complete was the one with CADO-NFS,
  thus we decided to stop the other run. \end {itemize} \par Bill Hart \end
  {quote} \par We verify the factorization above in SageMath as follows: \\
  \par {\sf sage: p90 = 1494743290443435945287842503336459830794974542928382\\
  48852612270757617561057674257880592603\\ sage: p115 =
  230988849487852221315416645031371036732923661613619\\
  2088115975953987911840431532723141985023484762629703896050377709\\ sage: c204
  = p90 * p115\\ sage: 389 * 691 * 5370056528687 * c204 ==
  -numerator(bernoulli(200))\\ True\\ sage: is\_prime(p90), is\_prime(p115),
  is\_prime(c204)\\ (True, True, False) }.

\bibitem{Note02}
\label {bibnote:factor} Given an integer $n$, there are many algorithms
  available for trying to write $n$ as a product of prime numbers. First we can
  apply {\em trial division}, where we simply divide $n$ by each prime $2, 3,
  5, 7, 11, 13, \ldots $ in turn, and see what small prime factors we find (up
  to a few digits). After using this method to eliminate as many primes as we
  have patience to eliminate, we typically next turn to a technique called {\em
  Lenstra's elliptic curve method}, which allows us to check $n$ for
  divisibility by bigger primes (e.g., around 10--15 digits). Once we've
  exhausted our patience using the elliptic curve method, we would next hit our
  number with something called the {\em quadratic sieve}, which works well for
  factoring numbers of the form $n=pq$, with $p$ and $q$ primes of roughly
  equal size, and $n$ having less than 100 digits (say, though the 100 depends
  greatly on the implementation). All of the above algorithms---and then
  some---are implemented in SageMath, and used by default when you type {\tt
  factor(n)} into SageMath. Try typing {\tt factor(some number, verbose=8)} to
  see for yourself. \par If the quadratic sieve fails, a final recourse is to
  run the {\em number field sieve} algorithm, possibly on a supercomputer. To
  give a sense of how powerful (or powerless, depending on perspective!) the
  number field sieve is, a record-setting factorization of a general number
  using this algorithm is the factorization of a 232 digit number called
  RSA-768 (see \url {https://eprint.iacr.org/2010/006.pdf}): \par $ n =
  12301866845301177551304949583849627207728535695953347921973224521\\
  517264005072636575187452021997864693899564749427740638459251925573263\\
  034537315482685079170261221429134616704292143116022212404792747377940\\
  80665351419597459856902143413 $ \par \noindent {}which factors as $pq$,
  where\par \noindent {} $ p =
  334780716989568987860441698482126908177047949837137685689124313889\\
  82883793878002287614711652531743087737814467999489 $ \par \noindent {}and\par
  \noindent {} $ q =
  367460436667995904282446337996279526322791581643430876426760322838\\
  15739666511279233373417143396810270092798736308917. $ \par \noindent {}We
  encourage you to try to factor $n$ in SageMath, and see that it fails.
  SageMath does not yet include an implementation of the number field sieve
  algorithm, though there are some free implementations currently available
  (see \url {http://www.boo.net/~jasonp/qs.html}).

\bibitem{Note03}
We can use SageMath (at \url {http://sagemath.org}) to quickly compute the
  ``hefty number'' $p = 2^{43,112,609}-1$. Simply type {\tt p = 2\^{}43112609 -
  1} to instantly compute $p$. In what sense have we {\em computed} $p$?
  Internally, $p$ is now stored in base $2$ in the computer's memory; given the
  special form of $p$ it is not surprising that it took little time to compute.
  Much more challenging is to compute all the base 10 digits of $p$, which
  takes a few seconds: {\tt d = str(p)}. Now type {\tt d[-50:]} to see the last
  50 digits of $p$. To compute the sum $58416637$ of the digits of $p$ type
  {\tt sum(p.digits())}.

\bibitem{Note04}
In contrast to the situation with factorization, testing integers of this size
  (e.g., the primes $p$ and $q$) for primality is relatively easy. There are
  fast algorithms that can tell whether or not any random thousand digit number
  is prime in a fraction of second. Try for yourself using the SageMath command
  {\tt is\_prime}. For example, if $p$ and $q$ are as in endnote~\ref
  {bibnote:factor}, then {\tt is\_prime(p)} and {\tt is\_prime(q)} quickly
  output True and {\tt is\_prime(p*q)} outputs False. However, if you type {\tt
  factor(p*q, verbose=8)} you can watch as SageMath tries forever and fails to
  factor $pq$.

\bibitem{Note05}
In Sage, the function {\tt prime\_range} enumerates primes in a range. For
  example, {\tt prime\_range(50)} outputs the primes up to $50$ and {\tt
  prime\_range(50,100)} outputs the primes between $50$ and $100$. Typing {\tt
  prime\_range(10\^{}8)} in SageMath enumerates the primes up to a hundred
  million in around a second. You can also enumerate primes up to a billion by
  typing {\tt v=prime\_range(10\^{}9)}, but this will use a large amount of
  memory, so be careful not to crash your computer if you try this. You can see
  that there are $\pi (10^9) = 50{,}847{,}534$ primes up to a billion by then
  typing {\tt len(v)}. You can also compute $\pi (10^9)$ directly, without
  enumerating all primes, using the command {\tt prime\_pi(10\^{}9)}. This is
  much faster since it uses some clever counting tricks to find the number of
  primes without actually listing them all. \par In Chapter~\ref
  {sec:tinkering} we tinkered with the staircase of primes by first counting
  both primes and prime powers. There are comparatively few prime powers that
  are not prime. Up to $10^8$, only $1{,}405$ of the $5{,}762{,}860$ prime
  powers are not themselves primes. To see this, first enter {\tt
  a~=~prime\_pi(10\^{}8); pp~=~len(prime\_powers(10\^{}8))}. Typing {\tt
  (a,~pp,~pp-a)} then outputs the triple {\tt (5761455, 5762860, 1405)}.

\bibitem{Note06}
Hardy and Littlewood give a nice conjectural answer to such questions about
  gaps between primes. See Problem {\bf A8} of Guy's book {\em Unsolved
  Problems in Number Theory} (2004). Note that Guy's book discusses counting
  the number $P_k(X)$ of pairs of primes up to $X$ that differ by a fixed even
  number $k$; we have $P_k(X)\geq \Gap _k(X)$, since for $P_k(X)$ there is no
  requirement that the pairs of primes be consecutive.

\bibitem{Note07}
If $f(x)$ and $g(x)$ are real-valued functions of a real variable $x$ such that
  for any $\epsilon >0$ both of them take their values between $ x^{1-\epsilon
  }$ and $ x^{1+\epsilon }$ for $x$ sufficiently large, then say that $f(x)$
  and $g(x)$ are {\bf good approximations of one another} if, for any positive
  $\epsilon $ the absolute value of their difference is less than $ x^{{1\over
  2}+\epsilon }$ for $x$ sufficiently large. The functions $\Li (X)$ and $R(X)$
  are good approximations of one another.

\bibitem{Note08}
This computation of $\pi (X)$ was done by David J. Platt in 2012, and is the
  largest value of $\pi (X)$ ever computed. See \url
  {http://arxiv.org/abs/1203.5712} for more details.

\bibitem{Note09}
In fact, the Riemann Hypothesis is equivalent to $|\Li (X) - \pi (X)| \leq
  \sqrt {X}\log (X)$ for all $X\geq 2.01$. See Section 1.4.1 of Crandall and
  Pomerance's book {\em Prime Numbers: A Computational Perspective}.

\bibitem{Note10}
For a proof of this here's a hint. Compute the difference between the
  derivatives of $\Li (x)$ and of $x/\log x$. The answer is $1/\log ^2(x)$. So
  you must show that the ratio of $\int _2^X dx/\log ^2(x)$ to $\Li (x)= \int
  _2^Xdx/\log (x)$ tends to zero as~$x$ goes to infinity, and this is a good
  calculus exercise.

\bibitem{Note11}
See \url {http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/} for the
  original German version and an English translation.

\bibitem{Note12}
We have $$\psi (X) =\sum _{p^n \le X}\log p$$ where the summation is over prime
  powers $p^n$ that are $\le X$.

\bibitem{Note13}
See \url {http://en.wikipedia.org/wiki/Fast_Fourier_transform}.

\bibitem{Note14}
\url {http://en.wikipedia.org/wiki/Distribution_\%28mathematics\%29} contains a
  more formal definition and treatment of distributions. Here is Schwartz's
  explanation for his choice of the word {\it distribution:} \begin
  {quote}``Why did we choose the name distribution? Because, if $\mu $ is a
  measure, i.e., a particular kind of distribution, it can be considered as a
  distribution of electric charges in the universe. Distributions give more
  general types of electric charges, for example dipoles and magnetic
  distributions. If we consider the dipole placed at the point $a$ having
  magnetic moment $M$, we easily see that it is defined by the distribution
  $-D_M \delta _{(a)}$. These objects occur in physics. Deny's thesis, which he
  defended shortly after, introduced electric distributions of finite energy,
  the only ones which really occur in practice; these objects really are
  distributions, and do not correspond to measures. Thus, distributions have
  two very different aspects: they are a generalization of the notion of
  function, and a generalization of the notion of distribution of electric
  charges in space. [...] Both these interpretations of distributions are
  currently used.''\end {quote}.

\bibitem{Note15}
David Mumford suggested that we offer the following paragraph from\\ \url
  {http://en.wikipedia.org/wiki/Dirac_delta_function}\\ on the Dirac delta
  function: \begin {quote} An infinitesimal formula for an infinitely tall,
  unit impulse delta function (infinitesimal version of Cauchy distribution)
  explicitly appears in an 1827 text of Augustin Louis Cauchy. Sim\'{e}on Denis
  Poisson considered the issue in connection with the study of wave propagation
  as did Gustav Kirchhoff somewhat later. Kirchhoff and Hermann von Helmholtz
  also introduced the unit impulse as a limit of Gaussians, which also
  corresponded to Lord Kelvin's notion of a point heat source. At the end of
  the 19th century, Oliver Heaviside used formal Fourier series to manipulate
  the unit impulse. The Dirac delta function as such was introduced as a
  ``convenient notation'' by Paul Dirac in his influential 1930 book {\em
  Principles of Quantum Mechanics}. He called it the ``delta function'' since
  he used it as a continuous analogue of the discrete Kronecker delta. \end
  {quote}.

\bibitem{Note16}
As discussed in\\ \url
  {http://en.wikipedia.org/wiki/Distribution_\%28mathematics\%29},
  ``generalized functions'' were introduced by Sergei Sobolev in the 1930s,
  then later independently introduced in the late 1940s by Laurent Schwartz,
  who developed a comprehensive theory of distributions.

\bibitem{Note17}
If the \RH {} holds they are precisely the {\it imaginary parts of the
  ``nontrivial'' zeroes of the Riemann zeta function}.

\bibitem{Note18}
{\it The construction of $ \Phi (t)$ from $\psi (X)$:} \par Succinctly, for
  positive real numbers $t$, $$\Phi (t): \ = \ e^{- t/2}\Psi '(t),$$ where
  $\Psi (t) = \psi (e^t)$ (see Figure~\ref {fig:psi_200}), and $\Psi '$ is the
  derivative of $\Psi (t)$, viewed as a distribution. We extend this function
  to all real arguments $t$ by requiring $\Phi (t)$ to be an even function of
  $t$, i.e., $\Phi (-t)= \Phi (t)$. But, to review this at a more leisurely
  pace, \begin {enumerate} \par \item Distort the $X$-axis of our staircase by
  replacing the variable $X$ by $e^t$ to get the function $$\Psi (t):= \psi
  (e^t).$$ No harm is done by this for we can retrieve our original $\psi (X)$
  as $$\psi (X) = \Psi (\log (X)).$$ Our distorted staircase has risers at ($0$
  and) all positive integral multiples of logs of prime numbers. \par \par \par
  \par \item Now we'll do something that might seem a bit more brutal: {\it
  take the derivative of this distorted staircase $\Psi (t)$.} This derivative
  $\Psi '(t)$ is a {\it generalized} function with support at all nonnegative
  integral multiples of logs of prime numbers. \par \vskip 20pt \ill
  {bigPsi_prime}{1}{$\Psi '(t)$ is a (weighted) sum of Dirac delta functions at
  the logarithms of prime powers $p^n$ weighted by $\log (p)$ (and by $\log
  (2\pi )$ at $0$). The taller the arrow, the larger the weight.} \par \item
  Now---for normalization purposes---multiply $\Psi '(t)$ by the function
  $e^{-t/2}$ which has no effect whatsoever on the support. \end {enumerate}
  \vskip 20pt\ill {psi_200}{.4}{Illustration of the staircase $\psi (X)$
  constructed in Chapter~\ref {sec:tinkering} that counts weighted prime
  powers.\label {fig:psi_200}} \par \par \par In summary: The generalized
  function that resulted from the above carpentry: $$\Phi (t) = e^{-t/2}\Psi
  '(t),$$ retains the information we want (the placement of primes) even if in
  a slightly different format.

\bibitem{Note19}
A version of the Riemann--von Mangoldt explicit formula gives some theoretical
  affirmation of the phenomena we are seeing here. We thank Andrew Granville
  for a discussion about this. \par \ill {granville}{0.25}{Andrew Granville}
  \par \par Even though the present endnote is not the place to give anything
  like a full account, we can't resist setting down a few of Granville's
  comments that might be helpful to people who wish to go further. (This
  discussion can be modified to analyze what happens unconditionally, but we
  will be assuming the \RH {} below.) The function ${\hat \Phi }_{\le C}(\theta
  )$ that we are graphing in this chapter can be written as: $${\hat \Phi
  }_{\le C}(\theta )= \sum _{n\le C}\Lambda (n)n^{-w}$$ where $w = {\frac
  {1}{2}}+i\theta $. This function, in turn, may be written (by Perron's
  formula) as \begin {align*} &{\frac {1}{2\pi i }}\lim _{T \to \infty }\int
  _{s=\sigma _o-iT}^{s=\sigma _o+iT}\sum _{n}\Lambda (n)n^{-w}\left ({\frac
  {C}{n}}\right )^{s}{\frac {ds}{s}}\\ &= {\frac {1}{2\pi i }}\lim _{T \to
  \infty }\int _{s=\sigma _o-iT}^{s=\sigma _o+iT}\sum _{n}\Lambda
  (n)n^{-w-s}C^{s}{\frac {ds}{s}}\\ &= -{\frac {1}{2\pi i }}\lim _{T \to \infty
  }\int _{s=\sigma _o-iT}^{s=\sigma _o+iT}\left ({\frac {\zeta '}{\zeta
  }}\right )(w+s){\frac {C^{s}}{s}}ds. \end {align*} \par Here, we assume that
  $\sigma _o$ is sufficiently large and $C$ is not a prime power. \par One
  proceeds, as is standard in the derivation of Explicit Formulae, by moving
  the line of integration to the left, picking up residues along the way. Fix
  the value of $w= {\frac {1}{2}}+i\theta $ and consider $$K_w(s,C):=\ \ {\frac
  {1}{2\pi i }}\left (\frac {\zeta '}{\zeta }\right )(w+s){\frac {C^{s}}{s}},$$
  which has poles at $$s= 0, \ \ \ 1-w,\ \ \ {\rm and }\ \ \ \rho -w, $$ for
  every zero $\rho $ of $\zeta (s)$. We distinguish five cases, giving
  descriptive names to each: \par \begin {enumerate} \item {\it Singular pole:}
  $s= 1-w$. \item {\it Trivial poles:} $s= \rho -w$ with $\rho $ a trivial zero
  of $\zeta (s)$. \item {\it Oscillatory poles:} $s= \rho -w = i(\gamma -\theta
  ) \ne 0$ with $\rho = 1/2 + i \gamma (\ne w)$ a nontrivial zero of $\zeta
  (s)$. (Recall that we are assuming the \RH {}, and our variable $w= {\frac
  {1}{2}}+i\theta $ runs through complex numbers of real part equal to ${\frac
  {1}{2}}$. So, in this case, $s$ is purely imaginary.) \item {\it Elementary
  pole:} $s=0$ when $ w$ is not a nontrivial zero of $\zeta (s)$---i.e., when
  $0=s\ne \rho -w$ for any nontrivial zero $\rho $. \item {\it Double pole:}
  $s=0$ when $ w$ is a nontrivial zero of $\zeta (s)$---i.e., when $0=s=\rho
  -w$ for some nontrivial zero $\rho $. This, when it occurs, is indeed a
  double pole, and the residue is given by $m\cdot \log C +\epsilon $. Here $m$
  is the multiplicity of the zero $\rho $ (which we expect always---or at least
  usually---to be equal to $1$) and $\epsilon $ is a constant (depending on
  $\rho $, but not on $C$). \par \end {enumerate} \par The standard technique
  for the ``Explicit formula'' will provide us with a formula for our function
  of interest ${\hat \Phi }_{\le C}(\theta )$. The formula has terms resulting
  from the residues of each of the first three types of poles, and of the {\it
  Elementary} or the {\it Double} pole---whichever exists. Here is the shape of
  the formula, given with terminology that is meant to be evocative: \par \par
  $$ {\bf (1)}\ \ \ {\hat \Phi }_{\le C}(\theta ) = {\Sing }_{\le C}(\theta ) +
  {\Triv }_{\le C}(\theta ) + {\Osc }_{\le C}(\theta ) + {\Elem }_{\le
  C}(\theta )$$ \par Or: \par $${\bf (2)}\ \ \ {\hat \Phi }_{\le C}(\theta ) =
  {\Sing }_{\le C}(\theta ) + {\Triv }_{\le C}(\theta ) + {\Osc }_{\le
  C}(\theta ) + {\Double }_{\le C}(\theta ),$$ \par \noindent the first if $w$
  is not a nontrivial zero of $\zeta (s)$ and the second if it is. \par The
  good news is that the functions ${\Sing }_{\le C}(\theta ), {\Triv }_{\le
  C}(\theta )$ (and also ${\Elem }_{\le C}(\theta )$ when it exists) are smooth
  (easily describable) functions of the two variables $C$ and $\theta $; for
  us, this means that they are not that related to the essential
  information-laden {\it discontinuous} structure of $ {\hat \Phi }_{\le
  C}(\theta )$. Let us bunch these three contributions together and call the
  sum ${\Smooth }(C,\theta )$, and rewrite the above two formulae as: $$ {\bf
  (1)}\ \ \ {\hat \Phi }_{\le C}(\theta ) ={\Smooth }(C,\theta ) + {\Osc }_{\le
  C}(\theta ) $$ \par Or: \par $${\bf (2)}\ \ \ {\hat \Phi }_{\le C}(\theta ) =
  {\Smooth }(C,\theta )+ {\Osc }_{\le C}(\theta ) + m\cdot \log C +\epsilon ,$$
  \par depending upon whether or not $w$ is a nontrivial zero of $\zeta (s)$.
  \par \par We now focus our attention on the Oscillatory term, ${\Osc }_{\le
  C}(\theta ) $, approximating it by a truncation: $$\Osc _w(C,X):= 2\sum
  _{|\gamma | < X} {{\frac {e^{i\log C\cdot (\gamma -\theta )}}{i(\gamma
  -\theta )}}}.$$ Here if a zero has multiplicity $m$, then we count it $m$
  times in the sum. Also, in this formula we have relegated the ``$\theta $" to
  the status of a subscript (i.e., $w = {\frac {1}{2}} +i\theta $) since we are
  keeping it constant, and we view the two variables $X$ and $C$ as linked in
  that we want the cutoff ``$X$" to be sufficiently large, say $X \gg C^2$, so
  that the error term can be controlled. \par At this point, we can perform a
  ``multiplicative version'' of a Ces\`aro summation---i.e., the operator $F(c)
  \mapsto ({\Ces } F)(C):= \int _1^CF(c) dc/c.$ This has the effect of forcing
  the oscillatory term to be bounded as $C$ tends to infinity. \par This
  implies that for any fixed $\theta $, \begin {itemize} \item ${\Ces }{\hat
  \Phi }_{\le C}(\theta )$ is bounded independent of $C$ if $\theta $ is not
  the imaginary part of a nontrivial zero of $\zeta (s)$ and \item ${\Ces
  }{\hat \Phi }_{\le C}(\theta )$ grows as ${\frac {m}{2}}\cdot (\log C)^2
  +O(\log C)$ if $\theta $ is the imaginary part of a nontrivial zero of $\zeta
  (s)$ of multiplicity $m$, \end {itemize} giving a theoretical confirmation of
  the striking feature of the graphs of our Chapter~\ref
  {ch:prime-to-spectrum}.

\bibitem{Note20}
A reference for this is: \par \vskip 10pt {\bf [I-K]: }H. Iwaniec; E. Kowalski,
  {\bf Analytic Number Theory}, {\em American Mathematical Society Colloquium
  Publications} {\bf 53} (2004). \par (See also the bibliography there.) \vskip
  10pt Many ways of seeing the explicit relationship are given in Chapter 5 of
  {\bf [I-K]}. For example, consider Exercise 5 on page 109. \par $$\sum _{\rho
  }{\hat \phi }(\rho ) = -\sum _{n \ge 1}\Lambda (n)\phi (n) + I(\phi ),$$ \par
  where \begin {itemize} \item $\phi $ is any smooth complex-valued function on
  $[1,+\infty )$ with compact support,\item ${\hat \phi }$ is its Mellin
  transform $${\hat \phi }(s):= \int _0^{\infty }\phi (x)x^{s-1}dx,$$ \item the
  last term on the right, $I(\phi )$, is just $$I(\phi ):= \int _1^{\infty
  }(1-{\frac {1}{x^3-x}})\phi (x)dx$$ (coming from the pole at $s=1$ and the
  ``trivial zeroes''). \item The more serious summation on the left hand side
  of the equation is over the nontrivial zeroes $\rho $, noting that if $\rho $
  is a nontrivial zero so is~${\bar \rho }$. \end {itemize} \par \par \par Of
  course, this ``explicit formulation'' is not {\it immediately} applicable to
  the graphs we are constructing since we cannot naively take ${\hat \phi }$ to
  be a function forcing the left hand side to be $G_C(x)$. \par See also
  Exercise 7 on page 112, which discusses the sums $$ x - \sum _{|\theta | \le
  C} {\frac {x^{{\frac {1}{2}} +i\theta }-1}{{\frac {1}{2}} +i\theta }}. $$.

\bibitem{Note21}
You may well ask how we propose to order these correction terms if RH is false.
  Order them in terms of (the absolute value of) their imaginary part, and in
  the unlikely situation that there is more than one zero with the same
  imaginary part, order zeroes of the same imaginary part by their real parts,
  going from right to left.

\bibitem{Note22}
Bombieri, {\em The Riemann Hypothesis}, available at \url
  {http://www.claymath.org/sites/default/files/official_problem_description.pdf}.

\end{thebibliography}