In this section we collect some numerical results verifying the second two hypotheses of Theorem \ref{ubc-0} for larger values of $X$, and smaller values of $t_0$, than were considered in Section \ref{newup-sec}.  This leads to improvements to the bound $\Lambda \leq 0.22$ conditional on the assumption that the Riemann Hypothesis can be numerically verified beyond the height $T \approx 1.2 \times 10^{11}$ used in Section \ref{newup-sec}. 

Having the freedom to assume the RH has been verified up to a certain height, allows a Barrier location $X$ to be the outcome of an optimised choice of $y_0, t_0$ where the Triangle bound becomes positive. This implies no further computations are required after the Barrier has been cleared from any zeros having passed through it. To bring the potential Barrier locations within reach of computation, it proves fruitful to mollify the Triangle bound with the first prime as follows:

$ $

The toy version of preventing $H_t(x+iy)$ from vanishing is that of establishing the inequality
$$ |\sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2}+\frac{t}{4} \log \frac{N^2}{n} - \frac{\pi i t}{8}}}| > N^{-y} |\sum_{n=1}^N \frac{1}{n^{\frac{1-y-ix}{2}+\frac{t}{4} \log \frac{N^2}{n} - \frac{\pi i t}{8}}}|.$$
Direct application of the triangle inequality lets us do this if $S < 2$, where S is the sum
$$ S := \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2}+\frac{t}{4} \log \frac{N^2}{n}}} + \frac{N^{-y}}{n^{\frac{1-y}{2}+\frac{t}{4} \log \frac{N^2}{n}}}.$$
One can estimate S using the estimate in the first section of this web page.

Now we use a Euler2 mollifier.  The lemma here is

\begin{lemma}\label{trib2}  Let $a_1,\dots,a_N$ be complex numbers, and let $b_2$ be a number such that whenever $1 \leq n \leq N$ is even, $b_n a_{n/2}$ lies on the line segment $\{ \theta a_n: 0 \leq \theta \leq 1\}$ connecting 0 with $a_n$.  Then we have the lower bound
$$ |1-b_2| |\sum_{n=1}^N a_n| \geq 2 |a_1| - (1-|b_2|) \sum_{n=1}^N |a_n| - 2 |b_2| \sum_{N/2 < n \leq N} |a_n|$$
and the upper bound
$$ |1-b_2| |\sum_{n=1}^N a_n| \leq (1-|b_2|) \sum_{n=1}^N |a_n| + 2 |b_2| \sum_{N/2 < n \leq N} |a_n|.$$
\end{lemma}

\begin{proof} The left-hand side can be written as
$$ |\sum_{n=1}^{2N} (1_{n \leq N} a_n - 1_{2|n} a_{n/2} b_2)|.$$
By the triangle inequality, this is bounded above by
$$ \sum_{n=1}^{2N} |1_{n \leq N} a_n - 1_{2|n} a_{n/2} b_2|$$
and below by
$$ 2 |a_1| - \sum_{n=1}^{2N} |1_{n \leq N} a_n - 1_{2|n} a_{n/2} b_2|.$$
We have
$$ \sum_{n=1}^{2N} |1_{n \leq N} a_n - 1_{2|n} a_{n/2} b_2| = \sum_{n=1}^N |a_n| - 1_{2|n} |a_{n/2}| |b_2| + \sum_{n=N+1}^{2N} 1_{2|n} |a_{n/2}| |b_2|$$
which we can rearrange as
$$ (1 - |b_2|) \sum_{n=1}^N |a_n| + 2 |b_2| \sum_{N/2 < n \leq N} |a_n|$$
and the claim follows.
\end{proof}

We apply this lemma with $a_n = \frac{1}{n^{\frac{1 \pm y-ix}{2}+\frac{t}{4} \log \frac{N^2}{n} - \frac{\pi i t}{8}}}$ and $b_2 = 
\frac{1}{2^{\frac{1+y-ix}{2}+\frac{t}{4} \log \frac{N^2}{2} - \frac{\pi i t}{8}}}$.  The hypotheses of the lemma are easily verified, and we conclude that
$$ |1-b_2| |\sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2}+\frac{t}{4} \log \frac{N^2}{n} - \frac{\pi i t}{8}}}| 
\geq 2 - (1-|b_2|) \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2}+\frac{t}{4} \log \frac{N^2}{n}}} - 2 |b_2| 
\sum_{N/2 < n \leq N} \frac{1}{n^{\frac{1+y}{2}+\frac{t}{4} \log \frac{N^2}{n}}} $$
and
$$ |1-b_2| |\sum_{n=1}^N \frac{1}{n^{\frac{1-y-ix}{2}+\frac{t}{4} \log \frac{N^2}{n} - \frac{\pi i t}{8}}}| 
\leq (1-|b_2|) \sum_{n=1}^N \frac{1}{n^{\frac{1-y}{2}+\frac{t}{4} \log \frac{N^2}{n}}} + 2 |b_2| 
\sum_{N/2 < n \leq N} \frac{1}{n^{\frac{1-y}{2}+\frac{t}{4} \log \frac{N^2}{n}}}.$$
This leads to a modified criterion
$$ (1-|b_2|) S + 2 |b_2| \sum_{N/2 < n \leq N} \frac{1}{n^{\frac{1+y}{2}+\frac{t}{4} \log \frac{N^2}{n}}} + \frac{N^{-y}}{n^{\frac{1-y}{2}+\frac{t}{4} \log \frac{N^2}{n}}}) < 2.$$
We gain a factor of $(1 - |b_2|)$ on the main term, but pick up an additional tail term summing over $N/2 < n \leq N$.  This tail term should be small for $N$ large.  The summand is monotone decreasing in $N$ so we can majorise it by N/2 times the value at N/2, thus we obtain the simplified criterion
\begin{equation}\label{triab2}
(1 - |b_2|) S + N |b_2| \frac{1 + 2^{-y}}{(N/2)^{\frac{1+y}{2}+\frac{t}{4} \log(2N)}} < 2.
\end{equation}

With this improved Triangle bound, more feasible Barrier locations and associated $\Lambda$ can now be established. $\Lambda\,\log_{10}X$ should be roughly constant, so a small percentage increase in $\log_{10}X$ should lead to a corresponding percentage decrease in $\Lambda$. Figure \ref{fig:lambdabarrier} illustrates that indeed $X$ has to increase exponentially to achieve smaller and smaller improvements in $\Lambda$.

\begin{figure}[h!]
  \includegraphics[width=0.9\linewidth]{Lambda_vs_x_barrierloc.png}
  \caption{The curve reflects all potential Barrier locations and their associated $\Lambda$, where the mollified Triangle bound becomes $\geq 0.03$. For this curve,  $t_0$ had been made equal to $y_0$. This may not always be the best choice, however should not be too far from the optimal.}
  \label{fig:lambdabarrier}
\end{figure}

$ $

Our conclusions are summarised in Table \ref{tab:table1}.

$ $

\begin{table}[h!]
  \begin{center}
    \caption{Conditional $\Lambda$ Results}
    \label{tab:table1}
    \begin{tabular}{l|r|r|r|c|r|c} % <-- Alignments: 1st column left, 2nd middle and 3rd right, with vertical lines in between
      $X$ & $t_{0}$ & $y_{0}$ & $\Lambda$ & $\textbf{Winding Number}$ & $N_{a}$ & $\textbf{Moll2 Bound}$\\
      \hline
      $2 \times 10^{12} + 129093$ & 0.198 & 0.15492 & 0.21 & 0 & 398942 & 0.0341\\
      $5 \times 10^{12} + 194858$ & 0.186 & 0.16733 & 0.20 & 0 & 630783 & 0.0376\\
      $2 \times 10^{13} + 131252$ & 0.180 & 0.14142 & 0.19 & 0 & 1261566 & 0.0349\\
      $6 \times 10^{13} + 123375$ & 0.168 & 0.15492 & 0.18 & 0 & 2185096 & 0.0377\\
      $3 \times 10^{14} + 188911$ & 0.161 & 0.13416 & 0.17 & 0 & 4886025 & 0.0369\\
      $2 \times 10^{15} + 122014$ & 0.153 & 0.11832 & 0.16 & 0 & 12615662 & 0.0532\\
      $7 \times 10^{15} + 68886$ & 0.139 & 0.14832 & 0.15 & 0 & 23601743 & 0.0350\\
      $6 \times 10^{16} + 156984$ & 0.132 & 0.12649 & 0.14 & 0 & 69098829 & 0.0307\\
      $6 \times 10^{17} + 88525$ & 0.122 & 0.12649 & 0.13 & 0 & 218509686 & 0.0347\\
      $9 \times 10^{18} + 35785$ & 0.113 & 0.11832 & 0.12 & 0 & 846284375 & 0.0318\\
    \end{tabular}
  \end{center}
\end{table}

The graphs in figure \ref{fig:meshbarrier} illustrate that for increasing $x$, the number of xy-rectangles to be evaluated within the Barrier as well as the number of mesh points required per rectangle (measured at $t=0$) increases exponentially. 

\begin{figure}[!ht]
  \includegraphics[width=1.0\linewidth]{mesh_rectangles_at_barrierlocs.png}
  \caption{The left graph shows how the number mesh points of the xy-rectangle at $t=0$ increases with $x$ for each Barrier. The graph on the right does the same, but now for the total number of xy-rectangles that need to be evaluated per Barrier. }
  \label{fig:meshbarrier}
\end{figure}

All Barrier runs generated a winding number of zero for each rectangle and the scripts completed successfully without any errors. For all Barrier locations, the computations of the mesh points where calculated at 20 digits accuracy except for the highest two where 10 digits where used (to be able to compute it within a reasonable time). Checks where made before each formal run to assure the target accuracy would be achieved. 

The computations for $X=2 \times 10^{20} + 66447$ and $X=9 \times 10^{21} + 70686$ in the above table were massive, and performed using a Boinc based grid computing setup, in which a few hundred volunteers participated. Their contributions can be tracked at http://anthgrid.com/dbnupperbound
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