Post #3142162

\subsection{Numerical Verification of the Expression} The following table shows the numerical verification of the expression \[ \frac{5}{3} \approx \frac{m+q}{m} \approx \frac{6 + (k^2 + 1)\epsilon' + k\epsilon}{4} \] on several abc triples with different values of \(k\). \begin{table}[h] \centering \begin{tabular}{|c|l|c|c|c|c|} \hline \(k\) & Triple (example) & Actual \(\frac{m+q}{m}\) & Right side \(\frac{6 + (k^2+1)\epsilon' + k\epsilon}{4}\) & Difference & Closeness to \(\frac{5}{3}\) \\ \hline 1 & 4 + 121 = 125 & 1.65760 & 1.65760 & $\sim 0$ & Very close \\ 1 & 1 + 2400 = 2401 & 1.63608 & 1.63608 & $\sim 0$ & Close \\ 1 & 263 + 3,442,688 = 151$^3$ & 1.66560 & 1.66560 & $\sim 0$ & Very close \\ 2 & 49 + 576 = 625 & 1.69901 & 1.70217 & 0.00316 & Moderate \\ 3 & 3 + 125 = 128 & 1.79620 & 1.8054 & 0.0092 & Moderate \\ 12 & 3$^5\times7$ + 5$^6\times67$ = 2$^{20}$ & 1.92743 & 1.92743 & $\sim 0$ & Far from 5/3 \\ \hline \end{tabular} \caption{Comparison of actual \(\frac{m+q}{m}\) with the derived expression. The match is excellent for several k=1 and k=12 triples, but shows small discrepancies for k=2 and k=3.} \label{tab:expression_verification} \end{table}