Numbering entries in a table by column

Question

I'm trying to make a large table that has four columns, with two columns that are numbered incrementally. Note that the real table will have around 40 numbered entries.

Here's roughly the desired behavior and a kludgy MWE, using two counters with the totcount package (rowA and rowB, note the use of \setcounter{rowB}{\totvalue{rowA}}).

\documentclass[11pt,letterpaper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{totcount}
\usepackage{array}

\newtotcounter{rowA}
\newcommand\rowA{\stepcounter{rowA}{\bf \arabic{rowA}.~}}

\newcounter{rowB}
\newcommand\rowB{\stepcounter{rowB}{\bf \arabic{rowB}.~}}

\renewcommand{\arraystretch}{2}

\begin{document}
\setcounter{rowB}{\totvalue{rowA}}
\[ \begin{array}{*2{>{\displaystyle}l}*2{>{\displaystyle}l}} 
  f(t) & \mathcal{L}\{f\}(s) & f(t) & \mathcal{L}\{f\}(s) \\ \hline 
 \rowA e^{at} & \frac{1}{s-a} &  \rowB t^n & \frac{n!}{s^{n+1}} \\
 \rowA \sin(bt) & \frac{b}{s^2 + b^2} & \rowB u(t-a), \, a\geq 0 & \frac{e^{-as}}{s}\\
 \rowA \cos(bt) & \frac{s}{s^2+b^2}  & \rowB \delta(t-a), \, a \geq 0 & e^{-as}
\end{array}\]
\end{document}

Questions

1. Is there a less kludgy way to do this? Ideally something that only requires one counter and numbering command?
2. I'm particularly annoyed with this solution in that I have to decide what information ends up in each column at the start---very WYSIWYG/non-TeX. I would love to be able to enter the formulas as a two column table and then wrap with multicols. I'm trying to do this with longtable and repeated headers, but no joy thus far.

Answer 1

I addressed the first question here, by applying automatic stepping in the column types (defining new column types, for this)

The reason why the A and B types are using \text{\bfseries ...} is that \bfseries is not allowed in math mode, which is active there.

\documentclass[11pt,letterpaper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{totcount}
\usepackage{array}

\newtotcounter{rowA}
\newcounter{rowB}

\newcolumntype{A}{>{\stepcounter{rowA}{\text{\bfseries\therowA.~}}}l}
\newcolumntype{B}{>{\stepcounter{rowB}{\text{\bfseries\therowB.~}}}l} 

\newcolumntype{M}{>{\displaystyle}l}

\renewcommand{\arraystretch}{2}

\begin{document}
\setcounter{rowB}{\totvalue{rowA}}
\[ \begin{array}{AMBM}
\multicolumn{1}{M}{f(t)} & \multicolumn{1}{M}{\mathcal{L}\{f\}(s)} & \multicolumn{1}{M}{f(t)} & \multicolumn{1}{M}{\mathcal{L}\{f\}(s)} \\ \hline 
    e^{at} & \frac{1}{s-a} &   t^n & \frac{n!}{s^{n+1}} \\
  \sin(bt) & \frac{b}{s^2 + b^2} &  u(t-a), \, a\geq 0 & \frac{e^{-as}}{s}\\
  \cos(bt) & \frac{s}{s^2+b^2}  &  \delta(t-a), \, a \geq 0 & e^{-as} \\
  & & & \\ % Fake line
\end{array}\]
\end{document}

Way with only one counter

\documentclass[11pt,letterpaper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{totcount}
\usepackage{array}

\newtotcounter{rowA}

\newcolumntype{A}{>{\displaystyle\stepcounter{rowA}{\mathbf{\therowA.~}}}l}
\makeatletter
% Using one counter only: add the current value of \rowA and the total counter value -- this will give the 'rowB' 
\newcolumntype{B}{>{\displaystyle\mathbf{\@arabic{\numexpr \c@rowA + \totvalue{rowA}}.~}}l}
\makeatother

\newcolumntype{M}{>{\displaystyle}l}

\renewcommand{\arraystretch}{2}

\begin{document}
\[ \begin{array}{AMBM}
  \multicolumn{1}{M}{f(t)} & \multicolumn{1}{M}{\mathcal{L}\{f\}(s)} & \multicolumn{1}{M}{f(t)} & \multicolumn{1}{M}{\mathcal{L}\{f\}(s)} \\ \hline 
  e^{at} & \frac{1}{s-a} &   t^n & \frac{n!}{s^{n+1}} \\
  \sin(bt) & \frac{b}{s^2 + b^2} &  u(t-a), \, a\geq 0 & \frac{e^{-as}}{s}\\
  \cos(bt) & \frac{s}{s^2+b^2}  &  \delta(t-a), \, a \geq 0 & e^{-as} \\
  \frac{1}{t} f(t) & \int_s^\infty F(u) ~\mathrm{d} u & \frac{1}{\sqrt{t}} & \frac{\sqrt{\pi}}{\sqrt{s}}\\
\end{array}\]
\end{document}