(Bonus / bounty follow-up challenges have been moved to Semiïnfinite autobinomonorownonomicrogram)
☆ Be the first next to make your own nontrivial autobinomonorownonomicrogram ☆
“Impossible!” you scoff? Might indeed be, except that your autobinomonorownonomicrogram may be infinitely wide and include leading 0s, as in 01 or 0010. $ \require{begingroup}\begingroup \def \l { \kern-.3em\cdots~ } \def \L { & ~\cdots\kern -.1em } \def \r { ~\cdots } \def \R { \kern-.2em\cdots~\\\hline } \def \p { \phantom{ \Rule {2.5ex}{2.0ex}{0.5ex}} } \def \X {\kern-.5em \Rule{2.5ex}{2.0ex}{0.5ex} \kern-.5em} \def \b {\kern-.5em \p \kern-.5em} \def \1 {\kern-.5em\rlap {\normalsize \bf \kern .2em 1 } \p \kern-.5em} \def \0 {\kern-.5em \rlap{ \scriptsize \kern.3em 0 } \p \kern-.5em} $
“Nontrivial  ?” you might ask. Well, trivial autobinomonorownonomicrograms are just too common.
$\small\begin{array}{c|c|} \sf\scriptsize Consecutive~counts~(in~binary) \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R \l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r \L &\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b& \R \end{array}$
is a trivial one that solves to
$\small\begin{array}{c|c|} \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R
\l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r
\L &\0&\1&\0&\1&\0&\1&\1&\0&\1&\0&\1&\0&\1& \R
\end{array}$
Note how the same infinite digit sequence ... 0 1 0 1 0 11 0 1 0 1 0 1... constitutes both the margins’ counts and the interior’s cells. (These counts are binary, so 10 = 2.) This example is called trivial as...
...nontrivial here means that multiple pairs of adjacent 1s occur among the cells. The example does not qualify because it has only one adjacent pair of 1s.
“Autobinomonorownonomicrogram?” It’s short for auto-bino-monorow-nono[micro]gram.
auto: Self-descriptive — cells’ contents match the margin counts’ actual digits.
bino: Binary numbers.
monorow: Exactly one row tall.
nonogram: This type of grid puzzle.
micro: No numbers greater than 2, which shows as binary 10 (or 010, 0010, ...).
(Thus 3 consecutive cells cannot be all 1s.)
Evolutionary path of autobinomonorownonomicrograms.
Begin with a familiar
nonogram
such as this 3×8,
where numbers at its left and top margins are
length counts of consecutive filled cells
in their respective rows and columns.
$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & 2 & 1 & 1 & 1 & 1 & 1 & 2 \kern.05em \\ \hline 2 ~~ 3 & \b & \b & \b & \b & \b & \b & \b & \b \kern.05em \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \kern.05em \\ \hline 3 ~~ 2 & \b & \b & \b & \b & \b & \b & \b & \b \kern.05em \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & 2 & 1 & 1 & 1 & 1 & 1 & 2 \kern.05em \\ \hline 2 ~~ 3 & \b & \b & \X & \X & \b & \X & \X & \X \kern.05em \\ \hline 1 ~~ 1 ~~ 1 & \b & \X & \b & \b & \X & \b & \b & \X \kern.05em \\ \hline 3 ~~ 2 & \b & \X & \X & \X & \b & \X & \X & \b \kern.05em \\ \hline \end{array}$$
Turn that into binary, where 0s and 1s indicate empty and full cells while binary numbers are used for counts. This already happens to be nontrivial as it has multiple sets of adjacent 1 cells.
$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & \! 10 \! & 1 & 1 & 1 & 1 & 1 & \! 10 \\ \hline 10 ~~ 11 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 11 ~~ 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & \! 10 \! & 1 & 1 & 1 & 1 & 1 & \! 10 \\ \hline 10 ~~ 11 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \1 \\ \hline 1 ~~ 1 ~~ 1 & \0 & \1 & \0 & \0 & \1 & \0 & \0 & \1 \\ \hline 11 ~~ 10 & \0 & \1 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline \end{array}$$
Empty a couple of corner-cell 1s to attain the micro quality, having counts only of 0, 1 and 10.
$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 &\bf 1 & 1 & 1 & 1 & 1 & 1 &\bf 1 \\ \hline 10~~\bf 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline{\bf 10}~~ 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 &\bf 1 & 1 & 1 & 1 & 1 & 1 &\bf 1 \\ \hline 10~~\bf 10 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline 1 ~~ 1 ~~ 1 & \0 & \1 & \0 & \0 & \1 & \0 & \0 & \1 \\ \hline{\bf 10}~~ 10 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline \end{array}$$
Almost there, pare down to just one row. The following 1×$\kern1mu\raise1mu\infty$ nonogram would be an autobinomonorownonomicrogram if only its digits were exactly matched. But its cells contain a 1 (circled) where the counts’ digits do not ($\,\scriptsize\wedge\,$).
$$\small\begin{array}{r|c|} \L & 0& 1& 1& 0& 1& 1& 0& 1& 0& 1& \R \l ~0 1~~1 0~~1 \rlap{\kern-.25em\scriptsize\raise-1.5ex\wedge} 0~~1~~0 1 \r \L &\0&\1&\1&\0&\1&\1 \rlap{\kern-.65em\Large\raise-.1ex\bigcirc} &\0&\1&\0&\1& \R \end{array}$$
This nonetheless qualifies as nontrivial because two pairs of adjacent cells contain 1s.
$\endgroup$
