(Scroll down for a bonus / bounty follow-up question)
You too can be the very first to make your very own
☆ semiïnfinitely significant autobinomonorownonomicrogram ☆
“But I already have my very own mid-infinite autobinomonorownonomicrogram,” you pule? Of course you have —yawn — by now, who doesn’t?
As contrast, here is a run-of-the-mill biïnfinite (bi-infinite) autobinomonorownonomicrogram. $ \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} $
$\small\begin{array}{c|c|} \sf\scriptsize Consecutive~counts~(in~binary) \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R \l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r \L &\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b& \R \end{array}$
And its solution.
$\small\begin{array}{c|c|} \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R
\l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r
\L &\0&\1&\0&\1&\0&\1&\1&\0&\1&\0&\1&\0&\1& \R
\end{array}$
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.) The present puzzle, however, seeks:
A semiïnfinite (semi-infinite) grid, which extends infinitely only leftward or rightward but not both.   (The example above is biïnfinite as it extends in both directions.)
A significant grid, meaning that it contains infinitely many row counts of 10 and thus infinitely many 1 1 adjacencies among cells.   (The example has only one row count of 10 and only one corresponding 1 1 cell adjacency.)
Here is a schematic of a right-extended semiïnfinitely significant autobinomonorownonomicrogram.
$$\small \def \s #1#2{{\scriptsize\raise1.3ex\matrix{ \sf #1 \\[-1ex] \sf #2 }}} \begin{array}{c|c|} &\s {some}{counts} & 1& 1&\cdots & 1& 1&\cdots & 1& 1&\R \s{some}{counts} ~ 10 ~ \cdots ~ 10 ~ \cdots ~ 10 \r & \s{some}{cells} &\1&\1& \cdots&\1&\1& \cdots&\1&\1& \R \end{array}$$
“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.
$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & \! 10 \! & 1 & 1 & 1 & 1 & 1 & \! 10 \\ \hline 10 ~~ 11 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 11 ~~ 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & \! 10 \! & 1 & 1 & 1 & 1 & 1 & \! 10 \\ \hline 10 ~~ 11 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \1 \\ \hline 1 ~~ 1 ~~ 1 & \0 & \1 & \0 & \0 & \1 & \0 & \0 & \1 \\ \hline 11 ~~ 10 & \0 & \1 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline \end{array}$$
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 ($\,\scriptsize\wedge\,$).
$$\small\begin{array}{r|c|} \L & 0& 1& 1& 0& 1& 1& 0& 1& 0& 1& \R \l ~0 1~~1 0~~1 \rlap{\kern-.25em\scriptsize\raise-1.5ex\wedge} 0~~1~~0 1 \r \L &\0&\1&\1&\0&\1&\1 \rlap{\kern-.65em\Large\raise-.1ex\bigcirc} &\0&\1&\0&\1& \R \end{array}$$
Bonus / bounty question:
 
Can a biïnfinite significant autobinomonorownonomicrogram
infinitely repeat an unchanging pattern?
(Doubtful at pose time,
but also a hint for the present puzzle.)
$$\small \def \s #1#2{{\scriptsize\raise1.3ex\matrix{ \sf #1 \\[-1ex] \sf #2 }}} \begin{array}{c|c|}\L & 1& 1& \cdots & 1& 1& \cdots & 1& 1& \cdots & 1& 1&\R \l 10 ~ \s{some}{counts} ~ 10 ~ \s {same}{counts} ~ 10 \r \L &\1&\1&\s{some}{cells}&\1&\1& \s {same}{cells}&\1&\1& \s{same}{cells}&\1&\1& \R \end{array}$$
$\endgroup$
