pattern - Logically linked numbers

The left and right numbers are linked. What should the last number on the right be, and why?

\begin{align} 135759&: 1 \\ 151364&: 4 \\ 255075&: 9 \\ 279422&: 36 \\ 292620&: 91 \\ 348777&: 135 \\ 398067&: 147 \\ 417894&: 265 \\ 459431&: 279 \\ 478926&: 307 \\ 609941&: 363 \\ 689245&: 435 \\ 814576&:\space? \\ \end{align}

Before anyone bothers, no right is not simply a polynomial of left (of course we could certainly fit one, but there is another way).

Hint

There is information somewhere in the community wiki post that may prove useful

Hint 2

The mapping from left to right is many to one and there are infinite possibilities for right.

Two more entries are
\begin{align}0&: 1\\1&: 0\\\end{align}

Hint 3

Nothing special about these, they are really just some more examples...

Since there were none, here are some with $8$s on the right: \begin{align}157388&: 8\\272813&: 18\\276276&:28\\291384&: 88\\\end{align}
To have a cube $\gt1$ on both sides, left must have at least $7$ digits, here are a few: \begin{align}1259712&: 27\\224755712&: 729\\559476224&: 1000\\1427628376&: 6859\\\end{align}

Answer

What's the $\bf ?$ in $~ \bf 814576 : \, ? ~$ ? $\require{begingroup} \begingroup \let \SSS \tiny \let \SS \scriptsize \let \S \small \def \T {\small\sf} \def \Q #1{{ \bf ? { \S\raise2mu ( } #1 { \S\raise2mu ) } }} \def \R #1{ \phantom{\Q0} \llap{ \bf #1 \kern 7mu } } \def \x #1{{\SSS\kern1mu\raise2mu \times \S\kern5mu 3 \kern1mu\SS\raise7mu #1}} \def \p { \kern8mu{ \S \raise1mu + }\kern8mu} \def \D { \kern6em } \def \E #1{ \D \llap{ #1 \kern4mu { \S\raise1mu = } \kern6mu } } $

$ \kern-12mu \E{ \bf ? } 514 $

Essential $\bf 1 \over 10\raise-4mu{\small\,3}$ of the story

$ \E{ \Q{\T digit} } $ how many enclosed regions, typically loops, in the drawing of the single $\T digit$
$ \E{ \Q{\T digits} } $ decimal value of the ternary number formed
$ \D \raise-6mu\strut $ when each $\T digit$ of $\T digits$ is replaced by $\Q{\T digit}$
$ \E{ \Q{814576} } \Q 8 \x 5 \p\Q 1 \x 4 \p\Q 4 \x 3 \p\Q 5 \x 2 \p\Q 7 \x 1 \p\Q 6 \x 0 $
$ \E{} \R2 \x5 \p\R0 \x4 \p\R1 \x3 \p\R0 \x2 \p\R0 \x1 \p\R1 \x0 $
$ \E{} 201001\raise-4mu{\small\,3} $
$ \E{} 514\raise-4mu{\scriptsize\,10} $ $\endgroup$

Middle $\bf 1 \over 10\raise-4mu{\small\,3}$ of the story

Notice the following retrieval from the community evidence locker. It originated from using various radices for the numbers being puzzled and displaying the only two radices with suggestive patterns.

    left  : right         binary        ternary
   135759 : 1                  1              1
   151364 : 4                1..             11
   255075 : 9               1..1            1..
   279422 : 36            1..1..           11..
   292620 : 91           1.11.11          1.1.1
   348777 : 135         1....111          12...

   398067 : 147         1..1..11          1211.
   417894 : 265        1....1..1         1..211
   459431 : 279        1...1.111         1.11..
   478926 : 307        1..11..11         1.21.1
   609941 : 363        1.11.1.11         11111.
   689245 : 435        11.11..11         121.1.

Watch what happens, starting with the dots, when ...

... the middle two columns are ignored and left-column digits without enclosures are dotted out.

  left                          right ternary
 .....9                                     1
 ....64                                    11
 ...0..                                   1..
 ..94..                                  11..
 .9.6.0                                 1.1.1
 .48...                                 12...
 .9806.                                 1211.
 4..894                                1..211

 4.94..                                1.11..
 4.89.6                                1.21.1
 60994.                                11111.
 689.4.                                121.1.   

No wonder the ternaries had so few $\small\tt 2$s, a peculiarity that had been noted early without direct result. Only the digit $8$ has two loops and thus transforms into $\small\tt 2$, while each of four digits$-0 \, 4 \, 6 \, 9-$has just one loop (or enclosed region) and transforms into $\small\tt 1$.

And no wonder that a few ternary numbers were exactly as wide as the left-column numbers and that none were wider, which was not noted along the way.

Final $\bf 1 \over 10\raise-4mu{\small\,3}$ of the story

How would anyone notice the above relationship? By following up on clues, such as those just mentioned, and allowing for dumb luck coincidence.

The records room at Puzzling HQ has a dossier on The Case of the Really, really, really hard sequence, which was reported just 11 days after the present puzzle and hinges on an eerily comparable ternary modus operandi. Sure enough, that case was cracked by the poser of the present puzzle, in barely 86 minutes and 48 seconds. To see a new puzzle that uses a similar device must have been precious! And then to write a solution that flaunts the very key to their own unsolved puzzle.

The present puzzle's poser's pathological ternary obsession did not go unnoticed, nor uncontracted, by the detective assigned to this case. Sooner or later, well, much after the hint ...

... $0 \kern2mu {:} \kern1mu 1$ revealed that two wildly different numbers, $135759$ and $0$, can produce the same degenerate result, $1$, it was time to consider that individual digits of the puzzling numbers might be nullifying each other or even being ignored. The binary and ternary patterns above seemed like the simplest leads to follow when embarking on this. (Not) one of the (first) possibilities to touch on, if only to shake up thought patterns, was the graphical way of looking at digits in that other case.

And it happened to work. Tickle me pink.