Color Logic - Balance

Question

^{Inspired by this puzzle}

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Find the rule that each color chain on the left follows, but none on the right follow.

Raw text if you're colorblind or something (O = orange, Y = yellow, G = green, C = cyan):

Left:

COGGYCYOC
GYGYG
O
YYGGGGYYY
OC
GYO
GYYYGOG
OCGCOYO
CYOYOOY
GCGCGCGC

Right:

CC
GCYO
YYYYYOOOO
YCGOYCG
OYCGCY
GOC
CCCGYOO
OOY
CYOGYYCG
GOCOYOGGCO

If you want to see more cases, you may suggest up to 5 additional color chains, and I will sort them accordingly. As a rule of thumb, try to keep the length at $10$ blocks or less. You may suggest chains in this chat.

Hint 1:

If you reverse a chain, it will stay on the same side.

Hint 2:

If you recolor all green to be orange, everything will stay on the same side.

Hint 3:

The rule can be described as the equality of two quantities.

Answer 1

The set on the left is defined by

having equal numbers of adjacent pairs of primary and secondary colored blocks. First, treating O/G and Y/C as equivalent gives a picture like this: Counting a group of n as (n - 1) pairs, all the left sequences have the same number of primary and secondary pairs, while the right sequences differ.

Another way to phrase the answer:

Count the number of primary and secondary colored blocks, weighting each end block .5 less than normal. A solitary block's weighted zero, then. The totals should be the same.

Answer 2

This is just putting Mr. Fish's answer into mathematics:

A sequence is valid if, and only if, it consists of two partitions of $n$ (the total number of blocks) into sub-sequences $a$ and $b$, with $\mid a\mid+\mid b\mid=n$. Furthermore, $\mid\operatorname{np}(a)-\operatorname{np}(b)\mid\le1$, where $\operatorname{np}(k)$ gives the number of parts of $k$. Also, the total sums of each part minus $1$ in each sequence are equal.