The uniqueness conjecture for subunit index-ratio classes

Complete counterexample and infinitely many distinct nonunique subunit classes.

For increasing positive divisors $d_1<\cdots<d_{\tau(n)}$, let

$$O(n)=\sum_{j\text{ odd}}d_j,\qquad E(n)=\sum_{j\text{ even}}d_j,\qquad R(n)=E(n)/O(n).$$

The distinct squares $1225=5^2\cdot7^2$ and $3025=5^2\cdot11^2$ have ordered divisor lists

$$\operatorname{Div}(1225)=(1,5,7,25,35,49,175,245,1225),$$ $$\operatorname{Div}(3025)=(1,5,11,25,55,121,275,605,3025).$$

Therefore

$$R(1225)=\frac{324}{1443}=\frac{108}{481}=\frac{756}{3367}=R(3025)<1.$$

Thus $1225$ and $3025$ are distinct members of $G_{108/481}$, disproving uniqueness.

In fact infinitely many different subunit classes are nonunique. Put $t=108/481$. For every prime $Q>3025$, the divisors of each $mQ^2$, with $m\in\{1225,3025\}$, occur in three consecutive blocks

$$\operatorname{Div}(m),\quad Q\operatorname{Div}(m),\quad Q^2\operatorname{Div}(m).$$

Each base list has nine divisors, so parity reverses in the middle block. Hence both distinct squares $1225Q^2$ and $3025Q^2$ have the same ratio

$$T_Q(t)=\frac{t(Q^2+1)+Q}{Q^2+1+Qt}<1.$$

For distinct eligible primes $Q,S$,

$$T_Q(t)-T_S(t)=\frac{(Q-S)(t^2-1)(QS-1)}{(Q^2+Qt+1)(S^2+St+1)}\neq0.$$

Consequently infinitely many distinct subunit ratio classes contain at least two integers.