Self-similarity of Fibonacci period structures at prime powers

Counterexample to the claimed prime-power multiplicity scaling

Consider the Fibonacci state transition

$$T(a,b)=(b,a+b) \quad\text{on}\quad (\mathbb Z/m\mathbb Z)^2. \tag{1}$$

We examine the transition from $m=11$ to $m=121$. The characteristic polynomial is

$$f(X)=X^2-X-1. \tag{2}$$

Modulo $11$, its roots are $4$ and $8$, with respective multiplicative orders

$$\operatorname{ord}_{11}(4)=5, \qquad \operatorname{ord}_{11}(8)=10. \tag{3}$$

Consequently, in eigenvector coordinates, the transition is

$$(x,y)\longmapsto(4x,8y) \pmod{11}. \tag{4}$$

The zero vector has period $1$. The $10$ states with $x\neq0$ and $y=0$ have period $5$, whereas the $110$ states with $y\neq0$ have period $10$. Therefore the complete orbit distribution modulo $11$ is

$$\begin{array}{c|c|c} \text{period}&\text{states}&\text{orbits}\\ \hline 1&1&1\\ 5&10&2\\ 10&110&11. \end{array} \tag{5}$$

The two roots lift modulo $121$ to

$$u=37, \qquad v=85, \qquad u^2-u-1\equiv v^2-v-1\equiv0\pmod{121}. \tag{6}$$

Because $v-u=48$ is invertible modulo $121$, the transition remains diagonalizable over $\mathbb Z/121\mathbb Z$:

$$(x,y)\longmapsto(37x,85y). \tag{7}$$

The root orders are exactly

$$\begin{aligned} 37^5&\equiv67\not\equiv1\pmod{121}, &37^{55}&\equiv1\pmod{121},\\ 85^{10}&\equiv111\not\equiv1\pmod{121}, &85^{55}&\equiv-1\pmod{121}, &85^{110}&\equiv1\pmod{121}, \end{aligned} \tag{8}$$

so

$$\operatorname{ord}_{121}(37)=55, \qquad \operatorname{ord}_{121}(85)=110. \tag{9}$$

Write $R=\mathbb Z/121\mathbb Z$. There are $110$ units in $R$, $10$ nonzero elements of $11R$, and $11$ elements of $11R$ altogether.

If both coordinates lie in $11R$, their periods are governed by the roots modulo $11$. They contribute one fixed point, $10$ states of period $5$, and $110$ states of period $10$. If $x$ is a unit and $y=0$, the resulting $110$ states have period $55$.

There are two disjoint families of period-$110$ states:

$$\begin{array}{c|c} \text{conditions}&\text{number of states}\\ \hline y\in R^\times,\quad x\in R&110\cdot121=13310\\ x\in R^\times,\quad y\in11R\setminus\{0\}&110\cdot10=1100. \end{array} \tag{10}$$

In the second family, the coordinate periods are $55$ and $10$, whose least common multiple is $110$. Hence the complete orbit distribution modulo $121$ is

$$\begin{array}{c|c|c} \text{period}&\text{states}&\text{orbits}\\ \hline 1&1&1\\ 5&10&2\\ 10&110&11\\ 55&110&2\\ 110&14410&131. \end{array} \tag{11}$$

The entries account for the entire state space:

$$1+10+110+110+14410=14641=121^2. \tag{12}$$

At modulus $11$, the period $10$ has multiplicity $11$. The conjectured scaling by the prime $p=11$ would therefore require its successor period $110$ to have multiplicity

$$11\cdot11=121. \tag{13}$$

Instead, its actual multiplicity is

$$\boxed{131\neq121.} \tag{14}$$

The additional $10$ orbits arise from the second family in (10); thus they cannot be removed by reinterpretation or by overlooking nonprimitive states. The middle-period multiplicities do remain $2$, exactly as predicted by the separate class-B2 assertion, so the failure concerns specifically the claimed universal multiplication rule for the longer periods.

The parity-transformed recurrence has roots $-u=v^{-1}$ and $-v=u^{-1}$, so inversion and interchange preserve the same orbit counts. Hence the counterexample applies to both recurrences in Marc T. Pudelko's Conjecture 6.