Repetend Rigidity at the Golden Scale
Abstract
A condition at the golden scale can force a repeating tail to have period one. We establish this for a denominator statistic that counts cyclic shifts of a reference tail, allowing arbitrary coprime integer bases. Let b,q\geq2 be coprime, put h=\operatorname{ord}_q(b), and let the positive integer m divide a power of b. Among the rows k/(qm) with 1\leq k\leq qm-1, give cyclic shifts of the 1/(qm) tail weight one and terminating rows weight \lambda\in[0,1]. This produces the score S_{b,q,\lambda}(m)=\frac{hm+\lambda(m-1)}{qm-1}. Write \tau=(\sqrt5-1)/2. For rational \lambda and integers r\geq1, its real interpolant satisfies \widetilde S_{b,q,\lambda}(\tau^{-r})=\tau exactly when h=1 and (q,\lambda,r)\in\{(2,0,2),(3,1,2)\}. The exclusion combines classical Fibonacci coordinates with the absence of unit orders in [q/2,2q/3) for q\geq5. Thus an equality at an irrational scale constrains the actual modular period, and the one-digit condition follows from the classification. For the one-digit profile, a unique admissible real weight exists exactly when q\in\{2,3\} and r\geq2; every later-power weight is irrational.
Three Blocks at Twelve
Changing the starting point of a repeating block rotates its digits without changing its period. Counting these rotations lets a denominator statistic retain the length of the underlying cycle. We ask whether that period can remain greater than one when the statistic’s real interpolant meets the reciprocal golden ratio at a golden-power scale. For the weighted count developed here, it cannot when the terminating-row weight is rational.
The fraction 1/12 has two decimal places before its repeating digit settles. It is a mixed cycle, with a nonrepeating prefix followed by a periodic tail. Throughout, the period is the length of the shortest repeating block. A separate return occurs when whole copies of that block have digit sum zero modulo the base minus one. Calling this cumulative closure, we can write \frac1{12}=0.08\lvert333\rvert. The bars denote indefinite repetition of the enclosed closure block. This is the first cumulative closure of the tail, since 3+3+3=9\equiv0\pmod9. Its primitive period is one and its closure length is three. The prefix lies outside the closure sum. Section (*) relates these two lengths exactly.
Advancing all eleven fractions k/12 to the same clearing depth separates their tails into three blocks.
| Numerator class | Numerators | Tail after the common depth | Size |
|---|---|---|---|
| k\equiv0\pmod3 | 3,6,9 | terminates | 3 |
| k\equiv1\pmod3 | 1,4,7,10 | repeats 3 | 4 |
| k\equiv2\pmod3 | 2,5,8,11 | repeats 6 | 4 |
The middle block matches the tail of 1/12. Canonical terminating expansions continue with zeros, so the first block does not literally match that nonzero tail. Literal synchronized digit matching counts four rows. Giving the terminating block full weight counts seven. If its weight is \lambda, the score is \frac{3\lambda+4}{11}.
The two endpoint counts answer different questions. The weight \lambda=0 compares the digits that are printed after the common depth. The weight \lambda=1 records membership in the union of the terminating and matching blocks.
The reciprocal golden ratio \tau=\frac1\varphi=\frac{\sqrt5-1}{2} satisfies \tau^{-2}=\varphi^2. Extending the displayed rational function from integer m to a real scale x, the full-weight curve meets \tau at x=\tau^{-2}. The actual score at m=4 is 7/11; the golden equality belongs to the interpolant. Varying the integer modulus, rational weight, and positive exponent leaves exactly two solutions, both at the square (Theorem 7).
The main extension permits longer repeating blocks. For any integer base coprime to q, the number of distinct cyclic shifts of the reference tail is the multiplicative order of the base. This order enters the numerator of the new score. Theorem 12 proves that the same two rational endpoints are the entire classification in this wider setting. Fibonacci coordinates would require every later even exponent to produce a unit order in an interval where no such order exists. This is the arithmetic consequence of the golden-scale condition. Period one becomes a conclusion even when it is allowed to vary.
One Digit After the Clearing Depth
Fix integers q\geq2 and b\geq2 with b\equiv1\pmod q. Write b=1+cq. A positive integer m is b-supported when every prime factor of m divides b. Equivalently, m divides b^D for some D\geq0.
Choose a b-supported m, put n=qm, and select a common depth D with m\mid b^D. Write u=\frac{b^D}{m}. The congruence b\equiv1\pmod q implies \gcd(b,q)=1, so u is a unit modulo q.
Theorem 1 (One-digit block profile). The numerators 1\leq k\leq qm-1 form one terminating block of size m-1 and q-1 periodic blocks of size m. More precisely, \begin{array}{c|c|c} \text{numerator condition}&\text{tail after depth }D&\text{size}\\ k\equiv0\pmod q&\text{terminating}&m-1\\ k\equiv1\pmod q&\text{matches }1/(qm)&m\\ k\not\equiv0,1\pmod q&\text{another one-digit tail}&(q-2)m \end{array} The partition and the matching relation are independent of the chosen common clearing depth.
Proof. At depth D, \frac{k b^D}{qm}=\frac{ku}{q}. Let r be the least nonnegative residue of ku modulo q. The fractional part at that depth is r/q. In the original denominator system, the corresponding remainder is mr modulo qm.
The residue r vanishes exactly when q\mid k. These are the m-1 numerators q,2q,\ldots,(m-1)q, and their fractions terminate. Every nonzero residue gives a one-digit periodic tail. Indeed, \left\lfloor\frac{br}{q}\right\rfloor =\left\lfloor cr+\frac rq\right\rfloor =cr, and the next remainder is br-crq=r. The digit cr therefore repeats. Distinct nonzero residues give distinct digits.
The reference tail is obtained from the residue u modulo q. A row has the same tail exactly when ku\equiv u\pmod q, which is equivalent to k\equiv1\pmod q. Each nonzero residue class contains m numerators between 1 and qm-1. The remaining q-2 classes contribute (q-2)m rows.
Increasing D multiplies u by a power of b. That power is one modulo q, so none of the residue labels changes. ◻
No primality hypothesis occurs in the argument. Every integer modulus q\geq2 has a one-digit setting, since the base b=q+1 is admissible. In decimal, the eligible moduli are the divisors of 9.
Repetend orbits, complementary blocks, and change of base have a long history. Midy-type theorems study sums of pieces within one periodic block [13, 5, 4, 11, 14]. Other treatments organize decimal expansions through subgroup sums, base changes, and complementarity [6, 3, 1, 15]. Theorem 1 instead inventories every numerator row at one fixed denominator after a shared clearing depth. Its residue count is elementary. These sources own the classical repetend, complement, and base-change setting. The Fibonacci coordinate identity used below is standard and is cited at its point of use. The result proved here is the simultaneous rigidity of the modulus, terminating-block weight, and reciprocal exponent inside the weighted fixed-denominator profile, including its extension to tails counted up to cyclic shift.
The Weighted Profile
Definition 2 (Weighted reference score). Under the hypotheses of Theorem 1, give every terminating row weight \lambda\in[0,1], every row matching the reference tail weight one, and every other row weight zero. Let C_{b,q,\lambda}(m) be the average over the qm-1 numerator rows.
Proposition 3 (Exact finite score). For every admissible b,q,m, and \lambda\in[0,1], C_{b,q,\lambda}(m) =\frac{\lambda(m-1)+m}{qm-1}. The value is independent of the admissible base, so it will also be written C_{q,\lambda}(m).
Proof. The terminating block contributes \lambda(m-1). The reference block contributes m. Every other block contributes zero. ◻
At \lambda=0, Proposition 3 is literal synchronized digit matching in the one-digit regime. At \lambda=1, it counts the union of the terminating and reference blocks.
Rational weights have an exact finite meaning. Write \lambda=a/d with integers d>0 and 0\leq a\leq d. Replace every numerator row by d labeled copies. Mark a copies of each terminating row, all d copies of each reference row, and no copies of any other row. The ordinary proportion of marked copies is \frac{a(m-1)+dm}{d(qm-1)} =C_{q,a/d}(m). For m>1, the converse is also operational. Each terminating row has a rational marked proportion in any uniform finite replication. The score itself determines the weight, since its coefficient of \lambda is the nonzero rational number (m-1)/(qm-1). An irrational weight therefore cannot give an ordinary finite marked proportion at such an ensemble. At m=1 there are no terminating rows and C_{q,\lambda}(1)=\frac1{q-1} for every real \lambda. That ensemble cannot distinguish the weights. Rational weights are precisely the weighting rules realized by uniform finite replication across the family of admissible ensembles; the real interval is their continuous relaxation.
The score belongs to the displayed denominator ensemble. The same rational number can occupy different blocks under different denominator presentations. For example, 1/3 is the reference row at denominator 3, while the equal number 2/6 lies in the other periodic block at denominator 6.
Definition 4 (Real interpolant). For q\geq2, \lambda\in[0,1], and x>1/q, define \widetilde C_{q,\lambda}(x) =\frac{(\lambda+1)x-\lambda}{qx-1}. At every admissible integer m, \widetilde C_{q,\lambda}(m)=C_{q,\lambda}(m).
The interpolant locates scales between integer denominators. Its value at a noninteger x is not a finite denominator score. The limit and direction make the role of the terminating-block weight explicit. They are \lim_{x\to\infty}\widetilde C_{q,\lambda}(x) =\frac{1+\lambda}{q} and \widetilde C_{q,\lambda}'(x) =\frac{\lambda(q-1)-1}{(qx-1)^2}. The profile is constant precisely at \lambda=1/(q-1).
Irrational weights can give a golden score at an integer denominator. For example, the decimal-twelve ensemble has score \tau when \lambda=(11\tau-4)/3\in(0,1). The classification below concerns the additional requirement that the scale be a positive integral reciprocal power of \tau. This requirement is a selected compatibility question; it does not follow from the denominator partition.
Golden Powers in Fibonacci Coordinates
Put \tau=\frac1\varphi=\frac{\sqrt5-1}{2}. Then \tau^2+\tau-1=0, \qquad \tau^2=1-\tau. Let F_0=0, F_1=1, and F_{j+1}=F_j+F_{j-1} for j\geq1.
Lemma 5 (Fibonacci coordinates). For every integer j\geq1, \tau^j=(-1)^{j+1}F_j\tau+(-1)^jF_{j-1}.
Proof. The formula holds at j=1. If it holds at j, multiplication by \tau and the relation \tau^2=1-\tau give \begin{aligned} \tau^{j+1} &=(-1)^{j+1}F_j(1-\tau)+(-1)^jF_{j-1}\tau\\ &=(-1)^{j+2}F_{j+1}\tau+(-1)^{j+1}F_j. \end{aligned} This is the formula at j+1. ◻
Fibonacci numbers are the rational coordinates of every power of \tau in the basis \{1,\tau\}. This standard identity is the mechanism that will decide the exponent. Beard gives this expansion explicitly for the reciprocal golden ratio [2]; see also [10]. Masáková and Pelantová use Fibonacci divisibility in a different Midy setting where the numeration base itself is \varphi [12]. Here the numeration base remains an integer and the golden quantity enters only through the real interpolant.
Lemma 6 (The power equation). Let q\geq2, let \lambda\in[0,1], and let r\geq1 be an integer. Then \widetilde C_{q,\lambda}(\tau^{-r})=\tau if and only if \tau^{r+1}-\lambda\tau^r-q\tau+\lambda+1=0.
Proof. The value \tau^{-r} exceeds one, so the denominator of the interpolant is positive. Substitution followed by multiplication by \tau^r gives Equation (1) without changing the equivalence. ◻
Rational Rigidity
Theorem 7 (Rational classification). Let q\geq2 be an integer, let \lambda\in\mathbb Q\cap[0,1], and let r\geq1 be an integer. Then \widetilde C_{q,\lambda}(\tau^{-r})=\tau if and only if (q,\lambda,r)=(2,0,2) \quad\text{or}\quad (q,\lambda,r)=(3,1,2).
Proof. Apply Lemma 5 to Equation (1). Its left side becomes A_r\tau+B_r, where \begin{aligned} A_r&=(-1)^r\bigl(F_{r+1}+\lambda F_r\bigr)-q,\\ B_r&=(-1)^{r+1}\bigl(F_r+\lambda F_{r-1}\bigr)+\lambda+1. \end{aligned} Both coefficients are rational. Since \tau is irrational, the power equation holds exactly when A_r=B_r=0.
If r is odd, then B_r=F_r+1+\lambda(F_{r-1}+1)>0. No odd exponent can occur.
Suppose that r is even and r\geq4. Then F_r\geq3 and F_{r-1}\geq2, while B_r=1-F_r+\lambda(1-F_{r-1})<0. No such even exponent can occur.
It remains to take r=2. Here B_2=0 and A_2=2+\lambda-q. Thus q=2+\lambda. The modulus q is an integer and 0\leq\lambda\leq1, so the only possibilities are (q,\lambda)=(2,0) and (3,1). Direct substitution proves both converses. ◻
Corollary 8 (The odd-modulus endpoint). Among odd integer moduli, rational weights in [0,1], and positive reciprocal powers, the equality with \tau holds only at (q,\lambda,r)=(3,1,2).
The theorem detects the integer modulus 3. Its primality is not used. The modulus 2 supplies the second rational endpoint under literal digit matching.
Real Flexibility
The rational endpoints sit on two longer real branches.
Theorem 9 (Complete real classification). Let q\geq2 be an integer, let \lambda\in[0,1] be real, and let r\geq1 be an integer. Then \widetilde C_{q,\lambda}(\tau^{-r})=\tau if and only if q\in\{2,3\}, \qquad r\geq2, \qquad \lambda=\lambda_{q,r}, where \lambda_{q,r} =\frac{q\tau-1-\tau^{r+1}}{1-\tau^r}. The values at r=2 are \lambda_{2,2}=0, \qquad \lambda_{3,2}=1. Every \lambda_{q,r} with r\geq3 is irrational.
Proof. Solving Equation (1) for \lambda gives Equation (2). Its denominator is positive. The condition 0\leq\lambda_{q,r}\leq1 is equivalent to 1+\tau^{r+1} \leq q\tau \leq 2-\tau^{r+2}.
If q\geq4, then q\tau\geq4\tau>2, while the right side of Equation (3) is less than 2. Hence no modulus q\geq4 occurs.
Take q=2. At r=2, the left inequality is an equality because \tau^3=2\tau-1. It remains true for every larger r because \tau^{r+1} decreases. The right inequality also holds for r\geq2, since 2-\tau^{r+2}\geq2-\tau^4=3\tau>2\tau. At r=1, the left inequality would require 3\tau\geq2, which is false.
Take q=3. The left inequality holds for every r\geq1. The right inequality is an equality at r=2 because \tau^4=2-3\tau, and it remains true for every larger r. At r=1, it would require 5\tau\leq3, which is false. The exact bounds \frac35<\tau<\frac23 justify the two strict failures. They follow by squaring 5\sqrt5>11 and 3\sqrt5<7.
Thus the real solutions are exactly the stated branches. At r=2, Theorem 7 gives the two endpoint weights. If a later weight were rational, the same theorem would force r=2. Every weight with r\geq3 is therefore irrational. ◻
The first later-power values are \lambda_{2,3}=\frac{\tau^3}{2}, \qquad \lambda_{3,3}=\frac{3\tau}{2}. The two branches move in opposite directions. Writing y=\tau^r gives \lambda_{q,r}=\frac{q\tau-1-\tau y}{1-y}. Its derivative with respect to y has the sign of (q-1)\tau-1. Since y decreases with r, the q=2 branch increases from zero toward 2\tau-1=\tau^3, while the q=3 branch decreases from one toward 3\tau-1.
Periods Under Cyclic Shift
Now let b,q\geq2 be arbitrary coprime integers. Retain a b-supported m and a common clearing depth D, and put u=\frac{b^D}{m},\qquad h=\operatorname{ord}_q(b),\qquad H=\langle b\rangle\subseteq(\mathbb Z/q\mathbb Z)^\times. Two purely periodic tails are counted together when their repeating digit blocks differ by a cyclic shift. The reference is still the tail of 1/(qm) at depth D.
The relation between cyclic shifts and multiplication by the base is classical. Kak and Chatterjee state the residue condition in their Theorem 7 [9]. We use it with the explicit hypothesis \gcd(b,q)=1 and prove the resulting row count here. Exact synchronized matching would still accept only one residue class; counting cyclic shifts makes the period length enter the score.
Proposition 10 (Cyclic-shift count). Among the numerators 1\leq k\leq qm-1, exactly hm rows have a tail that is a cyclic shift of the reference tail. They are exactly the rows with k\bmod q\in H. The terminating block has m-1 rows. These classes are independent of the common clearing depth. Giving the cyclic-shift rows weight one, terminating rows weight \lambda\in[0,1], and all other rows weight zero gives S_{b,q,\lambda}(m)=\frac{hm+\lambda(m-1)}{qm-1}.
Proof. At depth D, the tail is the canonical base-b expansion of the fractional part of ku/q. Both m and u are units modulo q. Every nonzero residue has a purely periodic expansion, since multiplication by b permutes the residues modulo q. Advancing the reference tail by j places replaces its residue u by ub^j modulo q.
The reference has least period h. Indeed, returning to its initial tail is equivalent to ub^j\equiv u\pmod q, hence to b^j\equiv1\pmod q. The canonical expansion identifies its fractional value uniquely; a nonzero residue over q cannot terminate because its reduced denominator is coprime to b. Thus its h cyclic shifts are exactly the tails with residues in uH. The row with numerator k has one of these tails precisely when ku\in uH, or k\in H, modulo q.
Each of the h nonzero numerator classes contains m rows. The zero class contains m-1 rows and consists exactly of the terminating fractions. This gives Equation (4). Increasing the clearing depth multiplies u by a power of b; it rotates the reference orbit and leaves the condition k\in H unchanged. ◻
The associated real interpolant is \widetilde S_{b,q,\lambda}(x) =\frac{(h+\lambda)x-\lambda}{qx-1},\qquad x>1/q. When h=1, it is the one-digit interpolant already classified. The finite-replication interpretation has the same boundary as before. For m>1 it detects rational weights, while at m=1 the score is h/(q-1) regardless of \lambda.
The obstruction to longer periods is an elementary restriction on unit orders. We write \varphi(q) for Euler’s totient.
Lemma 11 (An interval without unit orders). For every integer q\geq5, no unit modulo q has order h with \frac q2\leq h<\frac{2q}{3}.
Proof. Suppose such a unit exists. Its order divides \varphi(q), and \varphi(q)<q\leq2h. Therefore \varphi(q)=h, since any larger positive multiple of h would be at least 2h. The unit group is consequently cyclic.
If q is even and has an odd prime divisor, the totient product gives \varphi(q)<q/2, a contradiction. If q is a power of two, then q\geq8. The four distinct residues 1,\quad -1,\quad 1+q/2,\quad -1+q/2 all square to one modulo q. A cyclic group has at most two solutions of z^2=1, so this case is also impossible.
If q is odd with at least two distinct prime divisors, the Chinese remainder theorem gives at least four square roots of one by choosing signs independently on its prime-power factors. Again its unit group cannot be cyclic. The only remaining case is q=p^a for an odd prime p and an integer a\geq1. But then \frac hq=\frac{\varphi(q)}q=1-\frac1p\geq\frac23, contrary to the strict upper bound. ◻
Theorem 12 (Period rigidity). Let b,q\geq2 be coprime integers, let \lambda\in\mathbb Q\cap[0,1], and let r\geq1 be an integer. Then \widetilde S_{b,q,\lambda}(\tau^{-r})=\tau if and only if b\equiv1\pmod q and (q,\lambda,r)=(2,0,2) \quad\text{or}\quad (q,\lambda,r)=(3,1,2). In particular, the equality forces the reference tail to have period one.
Proof. Put h=\operatorname{ord}_q(b). Clearing the positive denominator in Equation (5) gives the equivalent equation \tau^{r+1}-\lambda\tau^r-q\tau+\lambda+h=0. By Lemma 5, its two rational coordinates are \begin{aligned} &(-1)^r(F_{r+1}+\lambda F_r)-q,\\ &(-1)^{r+1}(F_r+\lambda F_{r-1})+\lambda+h. \end{aligned} Both must vanish. For odd r, the second is F_r+h+\lambda(F_{r-1}+1)>0, which excludes that parity. For even r, the equations become q=F_{r+1}+\lambda F_r,\qquad h=F_r+\lambda(F_{r-1}-1).
Suppose r\geq4. Then q\geq F_5=5, and the Fibonacci recurrence in Equation (7) gives \begin{aligned} 2h-q&=F_{r-2}+\lambda(F_{r-3}-2),\\ 2q-3h&=F_{r-3}+\lambda(F_{r-4}+3). \end{aligned} The first expression is 1-\lambda\geq0 at r=4. For every even r\geq6, it is positive because F_{r-2}>0 and F_{r-3}\geq2. The second expression is positive for every r\geq4. Hence q/2\leq h<2q/3, contradicting Lemma 11.
Only r=2 remains. Equation (7) then gives h=1 and q=2+\lambda. Integrality of q and 0\leq\lambda\leq1 leave the two stated endpoints. The equality h=1 is equivalent to b\equiv1\pmod q. Conversely, under that congruence the interpolant is \widetilde C_{q,\lambda}, and both endpoints follow from Theorem 7. ◻
The order restriction is necessary. If h were a free integer parameter, q=5, h=3, \lambda=0, and r=4 would satisfy Equation (7). No base modulo 5 has order 3. The modular orbit removes an algebraically admissible solution. The theorem concerns rational weights; the real classification in Theorem 9 concerns the one-digit profile only.
Cumulative Closure
For a purely periodic tail, define its cumulative closure length L to be the least positive length consisting of complete primitive blocks whose combined digit sum is zero modulo b-1. Repeating a block introduces no new distinct cyclic shifts. Thus closure leaves the count hm in Proposition 10 unchanged.
Proposition 13 (Closure length). Let b,q\geq2 be coprime integers, and let 1\leq a<q with \gcd(a,q)=1. Write h=\operatorname{ord}_q(b) and let \sigma be the sum of the digits in one primitive block of a/q. Then L=\frac{h(b-1)}{\gcd(\sigma,b-1)} =\operatorname{ord}_{q(b-1)}(b). In particular, L is independent of the reduced numerator and of the starting phase. If h=1, then L=q.
Proof. After t complete blocks the digit sum is t\sigma. The least positive t with (b-1)\mid t\sigma is (b-1)/\gcd(\sigma,b-1), proving the first equality.
For any positive multiple N of h, the first N periodic digits, including any leading zeros, encode the integer P_N=\frac{a(b^N-1)}q. An integer and its base-b digit sum are congruent modulo b-1. Put R_N=(b^N-1)/(b-1). Closure at length N is equivalent to \begin{aligned} (b-1)\mid P_N &\quad\Longleftrightarrow\quad q\mid aR_N\\ &\quad\Longleftrightarrow\quad q\mid R_N\\ &\quad\Longleftrightarrow\quad b^N\equiv1\pmod{q(b-1)}. \end{aligned} The middle equivalence uses \gcd(a,q)=1. Conversely, every positive N satisfying the last congruence is a multiple of h, because it also satisfies b^N\equiv1\pmod q. Taking the least such N proves the order formula and its independence of numerator and phase. This includes b=2, where closure is modulo one and L=h.
If h=1, write b-1=cq. The single repeated digit is ca. Its closure after t digits requires cq\mid tca, equivalently q\mid t because \gcd(a,q)=1. The least such t is q. ◻
The proof identifies L with the least N\geq1 for which q divides the base-b repunit R_N=(b^N-1)/(b-1). Generalized repunits and their Lucas-sequence divisibility are classical; see Jaroma [8]. The expression \operatorname{ord}_{q(b-1)}(b) for this least length appears explicitly in Hasler’s SeqFan discussion [7]. Proposition 13 identifies that repunit length with digit-sum closure over complete primitive blocks.
Corollary 14 (Closure at the rational endpoints). Assume the hypotheses of Theorem 12 and the equality \widetilde S_{b,q,\lambda}(\tau^{-r})=\tau. For any b-supported m, the reference tail after a common clearing depth, and every accepted cyclic shift of that tail, has cumulative closure length L=q= \begin{cases} 2,&(q,\lambda,r)=(2,0,2),\\ 3,&(q,\lambda,r)=(3,1,2). \end{cases}
Proof. Theorem 12 gives exactly these two endpoints and h=1. The reference residue u=b^D/m and all its cyclic-shift residues ub^j are units modulo q. Their tails therefore have reduced denominator q, so Proposition 13 gives L=q. ◻
The Count and the Period
The decimal-twelve ensemble has finite full-weight score 7/11. Its associated curve meets \tau at the reciprocal square. The classification explains how far this crossing can extend. With rational weights in [0,1], allowing every integer modulus and every positive reciprocal exponent leaves only the two square endpoints. One counts literal matching at modulus two; the other includes the terminating block at modulus three.
Allowing the period to vary tests the same equality against modular arithmetic. The cyclic-shift count puts the order of the base into the profile. Fibonacci coordinates then prescribe the order required by each exponent. For later even powers this requirement falls into the gap between one half and two thirds of the modulus, where no unit order can lie. The residue count, the golden coordinates, and the unit group are all needed to reach the classification.
Rationality supplies a separate boundary. In the one-digit profile, real weights continue along two infinite branches, while rational weights stop at the square. Their finite-replication interpretation explains this restriction on weights; the choice to evaluate at golden powers remains the compatibility condition being studied.
At twelve, the repetend is one digit long and its cumulative closure is three digits long. Corollary 14 identifies the closure lengths of both rational endpoints. Once longer cycles are admitted, the modular-order obstruction forces the same one-digit structure throughout the stated rational classification. Within this profile, rational compatibility at the golden scale forces period one.
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Funding
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