The alignments at denominators 6 and 12 leave a gap that no integer denominator can fill. For every even base b not divisible by 3, we define repetend alignment at one common long-division depth and prove \alpha_b(n)\notin\left(\frac35,\frac7{11}\right) \qquad(n\ge2). Both endpoints are attained uniquely, at n=6 and n=12. With \varphi=(1+\sqrt5)/2, we have \frac35<\frac1\varphi<\frac7{11}, and the gap gives a complete three-tier classification. The inequality \alpha_b(n)\ge1/\varphi holds exactly when either n is b-supported, or n=3m with m b-supported and m\ge4.
The proof pairs every nonterminating fraction k/n with (n-k)/n. Their synchronized periodic digits sum to b-1 at every position. Because b is even, at most one member of the pair can match the reference digit. This converts a digitwise symmetry into a uniform bound controlled by the rough part of n. No primality assumption is placed on that rough part. The decimal theorem is the case b=10.
In base ten, the denominators 6 and 12 mark the two edges of an empty interval. Their alignments are \alpha_{10}(6)=\frac35, \qquad \alpha_{10}(12)=\frac7{11}, and no integer denominator n\ge2 has alignment strictly between them. The reciprocal golden ratio lies inside the gap, \frac35<\frac1\varphi<\frac7{11}. A threshold first visible in one exact family therefore separates every denominator into one of three ranges.
Repetend alignment compares the base-b digits of the fractions k/n with those of 1/n at the same long-division positions. Terminating fractions are counted as aligned. The common depth keeps the phase supplied by division instead of restarting each repeating word at a convenient place. The empty interval will be forced by one arithmetic datum, the part of the denominator not supported by the base.
Digit-Partitioning Primes and the Alignment Formula [6] establishes the exact formula \alpha_b(3m)=\frac{2m-1}{3m-1} for b-supported m when 3\nmid b. Why the Golden Ratio Selects the Prime Three [5] locates the crossing of the associated real profile at m=\varphi^2. This is a value of the profile, not a finite denominator count. Among admissible supported integers, the first value above the crossing is m=4. Denominators supported entirely by the base have alignment one.
The remaining question is whether any other denominator can enter between these families. For every even base not divisible by 3, no alignment value lies strictly between 3/5 and 7/11. Thus any denominator at or above the reciprocal golden ratio belongs to one of the two families already visible from the exact formula.
The mechanism is the radix complement. Write n=tm, where m contains every prime-power factor supported by the base and t is coprime to the base. Exactly m-1 fractions terminate. The remaining fractions pair under k\mapsto n-k, and the two members of each pair have complementary digits at every synchronized periodic position. In an even base no digit is its own complement, so each pair contributes total alignment at most one. It follows that \alpha_b(n) \le \frac{m(t+1)/2-1}{tm-1} \le \frac{t+1}{2t}. Outside the rough cores t=1 and t=3, an even base forces t\ge5, and the last bound is at most 3/5.
Complementary repetends are classical. Shrader-Frechette surveys their history among complementary rational numbers [1]. Modern treatments develop Midy’s property and repetend multiplication [4, 3]. Kak and Chatterjee use the same structure for full-period reciprocal sequences [2].
The result here is not the complement identity itself. It applies the numerator involution k\mapsto n-k at one synchronized depth across the complete denominator ensemble, including composite rough parts, and turns that pointwise identity into the sharp empty interval and the full three-tier classification.
The rough part therefore selects the structural case. The supported part changes the resolution within that case; for the rough core 3, it decides on which side of the gap the denominator falls. This distinction is the source of the three tiers.
Definition 1. An integer m\ge1 is b-supported if every prime factor of m divides b. For a positive integer n, its supported part s_b(n) is its largest b-supported divisor. Its rough part is t_b(n)=\frac{n}{s_b(n)}. Thus \gcd(t_b(n),b)=1. We write n=tm when the base is fixed, with m=s_b(n) and t=t_b(n).
Here, “supported part” replaces “smooth part” to avoid confusion with the usual size-based meaning of smooth numbers. In base 10, it absorbs all factors of 2 and 5. For example, 60=3\cdot20 has supported part 20 and rough part 3, while 28=7\cdot4 has supported part 4 and rough part 7.
Definition 2 (Synchronized repetend alignment). Let n=tm\ge2. If t=1, every fraction k/n terminates, and we set a_b(k/n,1/n)=1 for every k and \alpha_b(n)=1.
Suppose t>1. Choose a common clearing depth D\ge0 with m\mid b^D, and put L=\mathop{\mathrm{ord}}_t(b). Beginning immediately after the first D base-b digits, compare the next L digits of every k/n with the corresponding digits of 1/n. If k/n terminates, set a_b(k/n,1/n)=1. Otherwise let a_b(k/n,1/n) be the proportion of these L synchronized positions at which the two digits agree. Finally, define \alpha_b(n) =\frac1{n-1}\sum_{k=1}^{n-1}a_b(k/n,1/n).
This definition does not depend on the clearing depth. Increasing D advances every tail by the same number of digits. For a nonterminating numerator k, the reduced rough denominator is d=t/\gcd(k,t). Since d\mid t, its period \mathop{\mathrm{ord}}_d(b) divides L=\mathop{\mathrm{ord}}_t(b). The length-L equality pattern is therefore merely rotated and its number of matches is unchanged. In particular, the definition never resets the phases of individual repetends.
Lemma 3. The number of terminating fractions in \{k/n : 1 \le k \le n-1\} equals s_b(n)-1.
Proof. The fraction k/n terminates if and only if n/\gcd(k,n) is b-supported, which holds if and only if t\mid k. The multiples of t in \{1, \ldots, n-1\} are t,2t,\ldots,(m-1)t, giving m-1=s_b(n)-1 terminating fractions. ◻
The central tool is a symmetry between complementary fractions.
Lemma 4 (Complement pairing). Let b be even, and let n=tm\ge2. If k/n is nonterminating, then the synchronized periodic digits of k/n and (n-k)/n sum to b-1 at every position. Consequently, a_b(k/n, 1/n) \;+\; a_b((n-k)/n, 1/n) \;\le\; 1.
Proof. Choose D with m\mid b^D and put u=b^D/m. Since \gcd(u,t)=1 and t\nmid k, the least positive residue q_j\equiv b^juk\pmod t lies in \{1,\ldots,t-1\} for every j\ge0. After the common clearing depth, the digit of k/n at position j is \left\lfloor\frac{bq_j}{t}\right\rfloor. The corresponding residue for (n-k)/n is t-q_j, because b^ju(n-k)\equiv-b^juk\pmod t. Since \gcd(b,t)=1, the number bq_j/t is not an integer, and therefore \left\lfloor \frac{b(t-q_j)}{t} \right\rfloor \;=\; b - \left\lceil \frac{bq_j}{t} \right\rceil \;=\; b - 1 - \left\lfloor \frac{bq_j}{t} \right\rfloor. The two digits therefore sum to b-1 at every synchronized position.
Since b is even, no digit equals its own complement. At most one of the complementary digits can equal the corresponding digit of 1/n. Summing this pointwise inequality over the common L positions and dividing by L proves the score bound. ◻
Lemma 5. Let b be even, and let n=tm\ge2. Then \alpha_b(n) \;\le\; \frac{m(t+1)/2 - 1}{tm - 1} \;\le\; \frac{t+1}{2t}.
Proof. By Lemma 3, exactly m-1 fractions terminate, and each has score one. The remaining (t-1)m fractions are nonterminating. The map k\mapsto n-k preserves this set because t\mid k if and only if t\mid n-k.
This involution has no nonterminating fixed point. The rough part t is odd because it is coprime to the even base. If n is odd, there is no integral fixed point. If n is even, then m is even, and the only fixed point is n/2=t(m/2), which terminates.
The nonterminating fractions therefore form (t-1)m/2 distinct complement pairs. Lemma 4 bounds the contribution of each pair by one. Hence (tm - 1)\,\alpha_b(n) \;\le\; (m - 1) + \frac{(t-1)m}{2} \;=\; \frac{m(t+1)}{2} - 1. For the second inequality, \frac{t+1}{2t} -\frac{m(t+1)/2-1}{tm-1} =\frac{t-1}{2t(tm-1)}\ge0. ◻
Throughout this section, \varphi=\frac{1+\sqrt5}{2}.
Proposition 6 (The separating interval). The reciprocal golden ratio lies strictly between the two rational endpoints \frac35<\frac1\varphi<\frac7{11}.
Proof. The first inequality is equivalent to 3\varphi<5, and hence to 3\sqrt5<7; squaring gives 45<49. The second is equivalent to 11<7\varphi, and hence to 15<7\sqrt5; squaring gives 225<245. ◻
Theorem 7 (Three-tier gap theorem). Let b be even with 3\nmid b, and let n=tm\ge2. Exactly one of the following three cases holds.
t=1. Then n is b-supported and \alpha_b(n)=1.
t=3 and m\ge4. Then \frac7{11}\le\alpha_b(n)<\frac23. Equality at the lower endpoint holds only for n=12.
Neither of the preceding conditions holds. Then \alpha_b(n)\le3/5, with equality only for n=6.
Consequently, \alpha_b(n)\notin\left(\frac35,\frac7{11}\right), and \alpha_b(n)\ge\frac1\varphi if and only if n is b-supported, or n=3m with m b-supported and m\ge4.
Proof. If t=1, the denominator is supported by the base. Every fraction k/n terminates, so \alpha_b(n)=1.
Suppose t=3. Since 3\le b+1 and 3\nmid b, the synchronized alignment formula of [6] gives \alpha_b(3m)=\frac{2m-1}{3m-1}. This value is at least 7/11 exactly when m\ge4, because \frac{2m-1}{3m-1}\ge\frac7{11} \quad\Longleftrightarrow\quad m\ge4. It is strictly less than 2/3 for every finite m. This proves Tier II, and equality at 7/11 forces m=4 and n=12. If m<4, the supported-part condition and 3\nmid b leave only m=1 and m=2. The same formula gives \alpha_b(3)=\frac12, \qquad \alpha_b(6)=\frac35.
It remains to consider t\ne1,3. The rough part is coprime to the even base, so it is odd. Hence t\ge5. Lemma 5 and the strict form of its second inequality now give \alpha_b(n) \le\frac{m(t+1)/2-1}{tm-1} <\frac{t+1}{2t} \le\frac35. Thus equality at 3/5 occurs only in the preceding three-core case, where it forces m=2 and n=6. This proves Tier III and the forbidden interval. The final equivalence follows from Proposition 6. ◻
Corollary 8 (Decimal classification). For every integer n\ge2, \alpha_{10}(n)\ge\frac1\varphi if and only if either n is 10-supported, or n=3m with m 10-supported and m\ge4.
The same classification holds for every threshold strictly between 3/5 and 7/11. The reciprocal golden ratio is one such threshold, connected to the earlier three-core profile by its crossing at \varphi^2. The empty interval itself follows from the complement bound and the admissible integer values of the supported part.
Both edges of the forbidden interval are attained in base 10. \alpha_{10}(6)=\frac35, \qquad \alpha_{10}(12)=\frac7{11}. The first is the largest Tier III value. Indeed, the only other excluded denominator with rough part 3 is 3, whose alignment is 1/2. Every decimal rough part other than 1 and 3 is at least 7, and Lemma 5 gives \alpha_{10}(n)\le\frac47<\frac35. The second endpoint is the first value in Tier II, obtained from m=4 in the exact three-core formula.
The complement bound is uniform rather than exact for every rough part. For example, when m is 10-supported, the digit-partitioning formula gives \alpha_{10}(7m)=\frac{2m-1}{7m-1}\longrightarrow\frac27, whereas the universal upper-bound expression tends to 4/7.
The general theorem needs only t\ge5 outside the cores 1 and 3. Decimal notation gives the sharper estimate t\ge7 because 5 divides the base. Thus seven strengthens the decimal ceiling from 3/5 to 4/7, but it is not needed to create the three-tier gap.
Two facts do all the work. An even base makes every rough part odd, and it makes the digit complement b-1-d distinct from d. Once 3 is excluded from the base, the rough core 3 remains available and the formula of [6] applies. Every other nontrivial rough part is at least 5. The argument is uniform across these bases.
For odd b, the digit-complement identity remains true, but its alignment consequence fails. The middle digit (b-1)/2 equals its own complement, and the rough part may be even, allowing a nonterminating fixed point under k\mapsto n-k. The classification also acquires another family. The Alignment Limit for All Primes [7] gives \alpha_b(2m)=1 for every b-supported m and supplies the digit-bin method for prime rough parts. The middle digit is where the complement bound loses exclusivity. An odd-base extension must count that fixed digit, the possible fixed fraction, and composite rough parts directly.
For the even bases of the theorem, the count closes. Terminating rows contribute their exact number, and complement pairs cap the remaining contribution. Together with the three-core formula, these counts force both edges of the gap for every denominator.
[1]M. Shrader-Frechette, Complementary rational numbers, Math. Mag. 51 (1978), no. 2, 90–98. https://doi.org/10.1080/0025570X.1978.11976686.
[2]S. C. Kak and A. Chatterjee, On decimal sequences, IEEE Trans. Inform. Theory 27 (1981), no. 5, 647–652. https://doi.org/10.1109/TIT.1981.1056394.
[3]N. J. Armstrong and R. J. Armstrong, Some properties of repetends, Math. Gaz. 87 (2003), no. 510, 437–443. https://doi.org/10.1017/S0025557200173619.
[4]J. Lewittes, Midy's theorem for periodic decimals, Integers 7 (2007), Paper A02, 11 pp. https://eudml.org/doc/127778.
[5]A. S. Petty, Why the Golden Ratio Selects the Prime Three, research note, January 2020 (revised September 2026), DOI: https://doi.org/10.5281/zenodo.20399951.
[6]A. S. Petty, Digit-Partitioning Primes and the Alignment Formula, research note, April 2020 (revised September 2026), DOI: https://doi.org/10.5281/zenodo.21844074.
[7]A. S. Petty, The Alignment Limit for All Primes, research note, July 2020 (revised September 2026), DOI: https://doi.org/10.5281/zenodo.21844294.
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