Agreement with one marked fraction need not reflect agreement among the other fractions. For digit-partitioning prime cores, an exact count relates these two averages.
Fix a positional base b and an integer n\ge3. For the fractional field \mathcal{F}(n)=\{k/n:1\le k<n\}, we define the synchronized pairwise alignment \sigma_b(n) as the mean digit agreement over distinct unordered pairs. Reference alignment \alpha_b(n) instead compares every field member with 1/n and includes the reference self-match. Their signed difference F_b(n)=\alpha_b(n)-\sigma_b(n) is the focused alignment.
Let p\nmid b be prime, let p\le b+1, and let s be a positive b-supported integer with ps\ge3. Then \alpha_b(ps)=\frac{2s-1}{ps-1},\qquad F_b(ps)=\frac{s}{ps-1},\qquad \sigma_b(ps)=\frac{s-1}{ps-1}. The pairwise background average is therefore exactly the normalized terminating contribution to \alpha_b. What remains is the normalized size of the synchronized residue class containing the reference fraction. As s\to\infty through supported values, the three limits are 2/p, 1/p, and 1/p.
Within this prime-core family, the reciprocal-golden threshold lies strictly between the focused and reference-alignment limits exactly for p\in\{2,3\}. The case p=2 is degenerate; among odd primes, only p=3 has the gap. We also prove that for every prime p\ge3 not dividing the base, \sigma_b(p)=0 exactly when the base-b digit function is injective on the nonzero residues modulo p.
At n=12 in base ten, reference alignment splits exactly as \alpha_{10}(12)=\frac7{11} =\frac4{11}+\frac3{11} =F_{10}(12)+\sigma_{10}(12). After two decimal digits, the eleven rows divide into three terminating tails, four repeating threes, and four repeating sixes. Within these groups there are 3+6+6=15 agreeing unordered pairs out of 55. Their mean is 3/11, exactly the terminating-row contribution to the reference alignment. This equality of totals includes twelve pairs of nonterminating fractions. The residual 4/11 is the normalized size of the class containing 1/12. Each term lies below the reciprocal-golden threshold, while their sum lies above it. Counting pairs explains a feature that the reference score alone leaves unresolved.
Reference alignment \alpha_b(n) measures the finite fractional field \mathcal{F}(n) relative to 1/n. Pairwise alignment \sigma_b(n) measures the background coherence among all distinct unordered pairs. Here “fractional field” denotes this finite family of fractions. The two statistics average different sample spaces, so their difference F_b(n)=\alpha_b(n)-\sigma_b(n) is a signed residual. For digit-partitioning primes, the exact class count makes the pairwise average equal to the normalized terminating contribution to \alpha_b. The residual is the normalized reference residue class.
Normalized coordinate agreement is Hamming similarity, equivalently one minus normalized Hamming distance. Lempel and Greenberger organize cyclic Hamming correlation for finite sequence families [1], and Kak and Chatterjee apply it directly to full-period reciprocal digit sequences [2]. Repetend multiplication and its multiplicative-orbit organization are also classical [3]. The construction here averages that agreement over every distinct unordered pair in the complete denominator ensemble and compares the result with one marked reference row. The contribution is the exact prime-core decomposition and the zero-background characterization without a primitive-root hypothesis.
Why the Golden Ratio Selects the Prime Three [4] studies the full-weight alignment profile for one-digit repeating tails. Digit-Partitioning Primes and the Alignment Formula [5] supplies the synchronized prime-core formula, and The Three-Tier Theorem [6] carries the common-depth convention through the all-denominator classification. Here the exact question is what pairwise term is present inside the prime-core formula.
Definition 1. A positive integer s is b-supported if every prime factor of s divides b, equivalently if s divides a power of b. Every n\ge2 has a unique factorization n=qs, where s is its largest b-supported divisor and \gcd(q,b)=1. We call q the rough part of n.
Suppose first that q>1. Choose one clearing depth D with s\mid b^D and put L=\mathop{\mathrm{ord}}_q(b). The synchronized alignment \alpha_b(n) compares the next L digits of every k/n with the corresponding digits of 1/n, beginning after the same D digits. A terminating fraction receives score one. Otherwise its score is the proportion of matching synchronized positions. The mean over 1\le k<n is \alpha_b(n). If q=1, every fraction terminates and we set \alpha_b(n)=1. One common clearing depth is used for every row, whether the rough part is prime or composite.
Definition 2 (Synchronized pairwise alignment). Let n=qs\ge3. If q=1, set c_b(j,k;n)=1 for every 1\le j<k<n. If q>1, use the common depth D and window length L above. Set c_b(j,k;n)=1 when both j/n and k/n terminate, and set it to zero when exactly one terminates. If neither terminates, let c_b(j,k;n) be the proportion of the next L synchronized positions at which their digits agree. The pairwise alignment is \sigma_b(n)=\binom{n-1}{2}^{-1} \sum_{1\le j<k\le n-1}c_b(j,k;n). The focused alignment is the signed residual F_b(n)=\alpha_b(n)-\sigma_b(n).
Increasing D advances both nonterminating tails by the same power of b. Their length-L equality pattern is rotated, so its number of matches is unchanged. Individual repetends are never restarted at separate phases. The pairwise statistic begins at n=3 because \mathcal F(2) contains no unordered pair of distinct elements.
For a prime p\nmid b, write \delta_{p,b}(r)=\left\lfloor\frac{br}{p}\right\rfloor, \qquad 1\le r<p. The prime is digit-partitioning in base b when this digit function is injective. Digit-Partitioning Primes and the Alignment Formula [5] proves that this occurs exactly when p\le b+1.
Theorem 3. Let b\ge2, let p\nmid b be prime with p\le b+1, and let s\ge1 be b-supported. If ps\ge3, then \begin{aligned} \alpha_b(ps) &= \frac{2s-1}{ps-1}, \\[4pt] F_b(ps) &= \frac{s}{ps-1}, \\[4pt] \sigma_b(ps) &= \frac{s-1}{ps-1}. \end{aligned} Moreover, \sigma_b(ps) equals the normalized contribution of the terminating fractions to \alpha_b(ps), while F_b(ps) equals the normalized size of the synchronized nonzero residue class containing 1/(ps).
Proof. Equation (1) is the digit-partitioning formula of [5].
Choose D with s\mid b^D and put u=b^D/s. Since p\nmid b, both s and u are invertible modulo p. After the common depth D, the nonterminating tail of k/(ps) begins at the nonzero remainder ku\pmod p. Each nonzero remainder occurs for exactly s numerators k in \{1,\ldots,ps-1\}.
Thus the nonterminating fractions form p-1 synchronized classes, each of size s. Two members of one class have the same remainder at every synchronized position and match everywhere. Members of different classes remain at distinct remainders after multiplication by every power of b. Since p\le b+1, the digit function is injective, so their digits differ at every synchronized position.
There are also s-1 terminating fractions, namely the nonzero multiples of p below ps. Terminating pairs contribute one, mixed pairs contribute zero, and the preceding residue argument handles every nonterminating pair. The total pair score is therefore \Sigma = \binom{s-1}{2} + (p-1)\binom{s}{2}.
Because ps\ge3, the pair sample space is nonempty. Expanding the binomial coefficients gives \begin{aligned} \sigma_b(ps) &= \frac{\binom{s-1}{2} + (p-1)\binom{s}{2}}{\binom{ps-1}{2}} = \frac{(s-1)(s-2)/2 + (p-1)s(s-1)/2}{(ps-1)(ps-2)/2} \\ &= \frac{(s-1)\bigl[(s-2) + (p-1)s\bigr]}{(ps-1)(ps-2)} = \frac{(s-1)(ps - 2)}{(ps-1)(ps-2)} = \frac{s-1}{ps-1}. \end{aligned}
The s-1 terminating fractions contribute (s-1)/(ps-1) to \alpha_b(ps), exactly the value just found for \sigma_b(ps). The other aligned fractions are the s numerators congruent to 1 modulo p, the synchronized residue class of the reference. Hence F_b(ps)=\alpha_b(ps)-\sigma_b(ps)=\frac{s}{ps-1}. ◻
Corollary 4. As s \to \infty through b-supported integers, \alpha_b(ps) \to \frac{2}{p}, \qquad F_b(ps) \to \frac{1}{p}, \qquad \sigma_b(ps) \to \frac{1}{p}.
As the supported factor grows, the focused residual and the pairwise background approach equal shares of the limiting reference alignment.
Write \varphi=(1+\sqrt5)/2.
Theorem 5. Let p\nmid b be a digit-partitioning prime. The reciprocal-golden threshold lies strictly between the focused and reference-alignment limits, \frac1p<\frac1\varphi<\frac2p, if and only if p\in\{2,3\}. For p=2, the formula gives \alpha_b(2s)=1 for every admissible s, so the case is degenerate. Among odd primes, only p=3 has the gap.
Proof. Since \varphi<2, every prime satisfies p>\varphi, which is equivalent to 1/p<1/\varphi. Also 3<2\varphi=1+\sqrt5<4. Therefore 1/\varphi<2/p, equivalently p<2\varphi, holds exactly for the primes 2 and 3. ◻
For p=3, the limits are F_b\to1/3 and \alpha_b\to2/3, with 1/\varphi strictly between them. For p\ge5, the finite formula gives \frac2p-\alpha_b(ps) =\frac{p-2}{p(ps-1)}>0. Since 0\le F_b(ps)\le\alpha_b(ps) in this family, both quantities are below 2/p\le2/5<3/5<1/\varphi at every supported resolution. The last strict inequality is equivalent to 11<5\sqrt5, whose square is 121<125.
The focused residual by itself also stays below the threshold in every defined prime-core case. For p\ge3, F_b(ps)=\frac{s}{ps-1}\le\frac1{p-1}\le\frac12. For p=2 and 2s\ge3, the condition 2\nmid b makes every nontrivial supported s at least 3, and then F_b(2s)\le3/5<1/\varphi by the same inequality. Thus, in the nondegenerate odd-prime case, p=3 is the unique digit-partitioning prime core for which the pairwise background can carry the reference alignment across the threshold. In even bases not divisible by 3, this is the three-core tier classified in The Three-Tier Theorem [6]; the fully terminating tier is a separate mechanism.
Proposition 6. Let b\ge2 and let p\ge3 be prime with p\nmid b. Then \sigma_b(p)=0 if and only if p is digit-partitioning in base b.
Proof. Suppose first that p is digit-partitioning. For distinct j,k\in\{1,\ldots,p-1\}, the residues b^tj and b^tk remain distinct modulo p at every synchronized position t. Injectivity of \delta_{p,b} makes the corresponding digits distinct. Hence c_b(j,k;p)=0 for every pair, and \sigma_b(p)=0.
Conversely, if p is not digit-partitioning, there are distinct r,s\in\{1,\ldots,p-1\} with \delta_{p,b}(r)=\delta_{p,b}(s). The fractions r/p and s/p therefore match at the first synchronized position. With L=\mathop{\mathrm{ord}}_p(b), this gives c_b(r,s;p)\ge1/L>0, and consequently \sigma_b(p)>0. ◻
This provides an equivalent characterization of the digit-partitioning property. A prime p is digit-partitioning if and only if no two distinct fractions in \{k/p:1\le k<p\} share a digit at the same synchronized position.
For composite rough parts, the decomposition \alpha_b=F_b+\sigma_b remains a signed identity by definition. The common window L=\mathop{\mathrm{ord}}_q(b) handles residual periods dividing L. Every nonzero residue still occurs for exactly s numerators, but the residue orbits can have different lengths and distinct residues can emit the same digit. The sign of F_b is determined by the resulting reference and pairwise averages.
For a fixed base, the synchronized digit expansions of \mathcal{F}(n) carry three statistics. The value \alpha_b(n) is the mean alignment with the distinguished fraction 1/n. The value \sigma_b(n) is the mean background alignment over distinct pairs, and F_b(n) is their signed focused residual. In the digit-partitioning family, F_b is positive and equals the reference residue-class density. For primes p\ge3 not dividing b, \sigma_b(p) characterizes digit partitioning.
If n\ge3 is entirely b-supported, every fraction terminates. Then \alpha_b(n)=\sigma_b(n)=1 and F_b(n)=0. This is the terminating mechanism in The Three-Tier Theorem [6], distinct from the prime-three crossing.
For digit-partitioning prime cores, the limiting gap between F_b and \alpha_b has width 1/p. The pairwise term is larger than the distance from F_b to the reciprocal-golden threshold exactly when \frac1p>\frac1\varphi-\frac1p \quad\Longleftrightarrow\quad p<2\varphi. Among odd primes, this selects p=3. The pairwise term can therefore change the threshold outcome for the three-core family. At arbitrary composite rough parts, the signed identity remains exact while the residue interactions must be counted on their own common clock.
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[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]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.
[5]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.
[6]A. S. Petty, The Three-Tier Theorem, research note, October 2020 (revised September 2026), DOI: https://doi.org/10.5281/zenodo.21844347.
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