The Alignment Limit for All Primes

Divide 1 by 53. Then halve the answer.

 1/53 => 0.|0188679245283|
1/106 => 0.0|0943396226415|

The bars enclose the shortest repeating block. Both blocks have thirteen digits, but rotating the first never produces the second. The extra factor of two gives the decimal a prefix and puts its repeating tail into a different cycle.

Now compare all the fractions with each denominator against its first row. At denominator 53, compare $1/53,2/53,\ldots,52/53$ with $1/53$. Give each row the proportion of repeating positions that agree with the reference. Average those scores. This is the alignment.

At denominator 106, start every comparison after the first decimal place, once the factor of two has cleared. The one terminating row receives score 1. The other rows are compared over a full period on that same division clock.

Keep doubling the denominator. The average rises at first, then begins to alternate between two approaches. One tends to

$$\frac{80}{689}\approx0.1161,$$

the other to

$$\frac{82}{689}\approx0.1190.$$

At 7, the corresponding averages settle toward $2/7$. At 53, the two limiting values remain separate.

The gap is small. It is also exact. Increasing the denominator does not wash it away.

Four lanes through 53

Long division moves from one remainder to the next by multiplying by ten, dividing by the denominator, and keeping the new remainder. At 7, starting from 1 gives

1 → 3 → 2 → 6 → 4 → 5 → 1

Every nonzero remainder appears before the cycle closes. Starting elsewhere changes where we enter the same loop.

At 53, the cycle beginning at 1 runs through

1 → 10 → 47 → 46 → 36 → 42 → 49
  → 13 → 24 → 28 → 15 → 44 → 16 → 1

Thirteen of the fifty-two nonzero remainders. Start at 2 and division follows another thirteen, without touching the first group. The remaining remainders form two more cycles.

These are four lanes through the same set of remainders. Multiplication by ten stays within whichever lane it enters. Mathematically, they are the four cosets of the subgroup generated by ten modulo 53. Their common length is the multiplicative order of ten, here 13.

The factor we put beside 53 chooses the lane. For $106=53\cdot2$, advancing one decimal place leaves $5/53$ as the tail. The reference therefore enters the lane containing remainder 5. For $53\cdot2^a$, advancing $a$ places leaves remainder $5^a$ modulo 53. As $a$ increases, that remainder cycles through all four lanes.

The factor $m=2^a$ is supported on the base. It contains only primes that divide ten. Factors made from twos and fives can be cleared by enough decimal places, but the remainder left after clearing need not belong to the original cycle.

Two extra visits

There are fifty-two remainders and ten possible digits. The digit function sorts them into bins according to the next digit they produce. Eight bins contain five remainders each. The bins for digits 3 and 6 contain six each.

Look again at the block for $1/53$.

0188679245283

It visits digit 6 once and digit 3 once. Those are the two visits to a bin of size six. At each of the other eleven positions, the bin has size five. Adding the bin sizes over the period gives

$$T=11\cdot5+2\cdot6=67.$$

The block for $1/106$ visits 3 twice and 6 twice. Its total is

$$T=9\cdot5+4\cdot6=69.$$

Two extra visits to the crowded bins account for the difference.

The pairing is visible in the digits. The fractions $1/53$ and $52/53$ sum to one, and their periodic digits add to nine in every column. A 3 becomes a 6. A 6 becomes a 3. The two crowded bins exchange places, preserving the total. The same happens between $2/53$ and $51/53$.

Four cycles therefore give two bin totals, 67 and 69. The figure shows one starting point in each cycle. The block for $1/106$ is a rotation of the row for $51/53$.

Count a column, then a period

A crowded bin means more matches. Suppose the reference emits digit 3. Every row whose remainder lies in the bin for 3 emits it too.

At denominator $pm$, after clearing $m$, each nonzero remainder is represented by exactly $m$ rows. A bin containing six remainders therefore gives $6m$ matching rows at that position. A bin containing five gives $5m$.

Over a period of length $L$, the average contribution from the repeating rows is $mT/L$. The terminating rows contribute another $m-1$. Divide by the total number of rows, $pm-1$, and the formula follows.

$$\alpha_b(pm)=\frac{m-1+mT/L}{pm-1}.$$

Here $b$ is the base, $p$ is a prime that does not divide it, and $m$ is supported on its prime factors. The bin total $T$ is taken along the particular cycle selected by $m$. Every row uses the same starting column.

For denominator 106, $m=2$, $L=13$, and $T=69$. The exact mean is

$$\alpha_{10}(106)=\frac{1+2\cdot69/13}{105} =\frac{151}{1365}\approx0.1106.$$

As $m$ grows while selecting a fixed lane, the constant terms drop out and the limit is $(L+T)/(pL)$. At 53, the two possibilities are

$$\frac{13+67}{53\cdot13}=\frac{80}{689}, \qquad \frac{13+69}{53\cdot13}=\frac{82}{689}.$$

This also explains the simpler formula for digit-partitioning primes. Every occupied bin contains one remainder, so $T=L$. Subtracting that formula from the general one gives

$$\alpha_b(pm)-\frac{2m-1}{pm-1} =\frac{m(T-L)}{L(pm-1)}\ge0.$$

Each visited bin contains at least the reference remainder itself. Any additional remainder in it adds a match. Collisions can only increase the alignment above the earlier count.

A single orbit guarantees a single limit. Recovering the earlier formula requires singleton bins along the reference orbit. Even several orbits can share one limit if they encounter the same total crowding. In decimal, 53 is the first prime where their totals differ.

A bound that survives the split

The exact average depends on the lane. Its upper bound comes from the largest bin any lane can visit.

In a base $b\ge2$, no bin contains more than $\lceil(p-1)/b\rceil$ remainders. For an odd prime, this is at most $(p-1)/2$. Substituting that ceiling into the exact formula gives, for every prime $p\ge5$ not dividing the base,

$$\alpha_b(pm)<\frac{p+1}{2p}\le\frac35.$$

The constant $3/5$ cannot be lowered. Take prime 5 in base two. Both digit bins contain two remainders, and the alignment along $m=2^a$ is

$$\alpha_2(5m)=\frac{3m-1}{5m-1}\longrightarrow\frac35.$$

Every finite value stays below it, while the sequence approaches it as closely as we please.

The reciprocal golden ratio, $1/\varphi\approx0.618$, lies above $3/5=0.6$. So none of these primes can reach the golden threshold, however its orbits split. Among primes at least 3 that do not divide the base, only 3 can cross. In decimal, denominator 12 already does, with alignment $7/11$. The smaller denominators 3 and 6 remain below, at $1/2$ and $3/5$.

This bound comes entirely from the digit bins. Comparing it with $1/\varphi$ places the golden threshold above the whole range for primes $p\ge5$.

Beyond one prime

The restriction on the denominator is specific. After removing the factors supported on the base, exactly one prime must remain. It covers $106=2\cdot53$ and $120=40\cdot3$. It does not cover $91=7\cdot13$.

A denominator supported entirely on the base is already understood. Every fraction terminates and receives score 1, so decimal denominator 50 has alignment 1. The prime 2 also has alignment 1 in every base.

A composite part outside the base remains outside this formula. At $91=7\cdot13$, we cannot simply choose one prime cycle and reuse the count. The prime case gives a precise boundary to what has been resolved.

There is substantial prior work on the digits along these cycles. Kak and Chatterjee studied agreement with cyclic shifts through Hamming distances. Girstmair expressed full-period digit variance through Dedekind sums. Murty and Thangadurai studied digit means without requiring the base to be a primitive root.

The calculation here counts equal digits across the full denominator table. Its exact dependence on the reference cycle is carried by $T$. That gives both the splitting at 53 and the sharp bound across every prime order.

The choice left by the prefix

Return to the two decimals at the beginning. Their remaining prime is the same. Their periods have the same length. One cycle visits the crowded bins twice, the other four times. That difference survives in the averages even as the supported factor grows without bound.

The same bins that distinguish the routes also limit how much agreement any route can collect. We can retain the split at 53 and still bound every eligible prime beyond 3 below three fifths.

Long division clears the twos and fives from the tail denominator. It leaves a starting remainder behind. The prefix is finite, but the choice of cycle it leaves can remain in the limit.