Digit-Partitioning Primes and the Alignment Formula

Write the ten proper fractions of eleven in a column. Every decimal place contains all ten digits, each used once.

The first column runs from zero to nine. The second runs back from nine to zero. Then the two columns repeat. You can continue the division for a thousand places without finding two rows that share a digit in the same column.

Thirteen cannot keep this arrangement. It has twelve proper fractions and only ten decimal digits. Read their first digits in numerator order and you get

$$0,\ 1,\ 2,\ 3,\ 3,\ 4,\ 5,\ 6,\ 6,\ 7,\ 8,\ 9.$$

Three and six each appear twice. The failure is already visible before the repeating blocks have had a chance to do anything elaborate.

In Three and the Golden Ratio, the fractions of twelve fall into three groups, and their count gives an alignment of seven elevenths. The question here is how far that count travels. Seven has a six-digit cycle. Eleven has five separate pairs. Neither looks much like the single repeating digits at three.

One step of long division explains the property they share.

A digit for each remainder

Long division carries a remainder from one column to the next. Multiply it by the base, divide by the denominator, write the whole-number quotient as the next digit, and carry the new remainder forward.

For a prime denominator $p$ in base $b$, the digit written from remainder $r$ is

$$\delta(r)=\left\lfloor\frac{br}{p}\right\rfloor.$$

The floor brackets mean the whole-number part. In decimal, remainder 4 at denominator 11 gives $40=3\cdot11+7$. The digit is 3 and the next remainder is 7. That gives $70=6\cdot11+4$. The digit is 6, and we are back at remainder 4. Hence $4/11=0.363636\ldots$.

Each possible digit collects the remainders that produce it. Call that collection a digit bin. The next drawing puts the actual remainder numbers inside the bins, so we can see both the assignment and the first failure of separation.

At 3, the two nonzero remainders produce digits 3 and 6. At 7, six remainders produce six different digits. At 11, every digit has exactly one remainder. At 13, remainders 4 and 5 both produce digit 3, while remainders 8 and 9 both produce digit 6.

The boundary has a short proof. If $p<b$, increasing the remainder by one increases $br/p$ by more than one. Its whole-number part must increase too. Different remainders therefore give different digits.

There is one more possible endpoint. When $p=b+1$,

$$\delta(r)=\left\lfloor r-\frac{r}{b+1}\right\rfloor=r-1.$$

That is exactly the first column of the table of elevenths. Every remainder gets its own digit, with no gaps. Above this endpoint, the $p-1$ nonzero remainders outnumber the $b$ digits, so sharing is unavoidable.

Among primes that do not divide the base, different remainders produce different digits precisely when

$$p\leq b+1.$$

I call these primes digit-partitioning. In decimal they are 3, 7, and 11. Two and five divide the base and give terminating expansions, so they are excluded.

Now advance every row by one place. Multiplication by the base permutes the nonzero remainders modulo $p$. Distinct states remain distinct. If the digit function separates them now, it separates them after the next step, and after every step that follows. The whole infinite decimal inherits the separation from this single local rule.

Three different cycles

At three, each nonzero remainder returns to itself after one step. The digits are the familiar repeating 3 and repeating 6.

Seven takes the longer way around. Its six remainders form one cycle, and the digits read

$$1\to4\to2\to8\to5\to7\to1.$$

The fractions $1/7$ through $6/7$ enter that cycle at different places. Their repeating blocks are the six rotations of 142857. Read down any common column and the six digits are different.

Eleven breaks into five pairs. Since ten is one less than eleven, multiplying a remainder by ten sends it to its complement, $11-r$. The next step brings it back. The digits pair as 0 with 9, 1 with 8, and so on. Every pair sums to nine.

Two fixed states, one cycle of six, five cycles of two. The geometry changes completely. Yet in every case, knowing the digit tells us which nonzero remainder produced it. That is the property the alignment count needs.

1/3 = 0.|3|
2/3 = 0.|6|

1/7 = 0.|142857|
2/7 = 0.|285714|
3/7 = 0.|428571|
4/7 = 0.|571428|
5/7 = 0.|714285|
6/7 = 0.|857142|

 1/11 = 0.|09|
 2/11 = 0.|18|
 3/11 = 0.|27|
 4/11 = 0.|36|
 5/11 = 0.|45|
 6/11 = 0.|54|
 7/11 = 0.|63|
 8/11 = 0.|72|
 9/11 = 0.|81|
10/11 = 0.|90|

 1/13 = 0.|076923|
 2/13 = 0.|153846|
 3/13 = 0.|230769|
 4/13 = 0.|307692|
 5/13 = 0.|384615|
 6/13 = 0.|461538|
 7/13 = 0.|538461|
 8/13 = 0.|615384|
 9/13 = 0.|692307|
10/13 = 0.|769230|
11/13 = 0.|846153|
12/13 = 0.|923076|

Seven rows, three denominators

Return to the fractions of twelve. After two decimal places, three rows have terminated, four repeat the reference digit 3, and four repeat 6. A terminating row receives a score of 1 by definition. A repeating row receives the proportion of positions that agree with the reference. The average is therefore $(3+4)/11=7/11$.

Keep the factor four, but replace three by seven. We now have denominator 28 and twenty-seven fractions to compare with $1/28$.

The three terminating rows are $7/28$, $14/28$, and $21/28$. Four more rows match the reference tail after the second decimal place.

 1/28 = 0.03|571428|
 8/28 = 0.28|571428|
15/28 = 0.53|571428|
22/28 = 0.78|571428|

The remaining twenty rows disagree with the reference at every position. Some carry other rotations of the same six-digit block. A rotation is enough to make them different here. We compare the digits in the same column, without sliding one row against another to improve the fit.

There are still seven credited rows, now among twenty-seven. The score is $7/27$.

At denominator 44, the count is seven among forty-three. Three rows terminate, four match the reference, and thirty-six disagree throughout. Eleven’s five separate cycles require no extra term in the formula.

1/6 = 0.001010101…
1/3 = 0.010101010…
2/3 = 0.101010101…

The count in general

Write the denominator as $pm$, where $p$ is digit-partitioning and $m$ divides some power of the base. In decimal, $m$ may contain twos and fives, or be 1. These are the factors that enough steps of long division can clear away.

Start every comparison after a common number of places that clears $m$. The repeating tails then have denominator $p$, and the numerator’s class modulo $p$ determines its starting state relative to the reference.

The multiples of $p$ terminate. There are $m-1$ of them. The numerators one greater than a multiple of $p$ match the reference throughout. There are $m$ of those. Every other numerator gives a different remainder at every common position, and the digit function makes every resulting digit different too.

Average over all $pm-1$ proper fractions and the alignment is

$$\alpha_b(pm)=\frac{(m-1)+m}{pm-1} =\frac{2m-1}{pm-1}.$$

No cycle length appears. No count of separate cycles appears. We can leave those details out because distinct remainder states never become indistinguishable at the digit level.

The base determines which primes qualify and which factors $m$ it can clear. Once those conditions hold, the count is the same.

At a prime endpoint $p=b+1$, the complement pairs return as well. Their digits sum to five in base six and fifteen in base sixteen, just as they sum to nine in decimal.

Three keeps its place

As $m$ grows through values supported by the base, the alignment approaches $2/p$. For primes at least three, the finite scores lie below that limit. Exceeding the golden threshold within the digit-partitioning class therefore requires

$$\frac{2}{p}>\frac{1}{\varphi}, \qquad p<2\varphi\approx3.236,$$

where $\varphi=(1+\sqrt5)/2$.

Only three qualifies. Its limit is two thirds. Seven tends to two sevenths, eleven to two elevenths. The drawing with seven credited rows gives a finite view of the same arithmetic. The numerator stays put while the whole table grows around it.

Three belongs to the class whenever it does not divide the base. If the base is divisible by three, this case is absent. Odd bases also admit the exceptional prime two, whose alignment is always 1.

Three’s single repeating digits belong to a larger family. The digit boundary explains why the same count also works for much longer words. The golden comparison survives that enlargement.

Two shared digits at thirteen

Compare two rows from the table of thirteenths.

$$\frac{1}{13}=0.\overline{076923}, \qquad \frac{11}{13}=0.\overline{846153}.$$

The third digits are both 6. The sixth digits are both 3. This pair agrees in two of its six positions and repeats that pattern indefinitely. Its score is one third.

This is the first decimal prime beyond the boundary. Distinct fractions still have different remainder states at every step. But different states can now write the same digit. Complete agreement and complete disagreement no longer exhaust the possibilities.

The formula’s proof fails at a precise place. We can still count the terminating rows and the rows that follow the reference exactly. We can no longer assign zero to everything else. The partially matching rows contribute too.

Earlier bin and collision drawings

The original overview extends the remainder-to-digit assignment through 23. The first three primes have at most one remainder per occupied bin. Shared bins appear at 13 and become more numerous as the prime grows.

The second drawing extends the comparison through 499. Its lower chart counts unordered pairs of distinct remainder states that produce the same digit. That total describes the bins as a whole. An alignment score against one chosen reference also depends on the bins visited by that reference’s repeating cycle.

That distinction opens the next problem. The bins tell us where digits can agree. The remainder cycles tell us when a particular comparison reaches those bins. Beyond the boundary, the route through the table enters the count.

For now, watch the third place in the drawing. Remainders eight and nine both write a six. One step later, they write a one and a nine. The shared digit lasts one column. The distinction between the remainders survives.