Silent Primes

Divide one by thirteen. Write the digits as they come.

$$\frac{1}{13}=0.076923076923\ldots$$

Six digits, then back to the beginning. The block $076923$ is the repetend, the part of the decimal that loops.

Cut it out and paste copies onto a thin strip of paper, end to end. Make a second ribbon. Lay one on top of the other so the digits line up, then slide the top ribbon to the right by one position.

Count the columns where the two digits agree.

At thirteen, there are none. Each digit now sits above a different digit. Keep reading for six places or six thousand. The answer stays zero.

Try seventeen. Its repeating block is $0588235294117647$. This time two columns agree in each sixteen-place block. One holds a pair of eights. The other holds a pair of ones.

Nineteen gives zero again. So does twenty-three. Twenty-nine gives a positive count.

I call a prime silent at a given slide when its ribbon has no matches with the shifted copy. For a one-place slide in base ten, the complete list of silent primes greater than ten is

$$\{11,13,19,23,37,41,73\}.$$

Seven primes. After seventy-three, the list stops. There are infinitely many primes still to come, but none adds an eighth entry at this slide.

Nine numbers to factor

There is a recipe for the list.

Compute $100-9$. Then $100-18$. Then $100-27$. Keep subtracting another nine until you have made nine subtractions. Factor the results and keep the prime factors greater than ten.

Number	Factorization	Prime factor above ten
$91$	$7\times13$	$13$
$82$	$2\times41$	$41$
$73$	prime	$73$
$64$	$2^6$	none
$55$	$5\times11$	$11$
$46$	$2\times23$	$23$
$37$	prime	$37$
$28$	$2^2\times7$	none
$19$	prime	$19$

Read down the rightmost column. Sort the entries. There are the seven silent primes, recovered from nine small integers.

The recipe comes from Bin Derangements and the Gate Width Theorem. Long division assigns each nonzero remainder a digit. A multiplier moves the remainders around. Nine particular multipliers, in base ten, move every remainder into a bin with a different digit. They have the form

$$g\equiv-\frac{u}{10-u}\pmod p,$$

where $u$ runs from one to nine. The fractions are read modulo the prime. Dividing by a denominator means multiplying by its modular inverse.

Advancing long division by one digit multiplies the remainder by ten. Set $g=10$ in that condition and clear the denominator. The result says that $p$ divides

$$10(10-u)+u=100-9u.$$

That is where the nine numbers come from.

There is a distinction to keep in view. The recipe finds multipliers that produce no matches anywhere in the complete set of nonzero remainders. One ribbon may visit only some of those remainders. Silence on that ribbon alone could miss matches elsewhere.

For example, in base eight, $1/13$ repeats the block $0473$. Its one-place slide has no matches. Yet the full remainder count has two. Remainders two and three both produce digit one, and multiplying two by eight leaves remainder three modulo thirteen. The match exists outside the ribbon we chose.

In base ten at slide one, the two silent-prime lists do coincide. The paper proves this extra fact. For any prime above one hundred, the decimal of $1/p$ starts with at least two zeros, so a one-place slide already has a match. Below one hundred, the remaining possibilities are finite, and an explicit matching pair rules out each prime outside the seven. No assumption about a maximal repetend is needed.

The full-remainder recipe also works at every fixed slide in every base. For base $b$ and slide $\ell$, factor the $b-1$ integers

$$N(u)=b^{\ell+1}-u(b^\ell-1),$$

with $u$ from one to $b-1$, and keep their prime factors above $b$. These give exactly the globally silent primes. The ribbon question is finite too, since a ribbon-silent prime must satisfy $p\leq b^{\ell+1}$, though a shorter ribbon can require the extra check just illustrated.

The zero count has a finite answer. The positive counts leave a different question.

Hold the prime fixed

Take ninety-seven. Keep it fixed while changing the multiplier.

For each multiplier, move every nonzero remainder and count how many land in a bin with the same digit. Call the answer $C(g)$. Multiplication by ten gives the one-digit move from long division. Other multipliers give other rearrangements of the same ninety-six remainders.

Set aside the identity multiplier, which leaves everything where it was. Set aside the nine deranging multipliers, whose counts are zero. There are eighty-six multipliers left. These are the constructive multipliers, the ones that produce at least one match.

Their counts make the following distribution.

Most counts are eight or ten. Four multipliers produce sixteen matches. Twelve produce six. The average is about $9.628$.

Subtract that average from each count, square the difference, and average the squares. This gives the variance, about $6.001$. Its square root is the standard deviation, about $2.450$, a measure of the spread in the original units of matches.

This is the population the paper studies. Each prime has its own distribution over constructive multipliers and its own variance. The zero counts have been removed before that variance is calculated.

The silent-prime problem asks when the particular move supplied by long division lands among the zero counts. The variance problem asks how widely the positive counts spread. Both begin with the same digit comparison, but they ask different questions of it.

Strips inside the count

I wanted to know what makes one multiplier produce six matches and another sixteen. The answer becomes easier to see after relabeling the multipliers.

At ninety-seven, the relabeling assigns each constructive multiplier one integer $c$ between eleven and ninety-six. It is a one-to-one correspondence. The collision count can then be computed through nine yes-or-no comparisons.

Start at eighteen. For each failed comparison, subtract two. The result is $C(g)$.

The useful fact is that the labels where a comparison fails form intervals of consecutive integers. Each interval contains ten labels. For the second comparison, failure occurs from forty-nine through fifty-eight. The third comparison has two intervals. The fourth has three.

The endpoints come from ordinary division. Divide ninety-seven by two and round up to get forty-nine. For the third comparison, divide ninety-seven and twice ninety-seven by three, rounding up each time. That gives thirty-three and sixty-five. Start a ten-label strip at each answer.

Look at the second and fourth rows. They share the entire strip from forty-nine through fifty-eight. Every label in that strip fails both comparisons. Now look at the second and third rows. They share nothing.

Counting the length of each strip tells us how often each comparison fails. That is enough to calculate the mean. To calculate the variance, we also need to know how often two comparisons fail together. We have to count the overlaps.

This gives an exact formula for the variance whenever $p>b+1$, so there are constructive multipliers to average over. It is a finite sum of interval intersections, with no independence assumption. The figure shows three of the nine comparisons at ninety-seven. The formula includes all nine, and all their pairings.

The arithmetic has put the dependence on the page. The strips line up in some places and miss in others. Their lengths alone cannot describe the distribution.

The size of the spread

Let $Q$ be the smaller digit-bin size. In base $b$,

$$Q=\left\lfloor\frac{p-1}{b}\right\rfloor.$$

At ninety-seven in base ten, $Q=9$. At $9973$, it is $997$. As the prime grows, we get more remainders in each bin and more comparisons in the interval formula.

Here are exact variance calculations, rounded for display, at seven selected primes. Each row comes from a separate population of constructive multipliers.

Prime	Bin size $Q$	Variance	Variance divided by $Q$
$97$	$9$	$6.001$	$0.667$
$193$	$19$	$13.967$	$0.735$
$499$	$49$	$40.453$	$0.826$
$997$	$99$	$84.353$	$0.852$
$1999$	$199$	$173.099$	$0.870$
$4999$	$499$	$440.549$	$0.883$
$9973$	$997$	$884.827$	$0.887$

The proposed limit in base ten is $0.9$. In general, for each fixed base $b\geq3$, the conjecture is

$$\frac{\operatorname{Var}(C)}{Q}\longrightarrow 1-\frac1b.$$

This would make the standard deviation grow like $\sqrt{Q}$, with a base-dependent constant in front. In base ten that constant would be $\sqrt{0.9}$, about $0.949$.

The conjecture concerns variance on the scale of $Q$, or standard deviation on the scale of its square root. Those are two ways of describing the same proposed limit.

Part of the limit can be proved. In the overlap calculation, first pair each comparison only with itself. This is the diagonal contribution. Its total is $2Q/3$ plus a bounded remainder. After division by $Q$, it tends to two thirds.

The rest comes from pairing distinct comparisons. The gold strips in the interval figure give one small example. Controlling the combined contribution of all such pairings is the unresolved step. The numerical values support the proposed limit, but seven points cannot establish it, and the theorem for the diagonal part does not settle the full sum.

A reason to keep looking

The square-root scale catches my attention because it appears elsewhere in number theory. Under the Riemann Hypothesis, the error in the smooth approximation to the number of primes up to $x$ is bounded by a constant times $\sqrt{x}\log x$. That is a bound on prime-counting error, with a logarithmic factor, rather than an exact size for every fluctuation. Bombieri’s account explains the connection to the zeros of the zeta function.

I want to be careful here. Our variance averages over multipliers at one prime. Prime-counting error concerns how the primes themselves are distributed. A shared square-root shape does not identify the two, and this paper gives no theorem linking the collision variance to zeta zeros.

What I can follow directly is the arithmetic inside the count. A ribbon gives a question we can ask by eye. Nine factorizations settle its zero case in base ten. Once the count is positive, the intervals show where its comparisons reinforce or avoid one another.

The finite list is complete. The spread is still asking something of us. We can now point to the overlaps where the answer has to be found.