Silent Primes

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

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. Keep reading for six places or six thousand. The answer stays zero.

Try seventeen. Its repeating block is $0588235294117647$. 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

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

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

The recipe comes from Bin Derangements and the Gate Width Theorem. Group the nonzero remainders by the first digit of $r/p$. Exactly nine multipliers move every remainder out of its original digit bin. They are

Here $u$ runs from one through nine. Division means multiplication by a modular inverse. Long division advances a digit by multiplying the remainder by ten. Set $g=10$ and clear the denominator. The condition becomes

Nine small factorizations give the seven primes. But the recipe checks every nonzero remainder, while one ribbon may visit only some of them. We still have to rule out a ribbon that is silent while matches occur elsewhere.

In decimal at slide one, a prime above one hundred starts its reciprocal with at least two zeros. Those zeros already match across a one-place slide. Below one hundred, there are only fourteen primes outside our list to check. Each has a matching pair on the ribbon.

The factors and the fourteen witnesses

Number	Factorization	Above ten
91	7 × 13	13
82	2 × 41	41
73	73	73
64	2 × 2 × 2 × 2 × 2 × 2	none
55	5 × 11	11
46	2 × 23	23
37	37	37
28	2 × 2 × 7	none
19	19	19

For each remaining prime, the remainders $r$ and $s=10r\bmod p$ occur consecutively on the orbit of one and emit the same digit. This checks the ribbon itself, not merely the full table.

Prime	Remainder pair	Digit
17	15 → 14	8
29	13 → 14	4
31	7 → 8	2
43	24 → 25	5
47	21 → 22	4
53	47 → 46	8
59	52 → 48	8
61	27 → 26	4
67	37 → 35	5
71	16 → 18	2
79	18 → 22	2
83	64 → 59	7
89	10 → 11	1
97	75 → 71	7

The distinction is necessary. In base eight, $1/13$ repeats $0473$, with no one-place matches. Yet the full count is two. Remainders two and three both emit digit one, and $8\times2\equiv3\pmod{13}$. That match lies on another orbit.

For base $b$ and fixed slide $\ell$, the general recipe factors the $b-1$ integers

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

Their prime divisors above $b$ are exactly the globally silent primes. Ribbon silence also has a finite bound, $p\leq b^{\ell+1}$, but shorter orbits can require the extra check. The companion paper proves both statements.

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

Hold ninety-seven fixed

Keep the prime. Change the multiplier. For each multiplier $g$, move all ninety-six nonzero remainders and count how many keep their digit. Call the answer $C(g)$.

Set aside the identity, which leaves everything where it was. Set aside the nine zero counts. Eighty-six constructive multipliers remain.

Most give eight or ten matches. Four give sixteen. Twelve give six. The mean is about $9.628$.

Subtract the mean from each count, square the difference, and average the squares. The variance is about $6.001$. Its square root, the standard deviation, is about $2.450$ matches.

This population stays at one prime. The silent-prime question follows the particular multiplier ten across primes. The variance question keeps the prime fixed and measures the spread among its positive counts.

Why a failed test costs two

What makes one multiplier produce six matches and another sixteen?

Use the coordinate from the gate width theorem. It labels a nonidentity multiplier by the unique $c$ satisfying

Labels one through nine give the zeros. The constructive labels run from eleven through ninety-six. Label ten cannot occur.

There are nine tests. For test $m$, multiply $c$ by $m$, take the remainder modulo ninety-seven, and ask whether it exceeds $10m$.

A pass gives two matches. To see the pair, take $c=11$, which labels $g=53$. Test three passes because $33>30$. In the original digit bins, the two matches are

Multiplication by fifty-three takes remainder eighty-four to eighty-seven. Both emit eight. Reflect both remainders about ninety-seven and the second match appears. Thirteen goes to ten, and both emit one.

Every passing test gives a complementary pair like this. No other matches occur. Nine tests can contribute at most eighteen matches. Start at eighteen and subtract two for each failure.

From a passing test to its two remainders

The coordinate uses $x=[10r]_{97}$, where brackets mean the nonzero remainder. Two digits agree exactly when their transformed remainders agree modulo ten.

Put $t=[mc]_{97}$. When $t>10m$, the two transformed collisions are

$$97-t\longmapsto97-t+10m,$$ $$t\longmapsto t-10m.$$

They are complements, not reversals of the same arrow. For $c=11$ and $m=3$, they are $64\mapsto94$ and $33\mapsto3$. Undo multiplication by ten modulo ninety-seven to recover $84\mapsto87$ and $13\mapsto10$.

The number of tests in base $b$ is $Q=\lfloor(p-1)/b\rfloor$. Here it happens to be nine, the same as the number of zero multipliers. These are different counts. At $p=1999$ there are 199 tests and still nine zero multipliers.

In general, the collision-coordinate theorem gives

$$C(g)=2\#\{1\leq m\leq Q:[mc]_p>mb\}.$$

Where the strips cross

The labels that fail a test form intervals of consecutive integers. At ninety-seven, each interval contains ten labels. Test two fails from forty-nine through fifty-eight. Test three has two failure intervals. Test four has three.

Their starting points come from ordinary division. Divide ninety-seven by two and round up to get forty-nine. For test three, divide ninety-seven and twice ninety-seven by three, rounding up each time. Start a ten-label strip at each answer.

Tests two and four share the entire strip from forty-nine through fifty-eight. Tests two and three share nothing.

The lower grid compares every test with every other test. Gold means they fail together more often than independent tests with those same individual failure rates would. Purple means less often. We are measuring their dependence, not assuming it away.

The interval lengths give the mean. Their overlaps give the variance. Lengths alone cannot tell us how widely the counts spread.

Now draw every test. Put the constructive labels across the page and the tests down it. Color a cell when that test fails at that label. Read a column to recover the collision count.

The nine rows at ninety-seven grow to forty-nine at 499 and 199 at 1999. The strips still have width ten. Their starting points divide the prime into halves, thirds, quarters and finer parts. The curved bands are those rounded fractions laid one below another.

All 9,932,114 comparisons at 9,973

There are 997 tests and 9,962 constructive labels. Every cell is present in the full-resolution plate. Choose 100% in the image viewer to inspect the individual strips. The curve below gives the collision count for each column.

Color identifies a test’s height in the grid, not the size of a collision count. A blank cell passes. Each colored cell subtracts two matches from the starting count of 1,994.

The exact overlap calculation

Let $I_m$ be the failure labels for test $m$, and let $N=p-b-1$ count the constructive multipliers. Every interval has length $b$, with starts at $\lceil kp/m\rceil$ for $1\leq k<m$. Thus $|I_m|=b(m-1)$.

Write $H_{mn}=|I_m\cap I_n|$. The covariance pictured in each grid is

$$\operatorname{Cov}(Y_m,Y_n)=\frac{H_{mn}}{N}-\frac{|I_m||I_n|}{N^2},$$

where $Y_m$ is one for a failed test and zero otherwise. Since $C=2Q-2\sum_mY_m$,

$$\operatorname{Var}(C)=4\sum_{m,n=1}^{Q}\operatorname{Cov}(Y_m,Y_n).$$

The sum includes the diagonal and both orders of every off-diagonal pair. At ninety-seven, the second and fourth tests share ten of eighty-six labels. Independent failures with their individual rates would share only $10\times30/86$ labels on average. Tests two and three share none, against an independent reference of $10\times20/86$.

No random trial enters these counts. Independence supplies the reference product in a covariance, not a model imposed on the digit table.

Two thirds proved, nine tenths open

As the prime grows, so does the smaller bin size $Q=\lfloor(p-1)/b\rfloor$. Compare the variance with that size.

At ninety-seven, variance divided by $Q$ is about $0.667$. At 9973 it is about $0.887$. The proposed limit in decimal is $0.9$. For each fixed base $b\geq3$, the conjecture is

The values do not rise monotonically. After ninety-seven, the ratio falls to about $0.622$ at 101, then $0.595$ at 103. The fuller plot keeps those downward steps.

Part of the limit is proved. Pair each test only with itself, along the diagonal of the covariance grid. That contribution is $2Q/3$ plus a bounded remainder. Divided by $Q$, it tends to two thirds.

Everything left comes from distinct tests. Their contributions have both signs. The unresolved step is to control their combined effect as the prime grows. More computed points show the variation more faithfully. They do not prove a limit.

Exact populations behind the plotted values

Prime	$Q$	Variance	Variance / $Q$
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 seven checkpoints are the paper’s original selection. The expanded plot includes every prime from 97 through 1999, plus 4999 and 9973. Its rings identify the original seven. There are no connecting lines across the uncomputed gap above 1999.

For each prime, the variance is computed exactly over positive nonidentity multipliers. The displayed decimals are rounded. Direct digit-bin counts and the independent collision-coordinate formula agree for every multiplier in these populations.

A square root, in a different place

If the conjecture holds, the standard deviation grows like $\sqrt{Q}$, with a base-dependent constant. In decimal that constant is $\sqrt{0.9}$, about $0.949$.

The square-root scale catches my attention because it appears elsewhere in number theory. Under the Riemann Hypothesis, the error in the smooth prime-counting approximation is bounded by a constant times $\sqrt{x}\log x$. This is a bound, with a logarithmic factor, not an exact size for every fluctuation. Bombieri’s account explains the connection to the zeta zeros.

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.

The ribbons, distribution and interval plots

A ribbon gives a question we can ask by eye. Nine factorizations settle its zero case in decimal. 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.