The Collision Invariant

Divide 4 by 13. Then divide 5 by 13.

$$\begin{aligned} \frac4{13}&=0.\overline{307692},\\ \frac5{13}&=0.\overline{384615}. \end{aligned}$$

Both begin with a 3. Two different remainders give the same next digit. They belong to the same digit bin.

Now multiply the remainder 5 by 6 and divide by 13.

$$6\times5=30=2\times13+4.$$

The remainder changes from 5 to 4. The digit stays at 3. That agreement is a collision.

Try every nonzero remainder. Multiplication by 6 preserves the digit twice, at $5\mapsto4$ and $8\mapsto9$. The other ten moves leave their bins.

Six years, three papers

The research notes beginning in 2020 ask how much agreement long division permits. An alignment score gives one number. The pairwise matrix keeps the individual comparisons. Digit bins make it possible to count that agreement directly in the remainders. From there, the same arithmetic reaches finite tables, character sums, and $L$-functions.

The trilogy gathers the main results into a connected argument. The Collision Invariant establishes the finite object and the rules it obeys. The Collision Transform resolves the centered table into character components and studies what happens when primes sample it. The Collision Spectrum factors those components and identifies the $L$-values carried by the digits.

I submitted the three preprints to arXiv in March 2026. Six years of research notes now stand together as Invariant, Transform, and Spectrum. The familiar fractions return with more to answer for. A change of coordinates may reveal something new, but it still owes us the same count.

Here the work stays with the finite count. Four results belong together. They give its exact zero set, the number of digits needed to determine its deviation, the reflection of that deviation, and the balance of the individual crossings underneath it.

Change the labels. Keep the groups.

The next digit of $r/p$ in base $b$ is

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

The floor means round down. In decimal, $40/13$ and $50/13$ both round down to 3.

Multiply each remainder by ten and keep its remainder modulo 13. Our pair becomes

$$4\longmapsto1,\qquad5\longmapsto11.$$

The new labels end in the same digit. Every bin changes this way. Its members now share a remainder upon division by ten.

The bin called 3 has become the class ending in 1. Its name changes; its membership does not. Interval boundaries give way to a test of congruence.

Long division supplies the proof. Write $br=p\delta(r)+x$, where $x$ is the new remainder. Modulo $b$, this gives $x\equiv-p\delta(r)$. Since $p$ is invertible modulo $b$, knowing either side determines the other.

This linearization also works for a composite denominator $N$ coprime to the base. We use every remainder from 1 to $N-1$, including those that share a factor with $N$. A single repeating orbit need not visit them all.

Nine empty multipliers

At thirteen, multiplication by 6 gives two collisions. So does multiplication by 11. Apart from the identity, every other multiplier gives zero.

There are nine zeros. At 29, nine again. At 1009, with more than a thousand remainders to move, still nine.

The gate width theorem gives the whole list. For any prime $p>b$, the collision-free multipliers are

$$\begin{gathered} g\equiv-\frac{u}{b-u}\pmod p,\\ 1\leq u<b. \end{gathered}$$

These are $b-1$ distinct residues. The fractions are fixed before the prime is chosen. Only their residues change.

In decimal, $u=1$ gives $-1/9$, which is 10 modulo 13. The middle choice gives $-5/5=-1$. In an even base, that multiplier replaces digit $d$ by $b-1-d$. No digit is its own complement, so every remainder leaves its bin.

Why the list is complete

For a nonidentity multiplier $g$, let $c$ be the integer from 1 to $p-1$ satisfying

$$c(1-g)\equiv b\pmod p.$$

The value $c=b$ would require $g=0$, so it is excluded. Put $Q=\lfloor(p-1)/b\rfloor$. In the new coordinates, a collision has displacement $jb$ or $-jb$. Each successful positive displacement has a reflected partner. Counting both gives

$$C_p(g)=2\#\{1\leq j\leq Q:[jc]_p>jb\},$$

where $[jc]_p$ is the least nonnegative residue.

If $c<b$, then $jc\leq Q(b-1)<p$ and $jc<jb$. Every test fails. If $c>b$, the first test passes. Thus precisely $c=1,\ldots,b-1$ give zero. Solving for $g$ and setting $u=b-c$ gives the displayed rational family. The same count proves that every nonidentity collision count is even; the identity contributes the even number $p-1$.

What a hundred cannot change

Fix the multiplier at ten. One multiplication now advances long division by one digit. A collision means that the first two digits agree.

At denominator 109 there are 18 collisions. Subtract the smaller bin size, $\lfloor108/10\rfloor=10$, and 8 remain.

At denominator 209 there are 28. Subtract 20. Again 8. This denominator is composite, $209=11\times19$. The rule does not mind.

Cut the unit interval into a hundred equal pieces, labeled 00 through 99. A fraction in piece 37 begins with 37. The matching pieces are 00, 11, 22, and so on through 99.

Increasing a denominator coprime to ten by 100 adds one fraction to every piece. Ten pieces are selected, so the collision count grows by ten. The baseline grows by ten too. Their difference cannot notice the added hundred.

That difference is the collision invariant,

$$S(N)=C_N(10)-\left\lfloor\frac{N-1}{10}\right\rfloor.$$

The baseline sets the scale of a digit bin. It is not the exact average over multipliers.

For $N>100$ coprime to ten, the last two digits determine $S(N)$. All forty eligible endings fit in the collision periodic table. No primality test enters the lookup.

At lag $\ell$, the first and last digits of an $(\ell+1)$-digit block must agree. The table has modulus $b^{\ell+1}$. Those final $\ell+1$ base-$b$ digits of $N$ suffice, once $N>b^{\ell+1}$ and $N$ is coprime to $b$.

Nor can we discard the extra digit. In decimal at lag one, 01 and 91 both end in 1, but their values are 0 and $-9$.

The finite formula, including the endpoint

Put $B=b^\ell$ and $m=bB$. Let $G$ contain the integers $n$ from 0 to $m-1$ whose first and last base-$b$ digits agree. It has $B$ members. Write $N=mt+a$, where $a$ is a unit modulo $m$.

Each selected interval contributes

$$t+\left\lfloor\frac{(n+1)a}{m}\right\rfloor -\left\lfloor\frac{na}{m}\right\rfloor.$$

The last interval includes the excluded endpoint $N$, so subtract one. The $B$ copies of $t$ cancel those in the baseline, leaving

$$\begin{aligned} S_\ell(a)&=-1-\left\lfloor\frac ab\right\rfloor +\sum_{n\in G}d_n(a),\\ d_n(a)&=\left\lfloor\frac{(n+1)a}{m}\right\rfloor -\left\lfloor\frac{na}{m}\right\rfloor. \end{aligned}$$

The exact obstruction to a smaller base power is also uniform. The classes $1$ and $m-B+1$ agree modulo $B$, yet

$$\begin{aligned} S_\ell(1)&=0,\\ S_\ell(m-B+1)&=-(b-1)b^{\ell-1}. \end{aligned}$$

The companion proof derives both values from the same floor formula.

The other half of 09

The ending 09 carries $+8$. Reflect it across 100 and we get 91, carrying $-9$. Their sum is $-1$.

Every reflected pair has the same total. Twenty pairs give $-20$, and the forty entries have mean $-1/2$.

Read all forty values

The row gives the tens digit; the column gives the final digit. Read row 0, column 9 for ending 09.

Tens digit	Final 1	Final 3	Final 7	Final 9
0	0	2	0	8
1	-1	-1	1	-1
2	0	-2	6	0
3	-1	-1	-3	-1
4	-4	0	-2	0
5	-1	1	-1	3
6	0	2	0	0
7	-1	-7	1	-1
8	0	-2	0	0
9	-9	-1	-3	-1

This is the reflection identity. For every base and positive lag, with $m=b^{\ell+1}$,

$$S_\ell(a)+S_\ell(m-a)=-1.$$

There is a smaller version of the balance inside each entry. Take interval 11. At $a=9$, the step from $11a$ to $12a$ goes from 99 to 108 and crosses 100. At $a=91$, it goes from 1001 to 1092 and stops short of 1100. One crosses. Its partner does not.

Across the forty unit classes modulo 100, exactly twenty cross. Change the interval to any interior index from 1 to 98 and there are still twenty. The membership changes, sometimes intricately. The split stays equal.

Every crossing, from nine positions to a thousand

The half-group law holds for every modulus $m\geq3$, not just powers of a base. Reflection exchanges crossing and noncrossing on every interior interval. The two endpoint intervals are different. The first never crosses; the last always does. Keeping them in the count gives the offset $-1$ in the reflection identity.

The endpoint arithmetic behind the offset

For an interior index, complementary floors give $d_n(a)+d_n(m-a)=1$. At the first and last indices the totals are 0 and 2. Both endpoints belong to the selected diagonal, whose other $B-2$ members each contribute 1. Also

$$\left\lfloor\frac ab\right\rfloor+ \left\lfloor\frac{m-a}{b}\right\rfloor=B-1.$$

Add the two finite formulas,

$$\begin{aligned} &S_\ell(a)+S_\ell(m-a)\\ &\quad=-2-(B-1)\\ &\qquad+0+2+(B-2)\\ &\quad=-1. \end{aligned}$$

For the half-group law, the crossing test is $[(n+1)a]_m<a$. Neither equality nor a zero residue is possible when $a$ is a unit and $1\leq n\leq m-2$. Replacing $a$ by $m-a$ reverses the strict inequality. It pairs every crossing with exactly one noncrossing.

Three close-up views of the count

The first paper ends with a finite table whose size no longer grows with the denominator. Primes and composites read the same entries. A thousand more complete blocks add nothing to the deviation.

The Collision Transform takes up the next question. Center the table, sample it at primes, and give each contribution a weight. Its reciprocal-prime sum converges. Making the weights decay more slowly asks for cancellation that the finite symmetry alone cannot supply. The character coefficients say which $L$-functions enter that question; The Collision Spectrum identifies the special values already encoded in those coefficients.

The reflection pairs the cells. Nothing in it schedules the next prime.