The Character Structure of the Collision Fluctuation

I added up the collision deviations at the primes, giving each one a weight of $1/p$. Through $100$, the sum was about $-0.851$. Through $1,000$, it was $-1.149$. Through $10,000$, it was $-1.393$.

The sum was drifting downward. I wanted to know who was contributing.

In base ten, every prime in this calculation ends in $1$, $3$, $7$, or $9$. The total mixes those four streams together. A negative answer cannot tell us whether the streams are alike, whether one dominates, or whether differences between them disappear in the addition.

So I kept four running totals.

A reference level at each prime

First, the quantity being added.

Take the fractions $1/p,2/p,\ldots,(p-1)/p$ and look at their leading decimal digits. Each remainder belongs to the bin for the digit it produces. Multiply the remainder by ten, reduce modulo $p$, and look again. This is one step of long division. A collision occurs when the old remainder and the new remainder produce the same digit.

Count those collisions over all the nonzero remainders. At $p=13$, there are none. At $p=17$, there are two.

To give those counts a reference level, try all nonidentity multipliers. Some produce no collisions. The ones that do are called constructive. Average their positive counts. Bin Derangements and the Gate Width Theorem gives that mean exactly from the sizes of the digit bins.

At both $13$ and $17$, the constructive mean is $2$. Subtract it from the count for multiplication by ten.

Prime	Collisions for multiplication by ten	Constructive mean	Deviation
$13$	$0$	$2$	$-2$
$17$	$2$	$2$	$0$
$19$	$0$	$2$	$-2$
$23$	$0$	$7/3$	$-7/3$
$29$	$2$	$26/9$	$-8/9$

That last column is the collision deviation, written $\Delta_p$. It measures where multiplication by the base sits relative to the constructive mean at that prime. The reference level comes from the finite digit table itself.

Dividing by $p$ makes each prime’s contribution smaller. The first terms of the sum are therefore $-2/13$, $0/17$, $-2/19$, and $-7/69$. We include every prime greater than $11$. At $11$, the digit bins are singletons, no nonidentity multiplier is constructive, and there is no constructive mean to subtract.

This centering does not make the sum over primes zero. The mean was taken over multipliers at each individual prime. We then chose one multiplier and carried its deviation into a different sum.

Four running totals

Let $F_1$ collect the weighted deviations at primes ending in $1$. Let $F_3$, $F_7$, and $F_9$ do the same for the other endings. All four stop at the same cutoff.

Through $10,000$, the four contributions are approximately

Last digit	Weighted deviation sum
$1$	$-0.418680$
$3$	$-0.556673$
$7$	$-0.314742$
$9$	$-0.102500$

They are all negative at this cutoff, but they are far from equal. The stream ending in $3$ contributes more than five times the negative amount of the stream ending in $9$.

Keeping separate accounts already exposes that difference. There is also a way to combine the accounts that respects the arithmetic of their labels.

Weights that multiply correctly

Multiply two numbers ending in $3$ and $7$. Their product ends in $1$. Multiply two ending in $3$. Their product ends in $9$.

A Dirichlet character assigns a weight to each of these four endings, with one requirement. Multiplying the weights must give the weight of the product’s ending.

Giving every ending the weight $1$ works. A more useful recipe gives weight $+1$ to endings $1$ and $9$, and weight $-1$ to endings $3$ and $7$. Check the examples. For $3$ times $7$, the weights multiply to $(-1)(-1)=1$, the weight assigned to ending $1$. For $3$ times $3$, they also give $1$, the weight assigned to ending $9$.

Two more recipes use $i$, the square root of $-1$. It can be pictured as a quarter-turn. Four quarter-turns return to the start. The endings do the same under repeated multiplication by $3$, passing through $3,9,7,1$.

Assign those endings the weights $i,-1,-i,1$. Multiplication still works. In particular, the weights for $3$ and $7$ give $i(-i)=1$.

These are the four recipes. I have called the sums they produce $A_0,A_1,A_2,A_3$ to keep the notation short.

The all-ones recipe gives

$$A_0=F_1+F_3+F_7+F_9.$$

This is the principal character component. With this normalization, it is the total, not the average.

The real signed recipe gives

$$A_2=F_1-F_3-F_7+F_9.$$

It compares the combined contributions of endings $1$ and $9$ with those of endings $3$ and $7$.

The complex recipe collects two other differences into one number,

$$A_1=(F_1-F_9)+i(F_3-F_7).$$

Its real part compares $1$ with $9$. Its imaginary part compares $3$ with $7$. The fourth recipe reverses that imaginary part. Thus $A_3$ is the complex conjugate of $A_1$. These two components always have the same magnitude in this calculation. Their agreement is built into the weights.

The difference hidden in the total

Now apply the signed recipe to the four values at $10,000$.

The total is about $-1.393$. The contrast is about $+0.350$.

Endings $3$ and $7$ contribute the larger combined negative amount, so subtracting their contribution leaves a positive difference. This is a statement about the distribution of the deviation among the four classes. It is invisible in the total alone.

The full character calculation gives

Prime cutoff	$A_0$	Real part of $A_1$	Imaginary part of $A_1$	$A_2$
$100$	$-0.851414$	$-0.012410$	$-0.168966$	$0.300620$
$1,000$	$-1.149104$	$-0.208657$	$-0.217315$	$0.329630$
$5,000$	$-1.318994$	$-0.288890$	$-0.234378$	$0.354014$
$10,000$	$-1.392595$	$-0.316180$	$-0.241931$	$0.350235$

The total becomes more negative across these cutoffs. The signed contrast first rises, then falls a little. Even this small table resists the idea that all the components simply grow together.

Every stream comes back

Dirichlet characters are classical coordinates for multiplication on residue classes. Their usefulness here is exact. The four components retain all four running totals.

For example, the stream ending in $1$ comes back through

$$F_1=\frac{A_0+A_2+2\operatorname{Re}A_1}{4}.$$

Adding $A_0$ and $A_2$ cancels the streams ending in $3$ and $7$. Adding twice the real part of $A_1$ then cancels the stream ending in $9$. Four copies of $F_1$ remain. Division by four finishes the reconstruction.

The other streams return in the same way. This is finite Fourier inversion. It works at every cutoff. Parseval’s identity also says that the total squared size of the character components is four times that of the residue streams. The scale factor comes from the normalization of the transform.

There are four real quantities to recover. The two real components supply two of them. The real and imaginary parts of the conjugate pair supply the other two.

In another base, the admissible endings are the residue classes coprime to that base. There are $\varphi(b)$ of them, and exactly as many characters. We can also compare remainders several digit steps apart, or replace $1/p$ by $1/p^s$ to change the weight given to larger primes. The same finite reconstruction holds. The digit calculation supplies the coefficients. Character orthogonality supplies the change of coordinates.

Beyond the cutoff

The difficult question is still ahead of the calculation. As more primes enter, does a component settle, keep oscillating, or grow without bound?

A downward graph through $10,000$ cannot answer that. Neither can a positive contrast at four cutoffs. The finite theorem tells us exactly how to separate and recover the streams. It does not give their behavior at infinity.

But it gives the question a more precise form. At each prime, long division produces an exact deviation. Across primes, those deviations form a signal with a total, a signed contrast, and two further differences. We can follow each one without losing the ability to put the original streams back together.

I began with a sum that drifted downward. The calculation now preserves what that single number concealed. The primes ending in $3$ and $7$ pull more strongly in this range than those ending in $1$ and $9$. Whether that imbalance persists is a question I can finally ask on its own.