The Character Structure of the Collision Fluctuation

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

The total moves downward across those cutoffs. Which primes are pulling it there?

Every prime in this calculation ends in one, three, seven or nine. A single sum mixes those four streams together. I want to keep them apart long enough to see what the addition conceals.

Ten against the other multipliers

Take the fractions $1/p$ through $(p-1)/p$ and read their first decimal digits. Group the numerators by the digit they produce. Multiply each numerator by ten, take the remainder after division by $p$, and read the digit again. This is one step of long division.

Count the matches. At thirteen there are none. At seventeen there are two.

Now compare all the nonidentity multipliers. Some preserve no digits. Call those with positive counts constructive and take their mean. Bin Derangements and the Gate Width Theorem gives that mean from the bin sizes alone, without testing each multiplier.

Subtract it from the count for multiplication by ten. The difference is the collision deviation, $\Delta_p$.

A negative deviation means that multiplication by ten preserves fewer digits than the average constructive multiplier at that prime. Zero means it reaches the mean. At seventeen, the deviation is zero even though two digits match.

The five counts and their reference levels

Prime	Count	Mean	$\Delta_p$
13	0	$2$	$-2$
17	2	$2$	$0$
19	0	$2$	$-2$
23	0	$7/3$	$-7/3$
29	2	$26/9$	$-8/9$

In decimal there are $p-11$ constructive multipliers. A bin of size $n_d$ supplies $n_d(n_d-1)$ ordered pairs with distinct members. Each pair belongs to one multiplier, so the mean is

$$\overline C_{p,10}=\frac{\sum_{d=0}^{9}n_d(n_d-1)}{p-11}.$$

At thirteen, eight bins have one member and two have two. The crowded bins supply four pairs. Two multipliers share them, giving mean two. At twenty-nine, the pair count is fifty-two and there are eighteen constructive multipliers. The mean is $26/9$.

The prime must exceed eleven. At eleven the bins are singletons, every nonidentity count is zero, and the constructive mean is undefined.

Divide each deviation by its prime. The first contributions are $-2/13$, $0/17$, $-2/19$ and $-7/69$. Include every prime greater than eleven, up to the chosen cutoff.

The subtraction does not force this prime sum to vanish. The reference mean averages positive counts over multipliers at one prime. Here we choose one multiplier, ten, and carry its deviation into a sum over different primes.

Four running totals

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

Through $10,000$, their values are

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

All four totals are negative here. The stream ending in three contributes more than five times the negative amount of the stream ending in nine.

These are weighted totals, not per-prime averages. Each combines the number of primes in its class with the size and sign of their deviations. No random model enters the calculation.

Keeping separate accounts exposes the difference. There is also a way to combine them that respects the multiplication of their last digits.

Weights that multiply correctly

Multiply numbers ending in three and seven. Their product ends in one. Multiply two ending in three. Their product ends in nine.

A Dirichlet character assigns a nonzero weight to each of these four endings. Ending one gets weight one. Multiplying the weights must give the weight of the product’s ending.

Giving every ending weight one works. Another recipe gives $+1$ to endings one and nine, and $-1$ to three and seven. For three times seven, the weights give $(-1)(-1)=1$, the weight of ending one. For three times three, they also give one, the weight of ending nine.

Two more recipes use $i$, the square root of $-1$. Multiplication by $i$ is a quarter-turn. Four turns return to the start. Repeated multiplication by three makes the endings visit $1,3,9,7$, then one again.

Assign those endings the weights $1,i,-1,-i$. Both circuits obey the same multiplication.

The weight of ending three can only be $1$, $i$, $-1$ or $-i$, because its fourth power must be one. That choice fixes the other three weights. There are exactly four recipes.

Call the sums they produce $A_0,A_1,A_2,A_3$.

The all-ones recipe gives the principal component

With this normalization it is the total, not the average. The real signed recipe gives

It compares the combined contributions of one and nine with those of three and seven. The complex recipe collects the two remaining differences

Its real part compares one with nine. Its imaginary part compares three with seven. The fourth recipe reverses the imaginary part, so $A_3=\overline{A_1}$. Their equal magnitudes are built into the weights, not a coincidence in the data.

The difference hidden in the total

Apply the signed recipe at $10,000$. Keep the contributions from one and nine. Reverse those from three and seven.

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

Endings three and seven supply the larger combined negative contribution. Subtracting it leaves a positive difference. A downward total can contain this difference without revealing it.

Every step, through three cutoffs

Each row stops at a different cutoff. Within a row, the two panels use exactly the same primes. The left keeps the last-digit streams. The right keeps the total, the real and imaginary parts of $A_1$, and the signed contrast $A_2$. The vertical scales stay fixed within each column. Neither panel extrapolates.

The four numerical checkpoints are

Cutoff	$A_0$	$A_2$
100	-0.851414	0.300620
1,000	-1.149104	0.329630
5,000	-1.318994	0.354014
10,000	-1.392595	0.350235

Cutoff	$\operatorname{Re}A_1$	$\operatorname{Im}A_1$
100	-0.012410	-0.168966
1,000	-0.208657	-0.217315
5,000	-0.288890	-0.234378
10,000	-0.316180	-0.241931

There are twenty eligible primes through one hundred, 163 through one thousand, 664 through five thousand and 1,224 through ten thousand. The table includes all of them. The displayed decimals are rounded from rational sums.

The total becomes more negative at these cutoffs. The signed contrast rises and then falls a little. Following every intervening step does not turn that finite observation into a limit.

Every stream comes back

The four components retain all four running totals. To recover the stream ending in one, use

Add $A_0$ and $A_2$. The streams ending in three and seven cancel. Add twice the real part of $A_1$. The stream ending in nine cancels too. Four copies of $F_1$ remain.

The other streams return by changing the signs. This is finite Fourier inversion on the four unit classes, using the classical orthogonality of Dirichlet characters. The transform is standard. The digit table supplies the collision deviations we put into it.

Recover the other three streams

$$F_3=\frac{A_0-A_2+2\operatorname{Im}A_1}{4},$$ $$F_7=\frac{A_0-A_2-2\operatorname{Im}A_1}{4},$$ $$F_9=\frac{A_0+A_2-2\operatorname{Re}A_1}{4}.$$

There are four real quantities to recover. The real components $A_0$ and $A_2$ supply two. The real and imaginary parts of $A_1$ supply the other two. Its conjugate does not add a fifth independent quantity.

Parseval’s identity accounts for the squared sizes as well

$$\sum_{j=0}^{3}|A_j|^2=4\sum_{r\in\{1,3,7,9\}}|F_r|^2.$$

The factor four comes from using unnormalized character sums. Reconstruction and this identity hold at every finite cutoff, not just at the four checkpoints.

In another base, use the residue classes coprime to that base. There are $\varphi(b)$ classes and as many characters. We can compare remainders several digit steps apart, or replace $1/p$ by $1/p^s$. The finite reconstruction still works. For complex $s$, conjugating a component also conjugates $s$; the simple equal-magnitude pairing above uses real weights $1/p$.

These characters also define Dirichlet $L$-functions, where the coefficient of $n^{-s}$ is $\chi(n)$. Here we sum over primes and insert the additional coefficient $\Delta_p$. The shared characters do not make the two series the same. The companion paper proves the finite reconstruction and leaves the cutoff question open.

The running sums, weight table and signed contrast

At ten thousand, the primes ending in three and seven still supply the larger negative contribution. We can now follow that imbalance without confusing it with the downward total.

A positive contrast at four cutoffs does not say whether it stays positive. The full graph does not say whether it settles. Moving the cutoff adds new primes, and each arrives with an exact deviation of its own.

The four streams can be put back together. The question is where they go when we stop telling the primes where to end.