The Spectral Repulsion

Write these two rows of numbers above one another. Each row adds to one.

	First position	Second	Third	Fourth
First row	0.4	0.3	0.2	0.1
Second row	0.1	0.2	0.3	0.4

Multiply down each column and add the products. The overlap is $0.20$.

Now reverse the second row. The largest number sits below the largest, the smallest below the smallest. The overlap rises to $0.30$. Neither row has gained any weight. We have changed only which entries meet.

There are twenty-four ways to arrange the second row. Their average overlap is exactly $0.25$. So the first arrangement falls twenty percent below that average, while the second sits twenty percent above it.

This small example is the comparison I want to make with the collision spectrum and the primes. A low overlap needs a reference point. Keeping the weights fixed and changing their order supplies one.

The two rows supplied by arithmetic

The actual rows come from long division. Count the remainders that emit the same digit on two consecutive steps. Subtract the baseline count, then the mean deviation within the appropriate residue class. The resulting centered collision table assigns a signed weight to each unit residue modulo the square of the base.

A Dirichlet character is a multiplicative pattern on those residues. Comparing the table with each character separates it into channels. Some channels carry a large coefficient, others a small one. Squaring their magnitudes and dividing by the total gives the first row of nonnegative weights.

For the second row, use the same characters to combine primes below a fixed cutoff. Each prime contributes its character value divided by its square root. These character values can cancel, so different channels accumulate different totals. Square the magnitudes of those totals and normalize again. Both rows now add to one.

The cutoff is part of the calculation. In base $b$, the sums here use primes strictly between $b^2$ and $X$.

There is one restriction before the comparison begins. For an odd prime base, reflection removes every even character from the collision spectrum, and centering removes every character inherited from the base itself. The eligible channels are called primitive odd characters. Both rows are normalized on that same set.

The channels also come in conjugate pairs. The two members have equal weights in each row, so the shuffle keeps them together. In base five there are eight eligible characters, making four pairs. The opening example had the same number of positions, though its simple weights were chosen for illustration.

If there are $k$ pairs, their average shuffled overlap is $1/k$. Divide the observed pair overlap by that average and call the result $\rho$.

$$\rho=k\sum_{r=1}^{k}x_r y_r.$$

Here $x_r$ and $y_r$ are the two weights assigned to pair $r$. A value of one matches the shuffle average. A value below one means the actual pairing has less overlap. The mean-one statement is exact for any two fixed probability lists. It does not require a random model of the primes.

Below the shuffle average

At a cutoff of two million, base five gives $\rho\approx0.6216$. Its arithmetic pairing has about sixty-two percent of the overlap obtained by averaging all relabelings of those four pairs.

The other tested prime bases also fall below one. They do not give the same number. Base seven gives about $0.4801$, base nineteen about $0.7690$. Across the ten nontrivial bases, the mean is $0.5909$.

Increasing the cutoff changes the comparison. The mean across those ten bases is about $0.624$ at $X=250{,}000$, $0.591$ at two million and $0.546$ at five million.

All sixty nontrivial base-and-cutoff combinations are below one. The windows share many of the same primes, so these are not sixty independent trials. What they establish is a repeated deficit over the stated range. They do not yet identify a limiting constant.

The shuffle also has an exact variance. That lets us measure how far an observation sits from its mean. At two million, base five is about $0.71$ shuffle standard deviations below one; base thirty-seven is about $2.93$ below. Those values describe this particular relabeling comparison. They do not, by themselves, give a probability that the arithmetic pattern is accidental.

Base three needs separate treatment. It has just one eligible pair, so there is nothing to shuffle. Its ratio is exactly one at every cutoff where the normalization is defined. Including an excluded channel would change the question by mixing channel eligibility with placement.

I use spectral repulsion for the measured deficit after that distinction has been made. The open question is whether it persists as the cutoff grows, and what controls its size.

Concentration and placement

The two rows differ in another visible way. The collision weights are more concentrated.

A standard measure called effective support expresses spread as an equivalent number of equally weighted channels. Four equal weights give effective support four. A single nonzero weight gives one. Unequal weights can give a value between those integers.

At the two-million cutoff, the comparison is as follows.

Base	Eligible channels	Collision effective support	Prime effective support
5	8	4.59	5.21
13	72	17.97	49.71
29	392	63.95	290.02
37	648	92.05	493.73

At base twenty-nine, the collision distribution has the same entropy as roughly sixty-four equal weights. The prime distribution has the same entropy as roughly two hundred and ninety. These are measures of concentration, not counts of occupied channels.

The spotlight and floodlight comparison helps describe their spread. It does not explain their overlap. A narrow beam can fall on the brightest part of a broad one. The two rows at the beginning had identical shapes, yet reversing one changed their overlap. The shuffle holds both shapes fixed precisely so we can examine the placement.

Adding the residues

The same collision table allows a different experiment. Instead of matching character channels, pair residues that add to a chosen target.

In base three, the centered table modulo nine is small enough to write out in full. Residues that are not units receive weight zero.

Residue	0	1	2	3	4	5	6	7	8
Weight $f$	$0$	$2/3$	$4/3$	$0$	$2/3$	$-2/3$	$0$	$-4/3$	$-2/3$

Start with target zero. Each residue meets its negative modulo nine. One meets eight, two meets seven, four meets five. Their weights have opposite signs. Every product is a negative square.

Add those products over all nine starting positions. The result is $-16/3$.

Now do the same for each target from one through eight. This operation is called additive convolution. Its entire table is shown below.

The nine entries sum to zero. That average conceals a definite arrangement of positive and negative values. It would be a mistake to call the table blind to addition on the strength of its average.

The reflection gives the general statement directly. Writing $m=b^2$,

$$K(0)=\sum_{a\bmod m}f(a)f(-a) =-\sum_{a\bmod m}f(a)^2.$$

The value at zero is the negative of the table’s energy. The other targets collectively supply exactly the opposite amount. These identities hold for every base, including composite bases. In additive Fourier coordinates, the convolution has a further exact property. Every coefficient is nonpositive.

There is also a direct consequence for prime pairs. Seventeen and nineteen add to thirty-six, a multiple of nine. Their base-three table weights are $-2/3$ and $2/3$, so their product is $-4/9$. Whenever an even target is divisible by the base squared, every prime pair representing it has a nonpositive weight product. A nonunit prime receives zero weight. This fixes the sign of the weighted sum without asserting that a prime representation exists.

The additive sign law and the multiplicative overlap deficit answer different questions. Reflection proves the former. The finite character calculations record the latter. Neither needs to be made into an explanation of the other.

The four-column example can be understood by following each number to its new position. The arithmetic comparison asks for that same degree of explanation. Sixty small overlaps are enough to keep me looking, but they have not supplied it.