The Collision Transform

Take two cells from the decimal collision table. The cell 09 holds $8$. The cell 91 holds $-9$.

$$8+(-9)=-1.$$

The same addition works for all twenty pairs of cells whose labels sum to 100. The entries differ, sometimes sharply, but every pair adds to $-1$.

These are the deviations defined in The Collision Invariant. Count the remainders whose digit stays the same after one step of long division, then subtract the baseline $\lfloor(p-1)/10\rfloor$. For every prime $p>100$, its last two digits determine the answer. Every such prime ending in 09 receives $8$. Every one ending in 91 receives $-9$.

I want to add these values over primes. First I need to account for the bias already built into the table.

Four different averages

Read down the ten cells ending in 1 and take their average. Do the same for 3, 7 and 9.

Final digit	Mean collision deviation
1	$-17/10$
3	$-9/10$
7	$-1/10$
9	$7/10$

These are exact averages over the forty cells. No primes have been sampled to estimate them.

Adding $1/2$ to every entry would make the whole table average zero. It would leave these four families with different averages. I want to remove what the final digit alone contributes, so I subtract each family’s own mean.

The two opening cells become

$$\begin{aligned} f(09)&=8-\frac{7}{10}=\frac{73}{10},\\ f(91)&=-9-\left(-\frac{17}{10}\right)=-\frac{73}{10}. \end{aligned}$$

Now they are exact opposites. So is every other reflected pair. Each column also sums to zero.

The paper calls a column a reduction fiber. Its cells become the same residue when reduced modulo the base. Fiber centering means subtracting that column’s average. The construction works in every base and at every lag. Comparing digits $\ell$ steps apart gives a table modulo $q=b^{\ell+1}$.

I will stay with the forty decimal cells for the examples.

Giving the pair a weight

A Dirichlet character assigns a weight to each cell, consistently with multiplication of the labels modulo 100. There are forty such characters. Together they give a Fourier expansion of any table on these forty positions.

To find a character’s coefficient, multiply each centered entry by the conjugate of its character weight, add the results and divide by forty. A coefficient measures how much of that pattern is present in the table.

Look again at $73/10$ and $-73/10$.

An even character gives the same weight to 09 and 91. Whatever that weight is, the two contributions cancel. They cancel in every reflected pair. The entire coefficient is zero.

An odd character gives opposite weights to the two cells. The sign change reverses the negative entry, so this pair’s contributions add.

That proves why every even character disappears. It does not say every odd character survives. Contributions from different pairs can still cancel.

There is a second cancellation. A character that depends only on the final digit gives the same weight to every cell in a column. The column sums to zero, so that character also has coefficient zero.

In base ten, reflection removes the twenty even characters. Column centering removes two additional odd characters inherited from modulo 10. Eighteen character channels remain available. Their actual coefficients decide which are active.

This is the spectral gate. It follows from two operations the reader can check in the table, pairing the cells and summing the columns.

Adding the primes

For each prime greater than 100, look up its centered entry and divide by a power of the prime. The collision transform is the resulting series

$$F(s)=\sum_{p>100}\frac{f(p\bmod100)}{p^s}.$$

At $s=1$, the prime 109 contributes $73/(10\cdot109)$. The prime 191 contributes $-73/(10\cdot191)$. Their numerators are opposites, but their denominators differ. Pairing the finite table does not cancel the prime series term by term.

The question is whether the running total settles as more primes enter.

For $s>1$, convergence is straightforward because the entries are bounded and the weights decay fast enough. At $s=1$, the unweighted sum $\sum_p1/p$ diverges. The signs now have to do some work.

Mertens’ theorem in arithmetic progressions gives the needed fact. Add $1/p$ over primes ending in any particular eligible pair of digits. Each of the forty classes has the same growing term, $\frac1{40}\log\log x$, followed by its own constant and an error tending to zero.

Multiply those forty class sums by the forty entries of $f$. The common growing term is multiplied by zero, because the entries sum to zero. The forty constants give a finite value. The errors tend to zero.

Thus

$$F(1)\quad\text{converges}.$$

The convergence is unconditional, for every fixed base and lag. The finite centering has removed exactly the coefficient that would multiply the common divergence.

Where the L-functions enter

The character expansion lets us perform the same sum one pattern at a time. Each active character supplies a prime series of the form $\sum_p\psi(p)/p^s$, with a coefficient fixed by the digit table.

An $L$-function packages these character weights into an Euler product, one factor for each prime. Taking its logarithm produces the prime series, followed by terms for the squares, cubes and higher powers of primes. Subtract those extra terms and we recover the prime series exactly.

There is a small bookkeeping point. A character written modulo 100 may already come from a smaller modulus. We use its primitive character, the version at its smallest modulus, and call its collision coefficient $C_\psi$. At primes greater than 100, the character values agree.

The transform then has the form

$$F(s)=\sum_\psi C_\psi\log L(s,\psi)-R(s).$$

This identity initially holds for $\operatorname{Re}(s)>1$. Here $R(s)$ collects the higher prime powers and the finitely many primes excluded at the start. The higher powers give an absolutely convergent correction throughout $\operatorname{Re}(s)>1/2$. The finite correction causes no singularities.

The coefficients $C_\psi$ have been computed from the digit table before $s$ starts to move. The analytic question lies in the $L$-functions they multiply.

Lowering the exponent

At $s=1$, a prime contributes its centered value divided by $p$. At $s=1/2$, it is divided only by $\sqrt p$. Large primes contribute much more. Convergence would require stronger cancellation.

The paper proves a conditional statement about that cancellation. Fix $1/2\leq\theta<1$. Suppose every nontrivial zero of every active primitive $L$-function has real part at most $\theta$. Then the ordinary prime series converges throughout

$$\operatorname{Re}(s)>\theta.$$

The real part is the horizontal coordinate in the complex plane. The conclusion holds to the right of the specified line. It does not include the line itself.

Under the generalized Riemann hypothesis for the active functions, we may take $\theta=1/2$. This gives convergence for every real $s>1/2$, as well as throughout the corresponding complex half-plane. Convergence at $s=1/2$ is still a separate question.

It is useful to watch a finite piece of the sum as the exponent falls. The figure uses every prime between one million and ten million, a total of 586,081 primes in each base.

In base ten, this window contributes about $0.000065$ at $s=1$ and $-0.366499$ at $s=1/2$. The other bases have their own signs and sizes. These computations show what one declared interval contributes. They cannot decide whether the entire series converges.

A zero that cannot cancel

There is also a conclusion in the other direction. Suppose the ordinary collision series converges at a real point $\sigma\geq1/2$. What would a zero farther to the right have to do?

Near a zero $\rho$, the logarithmic derivative $L'/L$ blows up like $1/(s-\rho)$. This is a pole. The coefficient of that blow-up is the zero’s multiplicity. A simple zero contributes one. A double zero contributes two.

Differentiate the transform identity. If the collision series converges at $\sigma$, it defines an analytic function to the right of that line, and its derivative has no poles there. Neither does the correction $R'$. The weighted poles from the $L$-functions must therefore cancel.

At any zero $\rho$ with $\operatorname{Re}(\rho)>\sigma$, this requires

$$\sum_\psi C_\psi\,m_\psi(\rho)=0,$$

where $m_\psi(\rho)$ is the multiplicity of that zero in $L(s,\psi)$, or zero if the function does not vanish there.

A zero belonging to only one active function cannot satisfy this equation. Its multiplicity is positive and its collision coefficient is nonzero. There is nothing else at that point to cancel it.

If several active functions share exactly the same zero, their weighted multiplicities might cancel. The theorem leaves that possibility open. It gives a collective constraint, not a way to test each $L$-function separately.

I find this the most interesting part of the result. The entries $8$ and $-9$ began as differences between counts of digit matches. After centering, their finite table determines coefficients that a shared zero would have to balance exactly. The table does not tell us where that zero is. It tells us what would have to cancel there for the prime series to converge past it.