The Collision Transform and the Critical Strip

Take the prime $11$. Its centered collision weight in base three is $4/3$. Divide that by $11$ and its contribution to the prime sum is about $0.1212$. Divide by the square root of $11$ instead and the contribution becomes about $0.4020$.

The digits have not changed. Neither has the count. I have changed how much weight the prime receives.

Do this at every prime and the question becomes harder. Positive and negative contributions that balanced under one weighting may fail to balance under another. A prime near a million receives roughly a thousand times as much weight when the divisor changes from the prime to its square root.

The Centered Collision Sum proves convergence when the divisor is $p$. I want to know how much more weight the large primes can carry before that convergence fails.

Six numbers to start with

In base three, at lag one, the collision deviation depends on the prime’s remainder after division by nine. There are six possible remainders. Subtract the mean within each family modulo three and the centered table is

Remainder modulo 9	1	2	4	5	7	8
Centered weight	$2/3$	$4/3$	$2/3$	$-2/3$	$-4/3$	$-2/3$

Every prime greater than nine takes its weight from this row. Eleven takes $4/3$. Thirteen takes $2/3$. Seventeen takes $-2/3$. The table is finite. The list of primes drawing from it is infinite.

Write $f(p)$ for the weight assigned to the prime. The sum I am studying is

$$F(s)=\sum_{p>9}\frac{f(p)}{p^s},$$

where the sum runs over primes in increasing order. At $s=1$ we have the proved convergence result. Decrease $s$ and the divisor shrinks. More of each large prime’s weight remains in the sum.

The weights balance exactly within the finite table. Along the primes, they need not occur in perfectly balanced proportions. Convergence depends on how those imbalances accumulate.

The next nine million

Choose a range that can be checked. Take every prime greater than one million and no greater than ten million. There are $586{,}081$ of them. Add their centered contributions at six exponents, using each base’s own table.

At $s=1$, this entire window contributes about $0.000042$ in base three. At $s=1/2$, it contributes about $0.077496$. In base ten the corresponding contributions are about $0.000065$ and $-0.366499$.

The same primes are being counted in every curve. The changes come from their weights and from the exponent.

These numbers are useful because they measure a specific remaining piece of the sum. They do not tell us what every subsequent piece will do. A small contribution from this window does not prove convergence. A larger contribution does not disprove it. To decide the infinite question, I need something the finite computation cannot supply on its own.

Two channels survive

The six base-three weights have a particularly small character expansion. Dirichlet characters are the multiplicative patterns on the six remainders. They give another set of coordinates for the same table, just as a Fourier expansion describes a waveform by its component frequencies.

Multiply repeatedly by two modulo nine and the remainders occur in the order

$$1,\ 2,\ 4,\ 8,\ 7,\ 5,\ 1.$$

There are six characters on this cycle, but the centered table uses only two. Reflection removes the even characters. Centering within each family removes the remaining character that depends only on the remainder modulo three.

Call one surviving character $\chi$, with $\chi(2)=e^{\pi i/3}$. The other is its complex conjugate $\overline\chi$. The exact expansion is

$$f(a)=c\chi(a)+\overline c\,\overline\chi(a).$$

Here $c=(1-i\sqrt3)/3$. The complex quantities combine to give the real fractions in the table. Nothing has been approximated. We can now split the prime sum into two character sums, each with its own established analytic theory.

A different base changes the finite table and the characters that survive. It can therefore select different $L$-functions. The number of surviving characters alone does not tell us which finite computation will settle fastest.

The prime sum inside the logarithm

The connection to an $L$-function comes from its Euler product. Taking the logarithm gives

$$\log L(s,\chi) =\sum_p\frac{\chi(p)}{p^s} +\sum_p\sum_{k\ge2}\frac{\chi(p)^k}{k p^{ks}}.$$

The first sum is exactly the kind we need. The second contains the higher powers of each prime. This is the classical logarithmic expansion of the Euler product, initially valid to the right of $s=1$.

Subtract those extra terms and account for the finitely many small primes excluded from the collision sum. For any fixed base and lag, the result has the form

$$F(s)=\sum_{\chi}c_\chi\log L(s,\chi)-R(s).$$

The finite table determines the coefficients $c_\chi$. The correction $R(s)$ collects the higher prime powers and the small-prime adjustment.

Here the square root enters for a precise reason. The higher-power terms start with $p^{-2s}$. They converge absolutely whenever the real part of $s$ exceeds $1/2$. Throughout that region, the correction is an analytic function with no singularities. Possible singularities in the continued expression come from the logarithms of the selected $L$-functions.

That gives a formula beyond the region where we first derived it, wherever suitable zero-free domains allow the logarithms to be defined. It does not yet prove that adding the original prime terms still converges there. Analytic continuation of a formula and convergence of its defining series are separate claims.

To the right of one half

Allow $s$ to be complex. Its real part gives the horizontal coordinate. The critical strip lies between real parts zero and one, and the critical line runs vertically through $1/2$.

The Generalized Riemann Hypothesis places the nontrivial zeros of primitive Dirichlet $L$-functions on that line. Assume it for the primitive $L$-functions selected by this collision table. Standard estimates for their prime sums then prove that the ordinary collision series converges throughout

$$\operatorname{Re}(s)>\frac12.$$

This is a conditional theorem. Its conclusion covers every real exponent strictly greater than one half, however close. It does not include the boundary itself.

The reason is cancellation among the character values along the primes. Under GRH, the accumulated character sum has an estimate of square-root size, with logarithmic factors. Partial summation turns that estimate into convergence after division by $p^s$ for $s>1/2$.

The six-entry table tells us which estimates are needed. The analytic theorem supplies them under its hypothesis.

A zero that cannot hide

There is also a statement in the other direction.

Suppose we could prove that the ordinary collision sum converges at $s=0.7$. General Dirichlet-series theory would make its sum analytic everywhere to the right of that vertical line. It could have no singularity there.

Now imagine an active $L$-function with a zero whose real part is $0.8$. This is a hypothetical example. Its logarithm would produce a singularity inside the region where the collision sum is supposed to be analytic.

The only escape is exact cancellation with contributions from other active $L$-functions that vanish at the same point. If their zero multiplicities are $m_\chi$, the condition is

$$\sum_\chi c_\chi m_\chi=0.$$

A zero belonging to only one active primitive $L$-function cannot satisfy that condition. Its coefficient is nonzero, and its multiplicity is positive. It would obstruct convergence at $0.7$.

This is the part I find most interesting. The digit table supplies the coefficients in that cancellation equation. A statement about convergence of its prime sum becomes an exact restriction on the zeros of the functions it selects.

The restriction leaves work to do. Shared zeros could, in principle, cancel with the required weights. The paper does not locate the zeros or prove GRH.

A smaller split to examine

The finite computation leaves a more immediate arithmetic question too. Split the decimal window according to the prime’s actual remainder modulo three. At $s=1$, the two contributions are approximately

Prime remainder modulo 3	Window contribution
1	$+0.0002534203$
2	$-0.0001883766$
Combined	$+0.0000650437$

Their signs oppose each other at all six tested exponents. That opposition is a feature of this declared window. It does not establish a persistent bias in either class, or show that subtracting a further correction would destroy convergence. Those are questions about the infinite sums and the exact correction being applied.

For a nonzero centered table, the conjecture is that the boundary of ordinary convergence is exactly $1/2$. Even that statement would leave convergence on the boundary as a separate question.

The starting point was six fractions assigned to six remainders. Increasing the weight of large primes asks how strongly those fractions cancel along the prime sequence. The character expansion carries that question to particular $L$-functions. Their zeros can obstruct the sum, and the same six fractions determine whether the obstruction can cancel. That is a connection we can write down and prove, even while the location of the convergence boundary remains open.