The Collision Transform and the Critical Strip

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

The digits and the collision count stay put. Only the divisor changes.

Do this at every prime. 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 positive and negative terms now have more to cancel.

The Centered Collision Sum proves convergence with the divisor $p$. How much more weight can the large primes carry before the sum fails to converge?

Six numbers to start with

In base three, compare the first two digits of every fraction $r/p$. Count the matches, subtract the bin scale $\lfloor(p-1)/3\rfloor$, then subtract the mean for the prime’s family modulo three. Beyond nine, the answer depends only on the remainder modulo nine. The whole centered table fits on one row.

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

Eleven takes $4/3$. Thirteen takes $2/3$. Seventeen takes $-2/3$. Every larger prime consults the same six entries.

Write $f(p)$ for that weight. The sum is

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

with primes added in increasing order. At $s=1$, convergence is proved. Decrease the real exponent $s$ and more of each large prime’s contribution remains.

The six entries sum to zero. That alone will not settle the question. The primes visit these entries in their own order, and convergence depends on the imbalance left by those visits.

The next nine million

Take every prime in $10^6<p\leq10^7$. There are $586{,}081$. Start a fresh total at zero and add their contributions one at a time.

At $s=1$, the base-three path ends near $+0.000042$. At $s=1/2$, it ends near $+0.077496$. Base ten ends near $+0.000065$ and $-0.366499$, respectively. Each base uses its own centered table. Within a base, the same primes and the same table entries produce both paths.

The endpoint conceals a fair amount of travel. A window can finish near zero after substantial excursions on either side. Increasing the weights can also reverse the sign of its final contribution, as it does here in base ten.

These are finite sums. A small contribution from this window does not prove convergence. A larger one does not disprove it. The next window is not obliged to resemble this one.

Five bases, three exponents, every prime in the window

The full-size plate uses bases $3,5,7,10,12$ and exponents $1,3/4,1/2$. Every prime contributes to every path. Each panel has its own marked vertical scale. Enlarging a nearly flat path changes its display, not the size of the underlying sum.

The decimal window can also be split by the prime’s actual remainder modulo three. At $s=1$, its $292{,}963$ primes congruent to one contribute about $+0.0002534203$. The $293{,}118$ congruent to two contribute about $-0.0001883766$. Together they give $+0.0000650437$. The signs oppose each other at all six exponents checked in the companion paper. This is an observation about the declared window, not a theorem about an enduring bias.

Two turns around nine

Multiply repeatedly by two modulo nine. The remainders run through

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

A Dirichlet character assigns complex weights that multiply as the remainders do. Choose $\chi(2)=e^{\pi i/3}$. Each multiplication by two then turns its arrow through sixty degrees. Six turns bring it home. The conjugate character turns the other way.

Of the six characters on this cycle, only these two occur in the centered table. Reflection eliminates the even characters. Family centering eliminates the remaining odd character that depends only on the remainder modulo three. The exact expansion is

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

where

$$c=\frac{1-i\sqrt3}{3}.$$

At remainder one, the arrows are $c$ and $\overline c$. Their vertical parts cancel and their horizontal parts add to $2/3$. At remainder two, both arrows point along the positive real axis. Their sum is $4/3$. Continuing around the cycle recovers all six fractions exactly.

We can therefore split $F(s)$ into two prime sums, one weighted by $\chi$, the other by $\overline\chi$. In another base, or at another lag, a different finite table may select different characters. A character with a nonzero coefficient is called active.

The prime sum inside the logarithm

Each character has a Dirichlet $L$-function. Its Euler product has one factor for each prime. Take the logarithm and expand each factor using $-\log(1-z)=z+z^2/2+z^3/3+\cdots$.

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

The first sum is the one we need. The second brings along the squares, cubes and higher powers of each prime. Initially this identity holds for $\operatorname{Re}(s)>1$, where the sums converge absolutely.

Subtract the higher powers and the finitely many small primes excluded from the collision sum. For any fixed base and lag, we obtain

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

Here $R$ collects those corrections. The digit table supplies the coefficients $c_\chi$.

The higher powers start at $p^{-2s}$. Their absolute convergence requires $2\operatorname{Re}(s)>1$. Thus $R$ is analytic throughout $\operatorname{Re}(s)>1/2$, with no singularity to contribute there. Any singularity of the continued expression in this region must come from the selected logarithms.

On suitable zero-free domains, we can choose logarithms and continue the formula. That does not yet prove that adding the original prime terms converges there. Analytic continuation and ordinary convergence are different claims.

Moving the line left

Let $s$ be complex. Its real part gives the horizontal coordinate. The critical strip lies between real parts zero and one; the critical line passes through $1/2$.

The Generalized Riemann Hypothesis places the nontrivial zeros of primitive Dirichlet $L$-functions on that line. Assume GRH for the primitive $L$-functions selected by our table. Then the ordinary collision series converges, locally uniformly, throughout

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

This conditional theorem concerns the original prime sum, not just a continuation of its formula. It does not include the line itself.

Under the hypothesis, the accumulated character values along the primes have square-root bounds, with logarithmic factors. Partial summation combines that cancellation with the divisor $p^s$. For a real exponent $s=1/2+\varepsilon$, the resulting tail is bounded by an integral of the form

$$\int_X^\infty\frac{(\log t)^A}{t^{1+\varepsilon}}\,dt,$$

for a fixed logarithmic exponent $A$. Every positive $\varepsilon$ makes this integral finite and its tail tend to zero. At $\varepsilon=0$, the same bound no longer works. The companion paper gives the explicit-formula estimate and the partial-summation argument.

For a nonzero centered table, let $\sigma_c(F)$ denote the boundary of ordinary convergence. The conjecture is

$$\boxed{\sigma_c(F)=\frac12.}$$

There are two inequalities to prove. GRH for the selected functions gives $\sigma_c(F)\leq1/2$. Showing that the boundary cannot lie farther left is a separate problem. Even equality would leave the behavior on the boundary line undecided. Neither the finite paths above nor the conditional convergence theorem supplies that missing lower bound.

A zero that cannot hide

Now suppose we could prove ordinary convergence at $s=0.7$. General Dirichlet-series theory would make the sum analytic everywhere to the right of that line.

Put a hypothetical zero of an active $L$-function at $\rho=0.8+i\gamma$. Its logarithm has a singularity inside the supposed region of analyticity. The correction $R$ cannot remove it. Cancellation would have to come from other active $L$-functions vanishing at that same point.

Let $m_\chi$ be the multiplicity of each such zero, or zero if that function does not vanish at $\rho$. The necessary condition is

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

One way to see the condition is to differentiate the logarithms. A zero of multiplicity $m$ contributes $m/(s-\rho)$ to $L'/L$, plus a regular term. An analytic sum cannot retain that pole, so its total coefficient must be zero.

A zero belonging to just one active primitive $L$-function cannot cancel. In larger active sets, shared zeros could cancel only with exactly the required weights. This is a restriction on possible zeros, not a theorem locating them or proving GRH.

The finite window, character powers and theorem domains

These views place the six tested exponents, all six character powers and the two theorem directions side by side with the running paths and coefficient geometry above.

In base three we can go further. There are only two active functions. At a fixed point $\rho$, let their multiplicities be $m$ and $n$. Their weighted sum is

$$\begin{aligned} mc+n\overline c &=\frac{m+n}{3}\\ &\quad+i\frac{\sqrt3(n-m)}{3}. \end{aligned}$$

Its real part is $(m+n)/3$. Multiplicities are nonnegative integers. If either function vanishes, that real part is positive. Even a shared zero cannot cancel in this example.

Conjugating the character reflects a zero across the real axis. It does not automatically put a second zero at the original point. The calculation above allows either possibility. It needs no assumption that the two functions have different zeros.

At a shared simple zero, the vertical parts cancel and the horizontal parts give $2/3$. Making either zero multiple only pushes the point farther to the right. The cancellation equation asks for the origin.