The Neutrality Theorem

Take 107 and 307. Both are prime. Both end in 07. The collision table assigns them the same weight.

Now divide by three.

$$\begin{aligned} 107&=3\cdot35+2,\\ 307&=3\cdot102+1. \end{aligned}$$

One leaves two. The other leaves one. Sorting the primes this way separates two numbers that the collision table treats alike.

The centered collision table balances exactly. Its forty entries add to zero. Keep only the primes that leave one on division by three. Does the balance survive, or does it depend on the primes we have just thrown out?

It survives. Either group has enough arithmetic of its own to cancel its drift. That does not make its running sum zero. It means we can prove convergence at reciprocal weight without borrowing any cancellation from the other group.

The reason is in the hundreds.

The hundreds place returns

Start from each nonzero remainder in long division by $p$. Count the starts at which the next two decimal digits agree, then subtract $\lfloor(p-1)/10\rfloor$. This is the collision deviation. For primes above 100, the last two digits determine it.

At 107 the count and the amount subtracted are both 10. At 307 both are 30. The deviation is zero either way.

Centering subtracts one more quantity, the mean of the table entries with the same final digit. The ten entries ending in 7 have mean $-1/10$, so 07 has centered weight

$$f(07)=0-\left(-\frac1{10}\right)=\frac1{10}.$$

Both primes receive that weight. But their common ending does not determine their remainder on division by three. Write

$$p=100t+a.$$

Here $a$ holds the last two digits. Since a hundred leaves one on division by three,

$$p\equiv t+a\pmod3.$$

The quotient $t$ still matters. Keep the ending 07 and add a hundred at a time.

Integer	Centered weight	Remainder on division by 3
107	$1/10$	2
207	$1/10$	0
307	$1/10$	1

The middle number is composite, $207=3\cdot69$. The weight is defined there too. Primality has nothing to do with this little cycle.

Forty strands through three hundred

Use a block of 300. For each permitted ending $a$, put down

$$a,\qquad a+100,\qquad a+200.$$

One leaves one on division by three, one leaves two, and one is divisible by three. Each carries the same collision weight, $f(a)$.

Do this for all forty endings coprime to ten. Select remainder one. Every ending is still there, exactly once. Select remainder two. Every ending is there again.

Follow a strand from its two-digit ending to either side. The representative changes; the weight does not. The two outer lists each contain all forty weights, whose sum is zero. This is independent neutrality.

The lists contain residue classes, not forty consecutive primes. The representative 007 on the left stands for $7,307,607,\ldots$. Some members are prime and some are not. Every prime above 300 falls into one of these eighty classes.

The Chinese remainder theorem supplies the exact count. Since 100 and 3 are coprime, an ending modulo 100 and a remainder modulo three specify one class modulo 300. There are no missing endings and no duplicates.

At base $b$ and lag $\ell$, the collision table has modulus $m=b^{\ell+1}$. If three does not divide the base, the same three-copy construction works. Each of the two prime groups receives the whole centered table.

Three is not special to this part of the argument. For any auxiliary modulus $q$ coprime to $m$, fixing a remainder coprime to $q$ leaves one complete copy of the table on the joint modulus $mq$. Its weights still sum to zero. A modulus sharing factors with the base needs a separate argument.

Where the mirror sends an ending

Now return to the original forty cells, numbered below 100. There is another cancellation here, and it is easy to mistake it for the one we have just proved.

Reflection pairs $a$ with $100-a$. Their raw deviations add to $-1$; their centered weights are opposites.

$$f(100-a)=-f(a).$$

The endings 11 and 89 both leave two on division by three. So do 17 and 83, or 23 and 77. Reflection keeps this group together because

$$100-a\equiv1-a\pmod3.$$

Put in two, get two back. Put in zero, get one. Put in one, get zero.

The twelve endings in the remainder-two group cancel in six pairs. The other groups exchange partners. Their centered sums are $91/5$ and $-91/5$, not zero separately.

These are groups of endings. They are not the two groups of primes. The entry 07 belongs to the internal remainder-one group, yet the prime 107 leaves two. Moving to the block of 300 is what keeps the prime’s remainder instead of silently substituting its ending’s remainder.

When three divides the base

In base three, adding a power of the base cannot change a remainder modulo three. There are no three independent copies to sort. We have to use the centering itself.

At lag one the table is modulo nine. All six entries fit here.

Ending	Raw value	Family mean	Centered weight
1	0	$-2/3$	$2/3$
4	0	$-2/3$	$2/3$
7	$-2$	$-2/3$	$-4/3$
2	1	$-1/3$	$4/3$
5	$-1$	$-1/3$	$-2/3$
8	$-1$	$-1/3$	$-2/3$

The first three entries all leave one on division by three. The last three all leave two. Each family was centered separately, so each sum is zero before we do anything else.

The same reasoning works in a larger base divisible by three. In base twelve, for instance, the permissible final digits are $1,5,7,11$. Remainder one collects the complete families ending in 1 and 7. Remainder two collects those ending in 5 and 11. Every family already balances.

This covers every base and every positive lag. If three does not divide the base, each prime group contains a whole copy of the table. If three divides the base, it contains whole families that were centered separately. Neither operation cuts through the cancellation it needs.

Inspect the quotient layers at lags one, two and three

The decimal tables grow from forty to four hundred to four thousand entries. Each has three possible quotient layers. Selecting a true remainder modulo three keeps exactly one occurrence of every weight. The gray cells are the occurrences that selection removes.

What the primes do with the table

A complete table balances. A stretch of primes need not. They visit the entries at different times, and division by $p$ gives each visit a different size.

Take the same window of primes used to examine the collision transform, $10^6<p\le10^7$. Keep the two remainders separate.

Remainder modulo 3	Primes in the window	Sum of $f(p\bmod100)/p$
1	292,963	$+0.0002534$
2	293,118	$-0.0001884$
Both	586,081	$+0.0000650$

These are finite contributions, not values of infinite sums. The opposite signs do not contradict neutrality. Neutrality says which growing term must disappear; it does not prescribe the sign of what remains.

The passage to an infinite sum uses the classical Mertens theorem for arithmetic progressions. Each allowed class $A$ modulo 300 has

$$\begin{aligned} &\sum_{\substack{p\le X\\p\equiv A\ (\mathrm{mod}\ 300)}}\frac1p\\ &\quad=\frac{\log\log X}{80}+c_A+o(1). \end{aligned}$$

The constant $c_A$ depends on the class. The growing term does not.

Multiply by the collision weight of each class and add over just one remainder group. Its coefficient of $\log\log X$ is

$$\frac1{80}\sum_{a}f(a)=0.$$

Only forty weighted constants and an error tending to zero remain. Repeat for the other group. Both reciprocal-prime sums converge separately. Their limits need not be equal, and neither has to be zero. Omitting a fixed initial set of small primes changes a limit but not convergence.

The finite proof supplies the zero coefficient. Mertens’ theorem supplies the analytic conclusion. There is no assumption here that a finite sample of primes distributes itself evenly among the classes.

Three complementary views of the cancellation

These diagrams keep the small coordinate calculation, the full forty-weight table and the six-entry base-three example together. The first follows an integer as its hundreds change. The other two show the weights that the routing and family walks preserve.

Subtract the two ledgers

Call the two running totals $F_1$ and $F_2$. Keep their sum $F$ and their difference $G$.

$$\begin{aligned} F&=F_1+F_2,\\ G&=F_1-F_2. \end{aligned}$$

We can recover either ledger as $(F+G)/2$ or $(F-G)/2$. To compute $G$ directly, keep the signs of terms from primes leaving one and reverse the signs of terms from primes leaving two. That is the nontrivial Dirichlet character modulo three. Its twist measures a difference that the combined total can conceal.

In this window, the sum is about $0.0000650$. The difference is about $0.0004418$. Neither group needs the other to cancel its drift, yet their separation is nearly seven times the size of the total.

At weight $1/p$, both ledgers settle and so does their difference. Replace $p$ by $p^s$ with $s<1$, and the finite cancellation alone no longer proves convergence. The unresolved work is in how the primes deliver the positive and negative weights.

Return to 107 and 307. Each receives $1/10$ from the table. In the difference, 107 contributes $-1/1070$ and 307 contributes $+1/3070$. Equal table weights have become unequal terms with opposite signs. We can prove that the full reciprocal sum of such terms settles. Put a square root in the denominator, and how far can the difference wander?