The Neutrality Theorem

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

Now divide each by three.

$$107=3\cdot35+2,\qquad 307=3\cdot102+1.$$

One leaves two. The other leaves one. They belong to different groups, although the collision table cannot tell them apart.

That is the question I want to follow here. The centered collision table balances exactly. Its entries add to zero. If I keep only the primes that leave one on division by three, does that balance survive? Or was it relying on the primes I have just removed?

It survives in each group separately. To see why, we have to keep track of something the last two digits leave out.

The hundreds place returns

The collision count begins with long division. Start from each possible nonzero remainder and count the starts at which the next two decimal digits agree. Subtract the whole-number baseline $\lfloor(p-1)/10\rfloor$. The result is the collision deviation.

For primes above 100, this deviation depends only on the last two digits. At 107 the count is 10 and the baseline is 10. At 307 both are 30. The deviation is zero in either case.

Centering takes one further step. Within the table, the ten entries ending in 7 have mean $-1/10$. Subtract that mean, and the entry 07 has centered weight $+1/10$. That is the weight assigned to both of our primes.

But 07 does not determine the remainder on division by three. Write an integer as

$$p=100t+a,$$

where $a$ holds the last two digits. A hundred leaves one on division by three, so the remainder of $p$ is the remainder of $t+a$. The hundreds contribute too.

Keep 07 and add a hundred at a time. The remainders cycle through all three possibilities. After three steps they return to the start. The collision weight stays put throughout.

Two ways of sorting the same integers are crossing here. One reads their final two digits. The other reads their remainder on division by three. We need a table large enough to retain both.

Forty entries in each group

Use a block of 300. For any permitted ending $a$, the three representatives are

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

Exactly one leaves one on division by three. Exactly one leaves two. The third is divisible by three and cannot represent a prime greater than three.

This happens for every one of the forty endings coprime to ten. Select remainder one, and all forty collision entries are still present, once each. Select remainder two, and all forty are present again.

The diagrams show the allowed residue classes, not a claim that forty consecutive primes visit them equally often. For example, the representative 7 stands for integers $7,307,607,\ldots$. Some are prime and some are not. Every prime above 300 lies in one of the eighty allowed classes.

The centered table already sums to zero. Each group contains a complete copy, so each group sums to zero on its own. This is independent neutrality.

The counting argument is the Chinese remainder theorem in a form we can see. Because 100 and 3 have no common factor, specifying an ending modulo 100 and a remainder modulo three picks out exactly one class modulo 300.

The same argument works in any base not divisible by three. At a fixed lag, replace 100 by the appropriate power of the base. Each true nonzero remainder class modulo three still contains one copy of the whole collision table.

The group the mirror keeps

There is another exact cancellation inside the original forty-cell table.

Pair an ending $a$ with $100-a$. Before centering, their collision deviations add to $-1$. After centering, their weights are opposites.

Look at 11 and 89. Both leave two on division by three, and reflection exchanges them. The same is true of 17 and 83, or 23 and 77. Every table entry that leaves two has its reflection partner in that same group. Its centered sum is therefore zero. Its raw mean is $-1/2$.

The other two internal groups exchange partners. For example, 03 leaves zero and its partner 97 leaves one.

That reflection explains the neutral group inside the table. The prime classes require the larger block of 300. The distinction is easy to miss because both constructions use division by three. Our opening primes show why it has to be kept. Their common table entry 07 leaves one, while 107 itself leaves two.

Three as the base

The argument using three copies cannot work when three divides the base. There is no independent remainder coordinate to add. The centering itself supplies the cancellation.

Base three gives a small enough example to write out completely. At lag one the table is modulo nine. Its six allowed entries are $1,2,4,5,7,8$.

The entries $1,4,7$ all leave one on division by three. Their raw values are $0,0,-2$, with mean $-2/3$. Subtract the mean and they become

$$\frac23,\qquad\frac23,\qquad-\frac43.$$

They add to zero.

The entries $2,5,8$ have raw values $1,-1,-1$, with mean $-1/3$. Their centered values are

$$\frac43,\qquad-\frac23,\qquad-\frac23.$$

They add to zero too.

In a larger base divisible by three, a true remainder class collects several complete centering groups. Each already sums to zero. Collecting them preserves that sum.

So the theorem covers every base and every positive lag. Either a prime class contains the whole centered table, or it is assembled from complete groups that were centered separately.

Adding over the primes

A finite table with total zero need not give a zero sum over a finite stretch of primes. The primes visit its entries at different times, and division by $p$ gives each visit a different weight.

For the decimal primes between one million and ten million, the calculation gives

Remainder on division by three	Number of primes	Sum of centered weight divided by $p$
One	292,963	$+0.0002534$
Two	293,118	$-0.0001884$
Both	586,081	$+0.0000650$

These are contributions from one stated window. They show a positive imbalance in one group and a negative imbalance in the other. They are not the values of infinite sums.

There is, however, a convergence theorem. Mertens’ theorem for arithmetic progressions gives the sum of reciprocal primes in each allowed class as the same growing term, plus a class-dependent constant and an error tending to zero. In the decimal block of 300, that common term is $\log\log X/80$.

Multiply each class by its collision weight. Within either remainder group, the weights add to zero, so their copies of the growing term cancel. A finite sum of constants remains, with an error that vanishes.

Each of the two reciprocal-prime sums converges separately. They can converge to different numbers. Neither needs the other group to remove its logarithmic drift. The analytic step is the classical Mertens theorem in arithmetic progressions. Independent neutrality supplies its zero coefficient here.

The difference survives

Keep a running total for each group. Call their sum $F$ and their difference $G$. Then the two original totals are

$$F_1=\frac{F+G}{2},\qquad F_2=\frac{F-G}{2}.$$

Computing $G$ means retaining the signs of the remainder-one terms and reversing the signs of the remainder-two terms. That sign rule is the nontrivial Dirichlet character modulo three. The difference is a character twist of the collision sum.

It remains visible after the predictable drift has been removed. In the window above, the total is about $0.0000650$, while the difference is about $0.0004418$. Combining the groups hides much of their separation.

At reciprocal weight $1/p$, both totals and their difference converge. Give large primes more weight by using $1/p^s$ with $s<1$, and this argument no longer settles convergence. The finite table still balances. Controlling how the primes accumulate its positive and negative entries is another question.

I began with two primes that the collision table treats alike. Sorting them by division by three sent them in different directions, yet the balance survived the split. Each group kept enough of the table to cancel its own drift. What remains to understand is their difference, the part the combined sum can make difficult to see.