The General Neutrality Theorem

Take five primes that leave remainder one when divided by thirty.

Prime	Centered collision weight
151	$+0.7$
211	$+0.7$
241	$-2.3$
271	$+0.7$
331	$+0.7$

The weight measures the excess of digit matches after removing the baseline count and the average deviation for primes with the same last digit. We are working in base ten, at lag one.

These five weights add to $0.5$. Add more primes and the total changes. There is no claim that every sample balances.

The question is about a persistent bias. Centering removes the drift when we use all the primes. Could a restriction such as “remainder one modulo thirty” bring it back?

The general neutrality theorem says it cannot. The cancellation survives every nonzero prime remainder condition, and any finite collection of those conditions imposed together. The reason is visible in ten cells of the decimal table.

Ten places in the table

A centered collision weight depends on the last two digits. Our primes also have a specified remainder modulo thirty. Both pieces of information repeat after three hundred.

Within that span, the numbers leaving remainder one modulo thirty are

1, 31, 61, 91, 121, 151, 181, 211, 241, 271.

Read their last two digits. Every ending in the column $01,11,\ldots,91$ appears exactly once, in a different order.

That column already sums to zero. Centering was defined by subtracting the column’s mean from each entry. The additional remainder condition has kept the whole column.

The restriction can change which primes appear and how soon they appear. It has not changed the arithmetic average of the classes they are allowed to occupy.

This is also why the five primes at the opening need not balance. They have visited only part of the column, with some entries visited more than once.

More conditions, the same balance

Thirty shares factors with the decimal base. A different example shows the other possibility.

Ask for remainder one modulo twenty-one. That fixes a remainder modulo three and modulo seven simultaneously. Since twenty-one and one hundred have no common factor, every one of the forty decimal table entries is compatible with the condition.

For any last-two-digit ending, advance by hundreds. Exactly one of the twenty-one positions lands at remainder one modulo twenty-one. The whole table appears once in the combined cycle of length $2{,}100$.

So we have two concrete arrangements. Modulo thirty, the restriction keeps one complete column. Modulo twenty-one, it keeps the complete table. Both are balanced.

The proof follows this distinction in every base. An extra prime dividing the base selects whole centered columns. An extra prime not dividing the base gives a complete copy of the table. The Chinese remainder theorem joins several conditions without breaking those balanced pieces.

In the general statement, $Q$ is any product of distinct primes, and $r$ is a remainder coprime to $Q$. Among the compatible classes,

$$\sum_{\substack{A\ \mathrm{allowed}\\ A\equiv r\pmod Q}} f(A\bmod m)=0.$$

Here $f$ is the centered table and $m=b^{\ell+1}$ is its modulus at base $b$ and lag $\ell$. The classes $A$ run through one cycle long enough to carry both the table and the additional remainder condition.

This is the extension from neutrality at three. There is no preferred prime in the argument, and the conditions can be stacked.

The drift stays gone

The finite balance has a consequence for a sum over actual primes.

Weight each centered deviation by $1/p$. Then keep only primes in one of the specified groups, such as those leaving remainder one modulo thirty. That restricted sum converges.

The finite table alone does not prove this. Mertens’ theorem in arithmetic progressions supplies the analytic step. Each allowed class contributes the same leading $\log\log X$ growth to the sum of reciprocal primes up to $X$. Multiply those class sums by their collision weights and add. The coefficient of the common growth term is the sum of the weights.

It is zero.

The class-dependent constants remain, so different groups can converge to different values. Neutrality removes their steady drift independently. They do not have to borrow cancellation from another group.

This proves convergence with the weight $1/p$. Giving large primes more influence, by using $1/p^\sigma$ with $\sigma<1$, asks a further question. The finite balance still holds, but Mertens’ theorem no longer answers it.

An extra silence at four

There is another way to read the forty entries.

A Dirichlet character lays a multiplicative pattern of signs or complex phases across the residue classes. Averaging the table against that pattern gives its character coefficient. The coefficients are the table’s coordinates in this different basis.

One pattern is especially simple. Give $+1$ to odd numbers leaving remainder one modulo four, and $-1$ to those leaving remainder three. This is the character $\chi_4$.

Multiply each centered collision weight by its $\chi_4$ sign and add. In the decimal table, the result is exactly zero.

The four columns show the cancellation particularly well.

This silence needs a separate proof. Reflection removes the even characters, but $\chi_4$ is odd. Centering removes characters that depend only on the last digit, but $\chi_4$ distinguishes $1$ from $11$. Neither general rule has removed it.

The exact calculation uses the periodic sawtooth function associated with Bernoulli polynomials. It reduces the floor sums to a short list of possible nonzero terms, then shows that the decimal digit pattern samples none of them. The four bars display the conclusion of that calculation.

Of the twenty odd characters modulo one hundred, two disappear because they depend only on the last digit. This additional cancellation removes a third. Seventeen coefficients remain.

Seventeen contributions

Each remaining coefficient is paired with a prime sum carrying the same character.

The coefficient comes from the forty-entry collision table. The prime sum comes from adding the character’s values at primes, weighted by $1/p^\sigma$. Their product is one contribution to the centered collision sum.

For a finite cutoff $X$, Fourier inversion gives the exact identity

$$F_X(\sigma)=\sum_\chi \widehat f(\chi)\,P_X(\sigma,\chi).$$

The left side adds collision weights directly over primes. The right side computes the same answer through the character coordinates. Both are finite calculations.

I evaluated them using all $348{,}488$ primes between one hundred and five million. The real parts of the character contributions have both signs.

At $\sigma=1/2$, their net is approximately $+0.725464$. A positive total has emerged from contributions pointing in both directions.

There is a further numerical pattern. Among the seventeen surviving pairs, larger collision coefficients tend to meet smaller prime sums in magnitude. Pearson’s correlation is about $-0.4192$ at this cutoff. It is negative at the other three exponents in the figure too.

That is a relationship between two finite lists. Its persistence as the cutoff grows remains to be established.

The question after neutrality

The mixed signs prevent a simple positivity argument. A bound on the complete sum does not automatically bound each character contribution. Several terms can offset one another.

This does not settle whether the transform could constrain zeros of $L$-functions. In particular, finite mixed signs do not prove that singularities cancel, or that the weighting avoids directions sensitive to zeros. Those conclusions would need an analytic argument.

What is established is more concrete. We can impose several prime remainder conditions and prove that each resulting group has no steady logarithmic drift. We can also identify a character coefficient that vanishes for an additional arithmetic reason.

After those cancellations, the surviving terms are still there. Their signs differ. Their magnitudes are paired in a way the finite data invite us to examine.

The five primes at the opening belong to a group whose steady drift we can now account for. The theorem continues to hold as we add remainder conditions. That part of the question has an answer.

The weights are fixed by forty entries. The prime sums keep changing. Understanding how those fixed weights act on the growing sums is the arithmetic question I want to pursue.