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	$+0.7$

Four positive weights and one negative. The total is half.

The base is ten. Start long division from each nonzero remainder and count how often the first two digits agree. Subtract the baseline $\lfloor(p-1)/10\rfloor$, then subtract the mean deviation for the prime’s final-digit family. The result is the centered collision weight, $f(p\bmod100)$.

At 151 there are 14 matches. The baseline is 15, giving a deviation of $-1$. The family ending in 1 has mean $-17/10$. Subtract it and the weight is $7/10$.

The forty entries of the centered table add to zero. Our five primes do not. Nor should they. They visit only five entries.

The question is whether restricting the primes can bring back a persistent bias. Keep only those leaving one on division by thirty. Impose another remainder condition, then another. Does the cancellation eventually break?

For any collection of conditions modulo distinct primes, it survives. Each allowed group keeps whole pieces of the table that already balance. To see why, follow ten stops around a circle.

Add thirty. Go round once.

The collision table repeats after a hundred. The new condition repeats after thirty. Both repeat after three hundred.

Within that cycle, the numbers leaving one on division by thirty are

$$\begin{gathered} 1,\ 31,\ 61,\ 91,\ 121,\\ 151,\ 181,\ 211,\ 241,\ 271. \end{gathered}$$

Their last two digits visit $01,31,61,91,21,51,81,11,41,71$. Every entry ending in 1 appears once. Adding thirty changes the order, but it leaves no hole in the column.

Centering subtracts the column’s mean from each of its ten entries. The column therefore sums to zero, and so do these ten joint classes. They are residue classes, not ten consecutive primes.

The opening sample visits 51, 11, 41, 71 and 31. It misses five places, including 91 with its weight of $-7.3$. Nothing requires a short sample of primes to reproduce the balance of the whole column.

A whole column or the whole table

Now require remainder one modulo twenty-one. That imposes conditions modulo three and seven at once.

Twenty-one and a hundred are coprime. Fix any of the forty decimal endings and advance by hundreds. Exactly one of the next twenty-one positions satisfies the new condition. Every ending occurs once in the combined cycle of length $2{,}100$.

Modulo thirty, the condition keeps one complete column. Modulo twenty-one, it keeps the complete table. Both sums are zero.

The Chinese remainder argument works in every base. A prime dividing the base selects whole final-digit families. A prime not dividing the base adds an independent coordinate. Combining those conditions never cuts a centered family into unbalanced fragments.

Here is the general statement. At base $b$ and positive lag $\ell$, the centered table has modulus $m=b^{\ell+1}$. Let $Q$ be squarefree, let $r$ be coprime to $Q$, and put $M=\operatorname{lcm}(m,Q)$. Then

$$\sum_{\substack{1\le A<M\\(A,M)=1\\A\equiv r\pmod Q}} f(A\bmod m)=0.$$

Squarefree means that no prime factor occurs twice. That restriction matters. Modulo 25, the condition $A\equiv1$ keeps only endings 01 and 51 in the decimal table. Their weights are $1.7$ and $0.7$. The sum is $12/5$, not zero. The condition cuts through a column that centering balances only as a whole.

Open the congruence atlas, from forty entries to four thousand

The same three conditions act on the decimal tables at lags one, two and three. Conditions modulo 21 and 30 preserve complete balanced pieces at every lag. Modulo 25 provides the counterexample. Select the image to inspect every cell at full resolution.

Each group pays its own way

The finite balance tells us what happens when the weights are divided by the primes and added within one allowed group.

For each reduced class $A$ modulo $M$, Mertens’ theorem in arithmetic progressions gives

$$\begin{aligned} &\sum_{\substack{p\le X\\p\equiv A\pmod M}}\frac1p\\ &\qquad=\frac{\log\log X}{\varphi(M)}+c_A+o(1). \end{aligned}$$

Every class receives the same growing term. Multiply by its collision weight and add over the allowed classes. The coefficient of $\log\log X$ is the finite sum we have just proved to be zero.

The remaining constants have a finite weighted sum, and the finitely many error terms tend to zero. Thus the restricted collision sum converges at reciprocal weight. Starting beyond $m$ only removes finitely many terms.

Different groups may converge to different values. Neutrality does not say their sums vanish. It says each group cancels its own logarithmic drift.

Replace $1/p$ by $1/\sqrt p$ and the finite balance is unchanged. The convergence argument no longer applies. Larger primes now receive more weight, and the order in which they visit the table matters much more.

Four signs, one unexpected zero

Put a plus sign on odd numbers leaving one modulo four, and a minus sign on those leaving three. These signs multiply correctly, so they form the Dirichlet character $\chi_4$.

Multiply each collision weight by its sign. The four decimal columns now sum to $+9,-9,-9,+9$.

The total is exactly zero. Consequently

$$\widehat f(\chi_4) =\frac1{40}\sum_{a\in U_{100}}f(a)\chi_4(a)=0,$$

where $U_{100}$ consists of the forty endings coprime to a hundred.

This zero does not follow from either of the usual character restrictions. Reflection removes even characters, but $\chi_4(-1)=-1$. Centering removes characters that depend only on the final digit, but $\chi_4$ gives opposite signs to 1 and 11.

It needs a separate calculation. The calculation has only two places where anything can happen.

Twenty-five and seventy-five

Define a signed floor sum

$$B(k)=\sum_{a\in U_{100}}\chi_4(a) \left\lfloor\frac{ka}{100}\right\rfloor.$$

For $0\le k\le100$, its complete list of nonzero values is

$$B(25)=10,\qquad B(75)=-10.$$

All other values are zero. The proof is below. First see where the collision count asks this function for an answer.

Two adjacent decimal digits agree in the intervals for 00, 11, 22, and so on through 99. Numbering the hundred intervals from zero, their left endpoints are

$$\begin{aligned} G=\{&0,11,22,33,44,\\ &55,66,77,88,99\}. \end{aligned}$$

Counting fractions inside those intervals uses a floor at each end. After multiplying by the modulo-four signs and summing over the table, the raw collision deviation gives

$$-B(10)+\sum_{n\in G}\bigl(B(n+1)-B(n)\bigr).$$

The baseline supplies $B(10)$. The intervals supply $B(n)$ and $B(n+1)$. None asks for 25 or 75.

Every term is zero separately. Subtracting the family means changes nothing, since each column contains five plus signs and five minus signs. That proves the extra cancellation in the centered table.

Why the signed floor sum has only two nonzero values

The twenty positive-sign endings and the twenty negative-sign endings both sum to 1,000. Hence

$$\sum_{a\in U_{100}}\chi_4(a)=0, \qquad \sum_{a\in U_{100}}a\chi_4(a)=0.$$

Writing a floor as its argument minus its fractional part gives

$$B(k)=-\sum_{a\in U_{100}}\chi_4(a) \left\{\frac{ka}{100}\right\}.$$

If $k$ is coprime to 100, multiplication by $k$ permutes the forty endings. The last sum becomes $\chi_4(k)$ times the same signed fractional-part sum at $k=1$. It is zero by the two balances above.

If $k$ is odd and divisible by 5 but not 25, reduce to modulus 20. Each unit modulo 20 has five lifts in $U_{100}$, all with the same sign. The positive-sign units $1,9,13,17$ sum to 40. So do the negative-sign units $3,7,11,19$. The same permutation argument gives zero.

If $k$ is even, pair each ending $a<50$ with $a+50$. Their signs are opposite and their floors differ by the constant $k/2$. The signs among the twenty lower endings sum to zero, so the paired contributions cancel.

Only $k=25$ and $75$ remain. At 25, the positive-sign fractional parts are all $1/4$ and the negative-sign ones all $3/4$. Thus

$$B(25)=-20\left(\frac14-\frac34\right)=10.$$

At 75 the two fractional parts exchange places, giving $-10$. This is the finite floor-sum version of the sawtooth calculation in the companion paper.

The seventeen that remain

There are forty characters modulo a hundred. Reflection removes the twenty even ones. Two more depend only on the final digit and disappear under family centering. The modulo-four calculation removes one further character.

Seventeen coefficients are nonzero. They consist of eight conjugate pairs and one real coefficient. Their values come entirely from the forty-entry table.

For each active character, now form a prime sum

$$P_X(\sigma,\chi)=\sum_{100<p\le X}\frac{\chi(p)}{p^\sigma}.$$

Fourier inversion gives the exact finite identity

$$\begin{aligned} F_X(\sigma) &=\sum_{100<p\le X}\frac{f(p\bmod100)}{p^\sigma}\\ &=\sum_{\chi\ \mathrm{active}} \widehat f(\chi)P_X(\sigma,\chi). \end{aligned}$$

The left side adds weights directly. The right side adds the same weights in character coordinates. No infinite series is assumed in this equality.

Use every one of the $348{,}488$ primes between 100 and five million. At $\sigma=1/2$, ten of the seventeen contributions have positive real parts and seven have negative real parts. Their imaginary parts cancel in conjugate pairs. The net is approximately $+0.725464$.

The bars show contribution sizes, not merely the number of plus and minus signs. The three discarded odd characters are exact zeros before a single prime is added; they are not three more members of the seventeen.

Open the complex and magnitude views

The magnitude plot reveals another feature of this cutoff. Larger collision coefficients tend to be paired with smaller prime sums. Pearson’s correlation is about $-0.4192$. Conjugate partners have identical magnitudes, so this is a descriptive calculation on seventeen terms, not seventeen independent observations.

At exponents $1,0.8,0.6$, the correlations are also negative. That does not establish persistence as the cutoff grows, or a bound on the infinite sum.

Character labels and all four numerical checks

Every unit modulo 25 is a power of 2. Write $a\equiv2^{t(a)}\pmod{25}$. The labels in the figure mean

$$\chi_{e,j}(a)=\chi_4(a)^e e^{2\pi ijt(a)/20},$$

with $e=0,1$ and $j=0,\ldots,19$. Odd characters have $e+j$ odd. The family-constant zeros are $(0,5)$ and $(0,15)$; the additional zero is $(1,0)$. All other odd coefficients are nonzero, as exact reduction in twentieth roots of unity verifies.

$\sigma$	Signs $+/-$	Net	Correlation
$1$	10 / 7	$+0.077446$	$-0.4081$
$0.8$	12 / 5	$+0.185838$	$-0.5169$
$0.6$	10 / 7	$+0.444176$	$-0.4925$
$0.5$	10 / 7	$+0.725464$	$-0.4192$

All four rows use the same primes, $100<p\le5{,}000{,}000$. The three proved zeros are excluded from the positive and negative counts and from the correlations.

Three companion views of the calculation

The finite part is settled. Any squarefree collection of prime remainder conditions preserves the cancellation, and each resulting reciprocal-prime sum converges on its own. In decimal at lag one, the extra modulo-four zero also holds exactly.

At smaller exponents, the seventeen surviving terms still have to be controlled. Their mixed signs prevent a term-by-term positivity argument. They do not prove cancellation of the logarithmic singularities in the connection to $L$-functions.

Return to the circle. Adding thirty carries us through every place in ten steps. Keep only the primes and that tidy circuit disappears. A weight can recur while another waits hundreds of numbers for its turn.

Dividing by $p$ gives enough control to prove convergence. How much of that balance survives when we divide only by $\sqrt p$?