The Centered Collision Sum

Take the primes $1009$ and $1109$. Both end in $09$.

For each denominator, write out all the fractions between zero and one. Compare the first and third decimal places of every row. Count one collision whenever those two digits agree. A beginning such as $0.171\ldots$ counts. A beginning such as $0.173\ldots$ does not. The middle digit is free.

This is a lag-two comparison. Moving two places along a decimal expansion brings the third digit under the first.

At $1009$, there are $108$ collisions. At $1109$, there are $100$. Subtract the smaller digit-bin size to put the two counts on the same footing,

Prime	Collision count	Smaller bin size	Difference
1009	108	100	+8
1109	100	110	−10

The same final two digits, but opposite signs.

At lag one, two digits were enough to determine this difference. At lag two they are not. We need the last three. The addresses of these primes are $009$ and $109$, and they occupy different places in a larger finite table.

I want to understand what remains after the family bias is removed. The Collision Fluctuation Sum found the global drift at lag one and showed how to subtract it. Here the table extends to every fixed lag. Centering each family separately reveals more than convergence. It determines which arithmetic patterns can remain in the sum.

One more digit

Keep the denominator $p$ coprime to the base $b$. The smaller digit bins contain $\lfloor(p-1)/b\rfloor$ remainders. Call the collision count minus this quantity $S(p)$, with the base and lag understood.

The count looks farther into the expansion as the lag grows,

Lag	Places compared	Decimal table address
1	First and second	Last two digits of $p$
2	First and third	Last three digits of $p$
3	First and fourth	Last four digits of $p$

At lag $\ell$, the table modulus is

$$q=b^{\ell+1}.$$

For $p>q$, the remainder of $p$ on division by $q$ determines $S(p)$ exactly. The denominator can be prime or composite. The finite counting argument needs only coprimality with the base.

There is a simple reason for the growing address. At decimal lag two, a matching three-digit beginning has equal first and last digits. The middle digit can be anything. There are one hundred such beginnings, each occupying an interval of width $1/1000$ between zero and one.

Increase the denominator by one thousand. Each matching interval gains one fraction, so the collision count rises by one hundred. The smaller bin size rises by one hundred too. Their difference is unchanged. The proof accounts for the excluded endpoints, but this is the cancellation behind the table.

At any fixed lag, the same counting works. The table grows with the distance between the digits being compared. It does not grow with the primes that read it.

Subtract a column

Return to decimal lag one. The forty possible two-digit endings fall into four columns, according to the last digit. Average the ten entries in each column.

Last digit	Exact column mean of $S$
1	$-17/10$
3	$-9/10$
7	$-1/10$
9	$+7/10$

These are averages of finite lists of integers. No primes have to be sampled to obtain them.

The distinction between $S$ and the constructive-mean deviation $\Delta$ is useful here. The latter subtracts a slightly larger benchmark. These four numbers center the integer table itself, before that additional bin correction.

All four columns together have mean $-1/2$. Adding $1/2$ to every entry centers the table as a whole, but the columns still have different means. To center each column, subtract its own mean from each of its entries.

A smaller base lets us see the whole operation at once. In base three, lag one uses a table modulo nine. Its six allowed addresses are $1,2,4,5,7,8$. Group them by their remainder on division by three.

The first column contains the values $0,0,-2$. Its mean is $-2/3$, so subtracting the mean gives $2/3,2/3,-4/3$. They add to zero.

The other column contains $1,-1,-1$. Its mean is $-1/3$. After centering, its entries add to zero too.

There is another balance in the picture. Pair each address with its reflection around nine. The centered entries at $1$ and $8$ are opposites. So are those at $2$ and $7$, and at $4$ and $5$.

Both properties hold in every base and at every lag. Each family sums to zero. Reflection changes the sign of every centered entry.

For prime bases at lag one, the column means have a particularly short formula. If the column has remainder $s$, then

$$\mu(s)=\frac{s}{b}-1.$$

This is proved by summing the floor counts. In base seven the six means run from $-6/7$ to $-1/7$, evenly spaced. Composite bases still have exact finite means, but they need not follow that spacing. The decimal table above does not.

The patterns that disappear

The two balances tell us something about frequency before we sum over a single prime.

Imagine reading a table with a pattern of weights. Multiply each entry by its weight, then add. If the weights are constant within each column, the result is zero. Each column already sums to zero, and multiplying that zero by a fixed weight changes nothing.

A different cancellation happens when the weights agree at reflected addresses. The entries at $a$ and $q-a$ have opposite signs, so their weighted contributions cancel in pairs.

Dirichlet characters provide a complete set of such arithmetic patterns. Their weights respect multiplication of residue classes. Some characters give the same weight to reflected addresses. These are the even characters. Some can read only the remainder modulo the base, so they are constant within each family.

The centered table has zero coefficient in every one of those patterns. Only odd characters that distinguish entries within a family can survive. That is the spectral gate. It gives necessary conditions for survival, not a promise that every permitted coefficient is nonzero.

The six-entry example makes the distinction between global and family centering visible. Arrange its addresses by repeatedly multiplying by two modulo nine,

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

The six characters are the six Fourier modes around this cycle. The bars below show the squared magnitude of each normalized coefficient.

Mode zero assigns the same weight everywhere. Mode three assigns one sign to the first column and the opposite sign to the second. Family centering removes both. The two remaining modes vary inside the columns, where the individual entries still differ.

This is the extra information obtained from centering locally. We know which part of the table has been removed, and where to look for what remains.

Let the primes read the table

Now put the primes back in.

For decimal lag one, take each prime greater than one hundred. Read its two-digit table entry, subtract the mean belonging to its final digit, divide by the prime, and add. Writing the centered entry as $S^\circ(p)$, the sum is

$$H(x)=\sum_{100<p\leq x}\frac{S^\circ(p)}{p},$$

where the sum runs over primes. The computation gives the following values.

Primes included	Largest prime	Centered sum
1,000	8,167	0.081643173
10,000	105,019	0.077004672
100,000	1,300,051	0.077300691
664,554	9,999,991	0.077215575

The last entries are close, but they are still moving. None is being asserted as the exact limit.

The convergence proof begins with the finite character expansion. The centered table is a finite sum of character patterns, and its constant component is zero. For each remaining character $\chi$, the reciprocal-prime sum

$$\sum_{p\leq x}\frac{\chi(p)}p$$

converges as $x$ grows. This is a classical consequence of Mertens’ theorem in arithmetic progressions. Each allowed residue class receives the same leading share of the prime harmonic sum. The character weights cancel that common growth, leaving a constant and a term tending to zero.

There are only finitely many characters. The centered collision sum is a finite combination of their convergent sums. It therefore converges too.

The argument works for every fixed base and positive lag, starting beyond $b^{\ell+1}$. The limit need not be zero. There is also an unweighted statement. Average the centered entries over the primes up to $x$, without dividing each entry by its prime, and that average tends to zero. This uses the classical prime number theorem in arithmetic progressions.

The finite arithmetic supplies the table and its cancellations. The classical theorems supply the distribution of primes among its addresses.

Return to $1009$ and $1109$. Their lag-two deviations differ by eighteen. Since they belong to the same final-digit family, subtracting its mean leaves that difference exactly eighteen.

I find that a useful way to read the result. Centering has not made the primes indistinguishable. It has removed the average attached to each family while preserving the differences inside it. Those differences have a finite address, a restricted character expansion, and a convergent reciprocal-prime sum.

We can let the primes run indefinitely. The table tells us what each one is carrying.