The Centered Collision Sum

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

Write all the fractions between zero and one for each denominator. Compare the first and third decimal places of every row. A beginning such as $0.171\ldots$ counts as a collision. A beginning such as $0.173\ldots$ does not. The middle digit is free.

At $1009$, there are $108$ collisions. At $1109$, there are $100$. Subtract the smaller digit-bin size from each count.

Prime	Collisions	Bin size	Difference
1009	108	100	+8
1109	100	110	−10

The same final two digits. Opposite signs.

At lag one, the last two digits determine this reduced count. At lag two, we need three. The addresses are $009$ and $109$, and they occupy different places in a larger table.

Their shared last digit also puts them in the same family. That family has an exact mean. Subtract it from both entries and their eighteen-point difference survives untouched.

The Collision Fluctuation Sum removes one global drift. Here each family loses its own bias. The subtraction does more than balance the columns. It rules out whole patterns of variation, leaving the differences within a family a restricted set of ways to express themselves.

One more digit

Keep the denominator $p$ coprime to the base $b$. Call the collision count $C(p)$ and the smaller bin size $Q=\lfloor(p-1)/b\rfloor$. Their difference is

$$S(p)=C(p)-Q.$$

The base and lag are fixed while we vary $p$.

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

At lag $\ell$, the table modulus is $q=b^{\ell+1}$. Once $p>q$, the remainder of $p$ on division by $q$ determines $S(p)$ exactly. Primality is not needed for this finite statement. A composite denominator coprime to the base reads the same table.

The hundred cells in the opening figure explain why. Each cell is a matching three-digit beginning, an interval of width $1/1000$ between zero and one. Increase the denominator by one thousand. Each interval gains one fraction. The collision count rises by one hundred, and so does the smaller bin size. Their difference stays put.

The excluded endpoints require a little bookkeeping. They do not disturb the cancellation.

The exact count behind the cells

Let $\mathcal G$ contain the integers $n$ from $0$ through $q-1$ whose first and last base-$b$ digits agree, allowing leading zeros. There are $b^\ell$ such integers. The floor contribution from interval $n$ is

$$B_n(t)=\left\lfloor\frac{(n+1)t}{q}\right\rfloor -\left\lfloor\frac{nt}{q}\right\rfloor.$$

Summing the matching intervals gives

$$C(p)=-1+\sum_{n\in\mathcal G}B_n(p).$$

The minus one removes the endpoint contribution. Coprimality prevents an interior fraction from lying on an interval boundary.

Replacing $p$ by $p+q$ adds one to every summand. Both $C$ and $Q$ rise by $b^\ell$, proving the periodicity of $S$. For an allowed address $a$, the table entry is

$$T(a)=-1-\left\lfloor\frac ab\right\rfloor +\sum_{n\in\mathcal G}B_n(a).$$

Thus $S(p)=T(p\bmod q)$ in the stated range. The count includes every numerator from $1$ through $p-1$, even when $p$ is composite.

The address grows with the distance between the digits. It does not grow with the primes that read it.

Move the whole column

Return to decimal lag one. The forty allowed two-digit endings fall into four columns, according to their final digit. The forty-entry table gives their means directly.

Last digit	Mean of the ten entries
1	$-17/10$
3	$-9/10$
7	$-1/10$
9	$+7/10$

No prime sampling enters this calculation. These are averages of four finite lists.

All forty entries together have mean $-1/2$. Adding $1/2$ everywhere balances the table, but leaves the four columns with different biases. Subtracting each column’s own mean balances all four.

These means belong to $S=C-Q$. The constructive-mean deviation $\Delta$ uses a slightly larger benchmark. Its global drift is $-9/10$, not $-1/2$. We are centering the integer table here, before that additional bin correction.

Base three puts the whole operation within reach of six numbers. At lag one the addresses are $1,2,4,5,7,8$. Group them by their remainder modulo three.

Address	Raw entry	Centered entry
1	0	$2/3$
4	0	$2/3$
7	−2	$-4/3$
2	1	$4/3$
5	−1	$-2/3$
8	−1	$-2/3$

Every point in the first family moves up by two thirds. Every point in the second moves up by one third. Both families now sum to zero. No distance between members of the same family changes.

Pair the addresses around nine. The centered entries at $1$ and $8$ are opposites. So are those at $2$ and $7$, and those at $4$ and $5$.

These are the two balances we need. Every family sums to zero. Reflection changes the sign.

The two balances in any base and lag

For an allowed last digit $u$, average the $b^\ell$ entries with $a\equiv u\pmod b$. Write that mean as $\mu(u)$ and the centered entry as

$$T^\circ(a)=T(a)-\mu(a\bmod b).$$

The raw reflection law is $T(a)+T(q-a)=-1$. It also gives $\mu(u)+\mu(b-u)=-1$. Subtracting these equalities yields

$$T^\circ(q-a)=-T^\circ(a).$$

The family sums vanish by construction. These statements hold for every base and positive lag, not just prime bases.

For a prime base at lag one, summing the floor counts gives the shorter formula $\mu(u)=u/b-1$. In base seven the means run from $-6/7$ to $-1/7$. Composite bases need not follow that spacing. The decimal means above do not.

Weights that hear nothing

Read the centered table with a pattern of weights. Multiply each entry by its weight, then add.

Give every entry in a family the same weight. The answer is zero. Each family already sums to zero.

Now use weights that agree at reflected addresses. Again the answer is zero. Opposite entries cancel in pairs.

Dirichlet characters supply a complete set of arithmetic weighting patterns. Their weights respect multiplication. An even character agrees at $a$ and $q-a$. An odd character changes sign there.

Only odd characters that also vary within a family can have a nonzero coefficient in the centered table. That restriction is the spectral gate.

At modulus nine, repeatedly multiplying by two puts the six addresses in the order

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

The six characters are the six Fourier modes around this cycle. Mode zero gives every address the same weight. Mode three gives $+1$ to the family ending in one and $-1$ to the family ending in two. Neither can read the differences inside a family.

Removing the global mean kills mode zero. Removing the family means also kills mode three. Modes one and five remain, each with power $4/9$. They turn their weights inside the columns, where the entries still differ.

Eighteen places. Seventeen occupied.

Decimal lag one has forty characters modulo one hundred. Twenty are even, so reflection excludes them.

Four characters read only the final digit. Two are already among the even characters. Family centering excludes the other two. That leaves

$$40-20-2=18$$

permitted characters.

Permitted is not the same as present. Exact evaluation of this forty-entry table leaves seventeen nonzero coefficients. The eighteenth character reads an odd address as $+1$ if it is $1$ modulo four, and $-1$ if it is $3$ modulo four. It passes both tests. Its coefficient is nevertheless zero.

The gate tells us where a coefficient may occur. The entries decide whether it does.

These are the four weighting patterns from The Character Structure of the Collision Fluctuation. There they combine running sums of $\Delta$ over primes. Here they read the finite table of $S$, each repeating one weight down a whole family. Center the families and all four table coefficients vanish. The prime-weighted sums need not vanish with them. At a finite cutoff, the addresses carry different total prime weights.

Check all forty characters, including the empty place

Modulo twenty-five, every unit is a power of two. Write $a\equiv2^t\pmod{25}$, and put $s=0$ for $a\equiv1\pmod4$, $s=1$ for $a\equiv3\pmod4$. The forty characters are

$$\chi_{\varepsilon,j}(a)=(-1)^{\varepsilon s} e^{2\pi ijt/20},$$

with $\varepsilon=0,1$ and $j=0,\ldots,19$. These are the labels in the figure. A character is odd when $\varepsilon+j$ is odd. It reads only the last decimal digit when $\varepsilon=0$ and $j$ is a multiple of five.

The normalized coefficient is

$$\widehat T^\circ(\chi)=\frac1{40} \sum_{a\in(\mathbb Z/100\mathbb Z)^\times} T^\circ(a)\overline{\chi(a)}.$$

The empty permitted place is $(\varepsilon,j)=(1,0)$, the character modulo four. The twenty centered entries with $a\equiv1\pmod4$ sum to zero. The other twenty sum to zero too. Its weighted total is their difference.

For completeness, group the permitted characters by their conductor, the smallest modulus needed to specify their weights.

Conductor	Permitted	Nonzero
4	1	0
20	1	1
25	8	8
100	8	8

The gate means “not inherited from modulus ten,” not simply “conductor greater than ten.” The character modulo four is the useful warning.

The zero tests here are exact. The coefficients are integer polynomials in a twentieth root of unity, divided by $400$. Reducing those polynomials modulo $x^8-x^6+x^4-x^2+1$ distinguishes a true zero from a small numerical value.

Let the primes read the table

Take each prime greater than one hundred. Read its two-digit entry, subtract the mean belonging to its final digit, divide by the prime, and add. With $S^\circ(p)=T^\circ(p\bmod100)$, this gives

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

The three curves use exactly the same primes. The first retains the integer table’s bias. The second removes its single global mean. The third removes each family mean. Both centered sums converge, generally to different limits. Family centering earns its place by the additional character patterns it eliminates, not by being the only subtraction that permits convergence.

Primes included	Last prime	Family-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

Those are finite values. The last one is not an asserted value of the limit.

The proof begins with the finite character expansion. The centered table is a finite linear combination of nonprincipal characters. For each such character, the sum $\sum_{p\le x}\chi(p)/p$ converges. A finite combination of convergent sums converges.

The classical step is Mertens’ theorem in arithmetic progressions. Every allowed address receives the same leading share of the prime harmonic sum. Nonprincipal character weights add to zero over those addresses. Their common growing term cancels, leaving a constant and an error tending to zero.

This works for every fixed base and positive lag, starting beyond $b^{\ell+1}$. It does not require an estimate from the graph. Without the reciprocal weights, the average centered entry over primes tends to zero by the prime number theorem in arithmetic progressions. The reciprocal-prime sum itself need not tend to zero.

From six entries to 131,072

The same family balance and reflection law hold as the tables grow. These plates show every allowed address. Addresses increase from left to right, then continue on the next row. This compact arrangement makes reflection a half-turn of each rectangle. Gold turns to teal; a zero stays at the background color.

The common color scale divides each centered entry by the number of matching intervals, $b^\ell$, with symmetric logarithmic spacing to retain smaller variations. No values are clipped. Color does not encode character power. These are the entries before the character transform.

The upper field contains all forty thousand decimal addresses at lag four. The lower contains all 131,072 allowed addresses in base sixty-four at lag two. No cells are sampled or averaged. Select a plate to inspect it at full resolution.

Three compact views of the calculation

Return to $1009$ and $1109$. They share the final digit nine, but their lag-two table entries are eight and minus ten. No last-digit average can account for the eighteen between them.

Subtract that average. The two entries move together.

Eighteen stays eighteen.