The Collision Fluctuation Sum

Write the first two decimal digits of each fraction from $1/17$ through $16/17$. Put each fraction in a square. Its first digit chooses the row; its second digit chooses the column.

Sixteen fractions. Only two land on the diagonal.

They are $2/17=0.11\ldots$ and $15/17=0.88\ldots$. Each has the same digit in its first two places. The collision count is two.

The two fractions add to one. Their digits complement each other, turning eleven into eighty-eight. This is why collisions come in pairs. If multiplication preserves the digit of a remainder $r$, it also preserves the digit of $p-r$. Those remainders are distinct at an odd prime. Every match has a different partner.

We count every nonzero remainder, not just those visited by the repeating block of $1/p$. Multiplication by ten advances long division one place. The matching beginnings are $00,11,22,\ldots,99$.

A total that will not settle

A larger prime has more remainders available to match. Compare its count with a reference level. Test every multiplier except the identity, discard the zero counts, and take the mean of the positive ones. Bin Derangements and the Gate Width Theorem gives this constructive mean from the bin sizes.

Subtract that mean from the count for multiplication by ten. Call the difference $\Delta(p)$. A negative deviation means ten preserves fewer digits than the average constructive multiplier. The mean exists for decimal primes greater than eleven.

Now add $\Delta(p)/p$ over those primes. The first thousand terms total about $-1.368$. Ten thousand give $-1.597$. Through the primes below ten million, the total is $-1.895$.

Positive terms interrupt the descent. They cannot stop it. The exact drift coefficient is $-9/10$, and it comes from forty small integers. We can calculate it without adding a long list of primes.

The last two digits are enough

Write $p-1=10Q+R$. The ten digit bins contain either $Q$ or $Q+1$ remainders. Subtract the smaller size from the collision count.

At 191, the count is ten and the smaller bin size is nineteen. The difference is $-9$. At 109, eighteen matches minus ten gives $8$.

For every prime above one hundred, the last two digits determine this integer. End in $91$ and the answer is $-9$. End in $09$ and it is $8$, however large the prime becomes.

Each matching beginning occupies one hundredth of the interval from zero to one. Adding one hundred to the denominator adds one numerator to each of the ten matching slices. The count rises by ten. So does $Q$. Their difference stays put.

There are forty possible prime endings. Arrange them by their tens and units digits.

The forty integers, as data

Choose the tens digit at the left and the units digit across the top. A row labeled zero includes endings $01$, $03$, $07$ and $09$.

Tens	1	3	7	9
0	0	2	0	8
1	-1	-1	1	-1
2	0	-2	6	0
3	-1	-1	-3	-1
4	-4	0	-2	0
5	-1	1	-1	3
6	0	2	0	0
7	-1	-7	1	-1
8	0	-2	0	0
9	-9	-1	-3	-1

To compute an entry, let $a$ be the ending and let $n$ run through $0,11,22,\ldots,99$. Define

$$D_n(a)=\left\lfloor\frac{(n+1)a}{100}\right\rfloor-\left\lfloor\frac{na}{100}\right\rfloor.$$ $$F(a)=-1-\left\lfloor\frac a{10}\right\rfloor+\sum_nD_n(a).$$

Then $S(p)=F(a)$ for primes $p>100$ with ending $a$. The floor differences count the numerators in each slice. The final slice includes the endpoint $p$, which is not a nonzero remainder, so we subtract one. Writing $p=100t+a$ contributes $t$ to each of the ten slices and $10t$ to $Q$. Those contributions cancel.

One negative left in every pair

Turn the table halfway around. Ending $09$ meets $91$. Their entries are $8$ and $-9$. Ending $27$ meets $73$, giving $6$ and $-7$. Ending $43$ meets $57$, giving zero and $-1$.

Every pair falls one short of zero.

Twenty pairs leave twenty negatives. The forty entries sum to $-20$, so their mean is $-1/2$.

Why reflection leaves one

Use the floor differences $D_n$ above. For each of the eight interior matching slices, neither endpoint is an integer, and

$$D_n(a)+D_n(100-a)=1.$$

The slice at zero contributes zero to the combined count. The slice at ninety-nine contributes two. Eight ones, a zero and a two total ten. Meanwhile,

$$\left\lfloor\frac a{10}\right\rfloor+\left\lfloor\frac{100-a}{10}\right\rfloor=9.$$

Each table entry also contains its own minus one. Add the two formulas for $F$ and the result is $-2-9+10=-1$. This proves the reflection rule without inspecting the forty answers individually.

The extra places in the bins

The opening sum subtracts the constructive mean, not just $Q$. That mean is

The extra term tends to $R/10$. The four final digits $1,3,7,9$ give $R=0,2,6,8$. Their limiting corrections are $0,0.2,0.6,0.8$, with mean $0.4$.

The reflection contributes minus one half. The larger benchmark subtracts another four tenths.

That is the coefficient. The finite table supplies all of it.

Forty copies of the same growth

The table fixes the bias. We still need to know how much reciprocal-prime weight each ending receives.

Let $H_a(x)$ add $1/p$ over primes above one hundred, up to $x$, ending in $a$. Mertens’ theorem in arithmetic progressions gives

Each class has the same leading growth, a class-dependent constant $M_a$, and an error tending to zero. This is the classical input, not a conclusion drawn from the collision data. Williams’ theorem supplies it.

Give ending $a$ its limiting deviation $d_a=F(a)-R/10$. The forty values have mean $-9/10$, so they sum to $-36$. Multiply each class’s prime sum by its value and add. The growing parts contribute

The forty constants give one constant. The forty vanishing errors still tend to zero.

One correction remains. The exact deviation differs from $d_a$ by a quantity of order $1/p$. After the extra division by $p$, its terms are of order $1/p^2$. Their absolute sum converges. The finitely many primes below one hundred change only the constant too.

Write $\Phi(x)$ for the original weighted sum. We have proved

The raw sum tends to negative infinity. Its path need not be monotone.

The exact correction and the small primes

The finite correction splits as

$$\frac{QR}{p-11}=\frac R{10}+\frac{R(10-R)}{10(p-11)}.$$

Thus $\Delta(p)=d_a-\varepsilon(p)$, where $\varepsilon(p)=R(10-R)/(10(p-11))$. For $p>100$, dividing by $p$ gives a term bounded by a constant times $1/p^2$. This proves absolute convergence of the correction, not merely that its individual terms tend to zero.

The sum starts at thirteen. Its twenty terms through ninety-seven form a fixed finite contribution. No claim about those small primes is needed to establish the coefficient at infinity.

Nine tenths in every term

Add $9/10$ to each deviation before dividing by its prime. The forty limiting coefficients now sum to zero. Their copies of $\log\log x$ cancel exactly. Constants, vanishing errors and the convergent correction remain.

The centered sum therefore converges.

At the last plotted prime, the centered value is about $-0.298439$. That is a finite value, not a certified evaluation of the limit. Convergence comes from the cancellation above. It does not follow from how level the curve looks.

The checkpoints and two different constants

Terms	Last prime	Raw	Centered
1000	7951	-1.367764	-0.295916
10000	104779	-1.596580	-0.298594
100000	1299811	-1.773479	-0.298335
664574	9999991	-1.895354	-0.298439

All eligible primes below ten million are included. Integer collision counts and rational corrections are exact; the accumulated decimals use long-double arithmetic and are rounded here.

Adding $0.9\log\log x$ to the raw sum is a different centering. Its last value is about $0.606595$, not $-0.298439$. The difference is

$$\frac9{10}\left(\sum_{11<p\leq x}\frac1p-\log\log x\right),$$

which tends to a nonzero constant. Both procedures remove the drift. They do not produce the same limiting value.

At the largest plotted cutoff, $\Phi(x)/\log\log x$ is only about $-0.682$. The additive constant is still visible. Extrapolating that ratio is a poor substitute for counting the finite table.

The signed contrast loses its drift too

Average each column of the table and subtract its limiting correction.

Ending	Table mean	Correction	Deviation
1	-1.7	0.0	-1.7
3	-0.9	0.2	-1.1
7	-0.1	0.6	-0.7
9	0.7	0.8	-0.1

The signed recipe in The Character Structure of the Collision Fluctuation keeps endings one and nine and reverses three and seven. The combined family means on its two sides are equal.

$$-1.7-0.1=-1.1-0.7=-1.8.$$

Each final-digit family receives one quarter of the leading prime harmonic growth. These equal coefficients cancel in $A_2$. The same argument proves convergence of this particular contrast as well.

Its remaining value includes class constants, the finite correction and the initial primes. It is not merely a vanishing error. Nor does this calculation say that every character component converges. The real part of $A_1$, for example, compares the unequal family means at endings one and nine.

The decimal coefficient belongs to a general rule. In every fixed base $b\geq2$, reflection leaves a table mean of $-1/2$. The mean correction is $1/2-1/b$. The drift coefficient is therefore $-(b-1)/b$. The companion paper proves this for each fixed base.

Turn the table in other bases

For base $b$, write an allowed ending as $a=bq+u$, with $u$ coprime to $b$. Rows list $q$; columns list the allowed $u$ in increasing order. Negation reverses both orders. Every cell meets its partner under a half-turn.

Color represents $(F_b(a)+1/2)/b$. The added half centers the reflection, and division by the base puts all panels on the same scale. A symmetric logarithmic color scale brings out the small deviations without hiding the large ones. Opposite cells have opposite colors. Every entry is present.

Base 256 has 32,768 entries. Base 257 has 65,792. Their shapes differ, but the same half-turn reverses every centered value. Select the image for full-resolution inspection.

Seventeenths, the forty entries and the two sums

At $09$ and $91$, eight positive units meet nine negative ones. One negative remains. Every reflected pair leaves the same deficit.

The drift emerges over many primes. Its coefficient is fixed before the first one enters the sum.