The Collision Fluctuation Sum

I added up the collision deviations for the first thousand primes greater than eleven, dividing each deviation by its prime before adding it. The total was negative.

I kept going. Ten thousand primes. A hundred thousand. The total moved farther below zero.

Primes included	Largest prime	Weighted sum
1,000	7,951	−1.367764
10,000	104,779	−1.596580
100,000	1,299,811	−1.773479
664,574	9,999,991	−1.895354

There are positive terms in this sum. They interrupt the descent, but they do not stop it. The question is what keeps pulling the total down.

The answer fits in a table of forty integers. A reflection pairs its entries, and every pair falls one short of zero. After accounting for the unequal sizes of the digit bins, that small imbalance gives the sum its exact drift coefficient, $-9/10$.

To see where the table comes from, we need only the first two digits of a fraction.

Two digits at a time

Write out the fractions $1/17, 2/17, \ldots, 16/17$. Read the first two decimal places of each one.

Two rows begin with equal digits,

$$\frac{2}{17}=0.11\ldots, \qquad \frac{15}{17}=0.88\ldots.$$

The other fourteen do not. The lag-one collision count is two.

Multiplying a remainder by ten and continuing long division advances one digit. Asking whether the digit stays the same is therefore the same as asking whether the first two places agree. The acceptable beginnings are $00,11,22,\ldots,99$.

This count uses every nonzero numerator. At a prime whose remainders form one cycle, it is also the number of matches between that cycle and its one-place shift. If the remainders split into several cycles, we include all of them.

A large denominator naturally gives more opportunities for a match. To compare primes, I subtract a benchmark from the count. At each prime, take every multiplier except the identity, discard the ones producing no collisions, and average the remaining counts. This is the exact constructive mean used in Silent Primes.

The difference between the lag-one count and that mean is the collision deviation, $\Delta(p)$. Positive means more matches than the benchmark. Negative means fewer. The opening table adds $\Delta(p)/p$, so the contribution from a large prime receives less weight. In base ten this benchmark is defined for primes greater than eleven, which is where the sum begins.

Every collision has a partner

Look again at the two highlighted fractions. They add to one. Their decimal digits complement each other, place by place. The pair $11$ becomes $88$.

This gives a short proof that the collision count is even.

If the fractions associated with $r$ and $gr$ have equal leading digits, their complements have equal leading digits too. The complement sends each digit $d$ to $b-1-d$. It also sends the remainder $r$ to $p-r$, and multiplication carries that pair to another complementary pair.

For a prime $p>b\geq2$, the prime is odd. So $r$ and $p-r$ are distinct. Every collision has a different partner, for every nonzero multiplier. The count is twice the number of pairs.

That is the bilateral parity property. It explains the even counts. To explain the negative drift, we need to look at a second pairing, this time among the entries of a finite table.

Forty places for the primes

The ten digit bins divide the $p-1$ nonzero remainders as evenly as possible. If the smaller bins contain $Q$ remainders, the others contain $Q+1$.

Subtract $Q$ from the lag-one collision count and call the resulting integer $S(p)$. At $p=191$, for example, the bins all contain nineteen remainders. There are ten lag-one collisions, so

$$S(191)=10-19=-9.$$

At $p=109$, the smaller bins contain ten remainders and there are eighteen collisions. Here $S(109)=8$.

For every prime above one hundred, this integer is determined by the last two digits of the prime. All primes ending in $91$ give $-9$. All those ending in $09$ give $8$.

There are forty possible endings, because the last digit must be $1$, $3$, $7$, or $9$.

The reason is visible in the two-digit test. Each matching beginning occupies one interval of width $1/100$ among the fractions between zero and one. Increasing the denominator by one hundred adds one fraction to each of the ten matching intervals. The collision count rises by ten. The smaller bin size also rises by ten. Their difference stays put. The endpoint count needs care, but the cancellation is exact.

Now turn the table halfway around. The ending $09$ meets $91$, $27$ meets $73$, and $43$ meets $57$. Their entries add up as follows,

Reflected endings	Table entries	Sum
09 and 91	8 and −9	−1
27 and 73	6 and −7	−1
43 and 57	0 and −1	−1

Every reflected pair does this. In the notation of the paper,

$$F_{10}(a)+F_{10}(100-a)=-1.$$

Twenty pairs, each totaling minus one. The forty entries sum to minus twenty. Their average is exactly $-1/2$.

The mirror does not cancel the table. It leaves a deficit of one in every pair.

The other four tenths

We have found the mean of $S(p)$, the collision count minus the smaller bin size. The opening sum uses $\Delta(p)$, which subtracts the full constructive mean. These are slightly different benchmarks.

Write $p-1=10Q+R$. The number $R$ counts how many of the bins contain an extra remainder. The constructive mean exceeds $Q$ by

$$\frac{QR}{p-11}.$$

As the prime grows, this correction approaches $R/10$. A prime ending in $1$ has $R=0$. Endings $3$, $7$, and $9$ give $R=2,6,8$. The four limiting corrections are therefore $0,1/5,3/5,4/5$. Their average is $2/5$.

The table contributes $-1/2$. Subtracting the mean correction contributes another $-2/5$. Together,

$$-\frac12-\frac25=-\frac9{10}.$$

We can also read the four families separately. Average each column of the forty-entry table, then subtract its correction.

Prime’s final digit	Mean table entry	Limiting correction	Limiting mean deviation
1	−1.7	0	−1.7
3	−0.9	0.2	−1.1
7	−0.1	0.6	−0.7
9	0.7	0.8	−0.1

These terminating decimals are exact. Each column of the integer table has ten entries. The limiting prime averages follow because primes populate the ten endings in each family equally at leading order.

Every family has a negative mean deviation. Individual primes can still contribute positively. The table contains both signs, and the exact correction varies with the prime. A family mean describes the average of those contributions, not each member.

A finite bias with an infinite sum

The forty endings tell us what the deviations are. To understand their accumulation, we also need to know how much reciprocal-prime weight each ending receives.

This is where Mertens’ theorem in arithmetic progressions enters. For a fixed modulus, every allowed residue class receives the same leading share of the prime harmonic sum. The differences between classes contribute constants. Williams’ theorem supplies this classical step.

In base ten, the forty classes each receive one fortieth of the leading $\log\log x$ growth. Averaging their biases therefore gives the coefficient of the drift. Write $\Phi(x)$ for the weighted sum in the opening table. Then

$$\Phi(x)=-\frac9{10}\log\log x+\kappa+o(1).$$

Here $\kappa$ is a fixed constant, and $o(1)$ is a remainder tending to zero. The small difference between the exact bin correction and its limiting value contributes to the constant. It does not change the coefficient.

This proves that the raw sum tends to negative infinity. It can rise over some stretches. It cannot settle to a finite limit.

The descent is slow enough to disguise its coefficient. At the largest cutoff in the opening table, dividing the sum by $\log\log x$ gives about $-0.682$. The eventual limit is $-0.9$. Even after more than six hundred thousand terms, the additive constant is still large enough to separate those numbers. The forty-entry table gives a firmer answer than extrapolating the numerical ratio.

The same argument works in every fixed base $b\geq2$. Reflection still contributes $-1/2$. The mean correction is $1/2-1/b$. Their difference is $-(b-1)/b$.

Take out the drift

Once the coefficient is known, there is a direct operation to try. Add $9/10$ to each deviation before dividing by its prime.

The first thousand terms now total $-0.295916$. At ten thousand terms the total is $-0.298594$. At the last cutoff it is $-0.298439$.

The lower curve suggests convergence. The theorem proves it,

$$\lim_{x\to\infty}\sum_{11<p\leq x} \frac{\Delta(p)+9/10}{p} \quad\text{exists}.$$

Centering does not force that limit to be zero. It removes the unbounded drift. Nor does adding one global constant erase the differences between the four families, or between the ten entries inside each family. Those differences remain in the sum.

I began by watching a total move away from zero. The explanation led back to a much smaller object. Two decimal places pick out the collisions. Forty entries hold their integer deviations. Reflection fixes the mean, and the distribution of primes carries that mean into the sum.

The drift was visible only after adding many primes. Its coefficient was already fixed before the addition began.