The Collision Periodic Table

Every prime number past 100, in base ten, has a small integer attached to it. The integer is determined entirely by the prime’s last two digits.

For $109$, it is $+8$. For $1009$, it is still $+8$. Both end in $09$.

Here is what the eight measures. Write out all the fractions from $1/109$ to $108/109$ and compare the first two digits after the decimal point. Some of the rows begin like this.

 1/109 = 0.00917…    match
 2/109 = 0.01834…    no match
12/109 = 0.11009…    match
13/109 = 0.11926…    match

There are eighteen matches in the complete table. Subtract ten, the whole-number part of $108/10$, and eight remains.

At denominator $1009$, there are $1008$ rows and $108$ matches. Subtract the whole-number part of $1008/10$, which is $100$. Eight remains again.

The number of rows has grown. The number of matches has grown. The difference has stayed put.

Find the ending

A prime above $100$ can end in $1$, $3$, $7$, or $9$. There are ten possibilities for the tens digit, so forty possible endings. Here they are, with the collision deviation written in each cell. The small label identifies the ending. The large number is its value.

To find $109$, go to row $0$, column $9$. To find $191$, go to row $9$, column $1$. The first gives $+8$. The second gives $-9$.

The table is small enough to read in full. Eight entries are positive, twelve are zero, and twenty are negative. Some endings share a value. Ending in $09$ fixes the answer at $+8$, but a value such as zero appears in several different cells.

I call the integer a collision fingerprint. Its definition in base $b$ is

$$S_b(N)=C_b(N)-\left\lfloor\frac{N-1}{b}\right\rfloor.$$

Here $C_b(N)$ counts the fractions whose first two digits agree. The brackets mean take the whole-number part. The quantity being subtracted is the bin scale, a simple integer baseline for the count.

There is more scope here than the opening about primes suggests. The same rule holds for every denominator with no factor in common with the base. For example, $50{,}009=43\times1163$ is composite. It has $5008$ matches, a bin scale of $5000$, and the same $+8$ in cell $09$.

Primality never enters the finite calculation.

The hundred little intervals

Why should only two digits of the denominator survive?

The fractions we are testing lie between zero and one. Cut that interval into a hundred equal pieces. Each piece specifies the first two decimal digits. The first runs from $0.00$ to $0.01$, the second from $0.01$ to $0.02$, and so on.

A match lands in one of ten pieces, those beginning

00   11   22   33   44   55   66   77   88   99

Counting the fractions in those ten little intervals gives the collision count. Nothing beyond the second digit is needed.

Now increase the denominator from $N$ to $N+100$. Each of the hundred intervals gains exactly one fraction. Each of the ten matching intervals therefore gains one match. The collision count goes up by ten. The bin scale also goes up by ten. Their difference stays unchanged.

This is where the floor arithmetic earns its place. The number of fractions between two boundaries is a difference of floors. Adding $100$ to the denominator adds one to each such difference. Coprimality keeps the fractions off the interior boundaries, and the endpoint at one is excluded from both counts.

In base $b$, there are $b^2$ two-digit pieces and $b$ matching pieces. Increase the denominator by $b^2$ and both terms go up by $b$. The same cancellation gives

$$S_b(N)=T_b(N\bmod b^2).$$

The notation $T_b$ names the finite table. In decimal, the remainder modulo $100$ is simply the last two digits. The size of the denominator has disappeared from the deviation.

Add the opposite cell

Return to $09$ and $91$. Their addresses add to $100$. Their values add to $-1$.

Try $27$ and $73$. The values are $+6$ and $-7$. Try $41$ and $59$. The values are $-4$ and $+3$. Again, minus one.

Pair each ending $a$ with $100-a$. There are twenty pairs. The figure puts each pair on a number line so that you can see both the sum and the midpoint.

The exact identity is

$$T_b(a)+T_b(b^2-a)=-1.$$

The decimal pairs let us check it in one base. The proof uses the interval count to establish it in every base. Reflect the two floor counts and their interior contributions pair off. The first and last intervals have different endpoint corrections. Keep those corrections, subtract the two bin scales, and the remainder is exactly minus one.

The natural center of this table is therefore $-1/2$. The entry $+8$ is eight and a half units above it. The entry $-9$ is eight and a half units below it. Adding $1/2$ to every cell would put the reflection center at zero.

There is also an exact answer to the question of how many entries are negative. Two nonnegative integers cannot add to $-1$. Two negative integers cannot either, since their sum would be at most $-2$. Each pair has exactly one negative member.

Half the table is negative. The other half contains the positive entries and the zeros.

The mean follows just as directly. Twenty pairs contribute a total of $-20$ across forty cells, giving $-1/2$. This is an average over the cells themselves, available before any prime cutoff is chosen. It is part of the finite data that The Centered Collision Sum uses to separate fixed class biases from the sum over primes.

Change the base

In base three there are six admissible cells. In base twelve there are forty-eight. The individual values change, and so does the number of zeros. The negative half stays exactly half.

The extrema have a short formula too. The largest value is $b-2$, attained at the ending $b-1$. Its complementary ending, $b^2-b+1$, has the smallest value, $-(b-1)$.

In base ten those are $+8$ and $-9$. In base twelve they are $+10$ and $-11$. These are sharp bounds. They do not say that every integer between the bounds must appear.

The reflection alone does not determine all the cells. It tells us how they pair, where their mean lies, and how many are negative. The floor count supplies the individual values. Together they give a table that can be computed without searching for a prime in any of its classes.

Eight, before the primes

There is an especially small way to see the $+8$ at ending $09$.

Use denominator $9$. Its eight fractions begin $0.11\ldots$, $0.22\ldots$, all the way through $0.88\ldots$. Every row matches. The count is eight and the bin scale is zero. The deviation is already $+8$.

At $109$, the count is eighteen and the baseline is ten. At $1009$, they are $108$ and $100$. The table records what survives those equal increases.

For denominators coprime to ten, adding $100$ leaves the deviation unchanged. That is the periodicity. It holds whether the denominator is prime or composite.

So those eight fractions at denominator $9$ already give the deviation for every larger denominator ending in $09$.