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.

At $109$, it is $+8$. At $1009$, it is $+8$ again.

To find it, write the fractions $1/109$ through $108/109$. Read the first two decimal digits of each fraction and count the rows where they agree.

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

Eighteen rows match. Now group the rows by their first digit. These ten bins have ten or eleven rows apiece. Subtract the smaller size, $\lfloor108/10\rfloor=10$. That leaves eight.

At $1009$, there are $108$ matches. The bin size is now $100$. Eight again.

The fractions have changed. There are nine hundred more of them. The subtraction leaves the same integer because the denominator still ends in $09$.

The Collision Fluctuation Sum uses the mean of a finite table to explain a drift across primes. The Centered Collision Sum subtracts its family means. Here I want to put the table itself on the page, including the endpoint arithmetic that fixes its balance.

Find the ending

Call the number of matching rows $C_b(N)$, where $b$ is the base and $N$ the denominator. Subtract the bin scale to get the reduced count

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

The denominators in this article are coprime to the base. In decimal, they end in $1$, $3$, $7$ or $9$. Combine those four endings with ten possible tens digits and there are forty places to look.

For $109$, go to row zero, column nine. The entry is $+8$. For $191$, go to row nine, column one. The entry is $-9$.

The table also handles $50{,}009=43\times1163$. Its $5{,}008$ matches exceed the bin scale of $5{,}000$ by eight. Primality never enters this calculation.

All forty entries as numbers

The row supplies the tens digit. The column supplies the units digit. These are the reduced counts, not the collision counts themselves.

First digit	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

There are eight positive entries, twelve zeros and twenty negative entries. Cell $01$ is zero. The finite floor formula below defines that cell directly; equivalently, the empty fraction list at denominator one has count zero.

The hundred little intervals

Put the fractions on the interval from zero to one. Cut it into a hundred equal pieces. A fraction in the first piece begins $0.00$, one in the next begins $0.01$, and so on.

A match lands in one of ten pieces

$$00,\quad11,\quad22,\quad\ldots,\quad99.$$

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

The dots do not stay put. They move to the positions specified by the new denominator. It is the number in each piece that increases by one.

You can check this with floors. Piece $j$ has count

$$\left\lfloor\frac{(j+1)N}{100}\right\rfloor -\left\lfloor\frac{jN}{100}\right\rfloor,$$

except in the last piece, where we subtract one to exclude the endpoint $1$. Replacing $N$ by $N+100$ adds $j+1$ to the first floor and $j$ to the second. The difference gains one. The excluded endpoint remains excluded.

Coprimality keeps the fractions off every interior boundary, so there is no rounding ambiguity to settle.

In base $b$, there are $b^2$ pieces and $b$ matching ones. Adding $b^2$ to the denominator adds $b$ to both counts. One finite function $T_b$ therefore supplies every reduced count

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

Forty entries suffice in decimal, however many digits the denominator has.

Seventeen fills what eighty-three leaves empty

Take two addresses that add to one hundred. Their table values always add to minus one.

$$\begin{aligned} T_{10}(09)+T_{10}(91)&=8-9=-1,\\ T_{10}(27)+T_{10}(73)&=6-7=-1,\\ T_{10}(41)+T_{10}(59)&=-4+3=-1. \end{aligned}$$

To see where the missing one comes from, use $17$ and $83$. Neither denominator reaches a hundred, so its fractions are more than one hundredth apart. No piece can hold two of them.

The matching pieces $00$ and $99$ are empty at both denominators. The first positive fraction lies above $0.01$; the last lies below $0.99$.

That leaves eight interior pieces. In each, exactly one of the two denominators supplies a fraction.

Seventeen supplies two matches. Eighty-three supplies six. Together, eight. Their bin scales are one and eight. Together, nine.

The reduced counts must add to $8-9=-1$.

The endpoint arithmetic in any base

Put $q=b^2$ and choose an address $a$ coprime to $b$, with $1\leq a<q$. For a piece numbered $j$, write its floor increment as

$$D_j(a)=\left\lfloor\frac{(j+1)a}{q}\right\rfloor -\left\lfloor\frac{ja}{q}\right\rfloor.$$

For every interior boundary $1\leq j<q$, coprimality gives

$$\left\lfloor\frac{j(q-a)}q\right\rfloor =j-1-\left\lfloor\frac{ja}q\right\rfloor.$$

Subtract successive instances. For an interior piece, where $1\leq j\leq q-2$, the two increments add to one. At the first and last pieces they instead add to zero and two

$$\begin{aligned} D_0(a)+D_0(q-a)&=0,\\ D_j(a)+D_j(q-a)&=1,\\ D_{q-1}(a)+D_{q-1}(q-a)&=2. \end{aligned}$$

The matching pieces have indices $d(b+1)$ for $d=0,\ldots,b-1$. They include both end pieces and $b-2$ interior pieces. Each last-piece increment counts the excluded endpoint once. Removing it from each denominator leaves

$$\begin{aligned} &C_b(a)+C_b(q-a)\\ &\quad=0+(b-2)+2-2\\ &\quad=b-2. \end{aligned}$$

The two bin scales add to $b-1$. Indeed, write $a=bh+s$ with $1\leq s<b$. The scales are $h$ and $b-h-1$. Subtracting gives

$$\begin{aligned} &T_b(a)+T_b(b^2-a)\\ &\quad=(b-2)-(b-1)\\ &\quad=-1. \end{aligned}$$

The finite table itself can be calculated without choosing any larger denominator

$$T_b(a)=-1-\left\lfloor\frac ab\right\rfloor +\sum_{d=0}^{b-1}D_{d(b+1)}(a).$$

At address one, the same formula uses the empty count $C_b(1)=0$.

The center of the table is therefore $-1/2$. The extremes $+8$ and $-9$ are equally far from it.

There are twenty reflected pairs in decimal. Each pair contributes minus one, so the forty entries have mean $-1/2$. Exactly twenty entries are negative. Two nonnegative integers cannot add to $-1$, and two negative integers add to at most $-2$. Each pair must have one of each kind.

This counts cells, not primes. How the primes visit those cells is a separate question.

Change the base

In base three the whole table fits into six places. Addresses below are written in base three; the entries are ordinary signed integers.

First digit	Last digit 1	Last digit 2
0	0	+1
1	0	−1
2	−2	−1

Rotate the rectangle through half a turn. Each cell meets its reflected partner. The pair sums are $0-1$, $1-2$ and $0-1$. Three pairs, three deficits.

Base twelve has forty-eight cells, of which twenty-four are negative. Its largest value is $+10$ and its smallest $-11$. The gallery follows these small tables into larger bases. Every cell is calculated, including the fine bands in the largest views.

From six cells to 123,462

Rows give the first address digit. Columns give the allowed final digits in increasing order; digits sharing a factor with the base are omitted. For comparison across bases, the color shows $(T_b+1/2)/b$ on one symmetric logarithmic scale. Gold and teal exchange under a half-turn. This centers the colors on the reflection midpoint; it does not subtract a family mean.

Base $360$ has $96$ allowed final digits and $34{,}560$ cells. Base $361=19^2$ has $342$ allowed final digits and $123{,}462$ cells. Both plates retain every cell. Select either figure to inspect the full-resolution image.

The forty-eight base-twelve entries

Row and column labels give digit values in ordinary decimal notation, so the last row is digit eleven. Only final digits $1$, $5$, $7$ and $11$ are coprime to twelve.

First digit	1	5	7	11
0	0	0	0	+10
1	−1	−1	−1	−1
2	−2	0	+2	0
3	−3	−1	+1	+1
4	0	0	+6	+2
5	−5	+1	−5	−1
6	0	+4	−2	+4
7	−3	−7	−1	−1
8	−2	−2	0	+2
9	−1	−3	−1	+1
10	0	0	0	0
11	−11	−1	−1	−1

The reflection law fixes the mean and the negative half in every base. To fix the extremes, we need one more count. At an address below $b^2$, each matching piece contains at most one fraction. The first and last pieces are empty. There can be at most $b-2$ matches, and subtracting a nonnegative bin scale cannot raise that ceiling.

Eight, before the primes

Nine reaches the ceiling in decimal. Write its eight fractions

1/9 = 0.111…       5/9 = 0.555…
2/9 = 0.222…       6/9 = 0.666…
3/9 = 0.333…       7/9 = 0.777…
4/9 = 0.444…       8/9 = 0.888…

Eight matches. Bin scale zero. Reduced count $+8$.

In base $b\geq3$, denominator $b-1$ does the same thing. Its $b-2$ fractions each repeat a single digit, attaining $b-2$. Reflection forces the other extreme, $-(b-1)$, at address $b^2-b+1$. Base two has just the two entries $0$ and $-1$.

Adding a hundred to nine gives $109$. The matching count rises to eighteen and the bin scale to ten. Add another nine hundred and they become $108$ and $100$.

Now write a thousand digits to the left of $09$. Nobody is going to list all those fractions. The excess count is already settled by the eight fractions of nine that fit on these four lines.

The decimal ledger, its twenty pairs, and the sign counts