The Collision Periodic Table
Abstract
Fix a base b\geq2. Partition the nonzero residues modulo an integer N coprime to b into b digit bins and count the residues that remain in the same bin after multiplication by b. After subtracting the bin scale \lfloor(N-1)/b\rfloor, the resulting lag-one collision deviation depends only on N\bmod b^2. One finite function on the unit classes modulo b^2 therefore contains every such deviation, including all deviations indexed by primes not dividing the base.
The finite function satisfies T_b(a)+T_b(b^2-a)=-1. Its exact minimum is -(b-1), its exact maximum is b-2, exactly half of its values are negative, and its mean is -1/2. The complete forty-cell table in base ten is given explicitly. All structural claims follow from finite floor arithmetic and endpoint counting.
Forty cells behind every prime
The primes 109 and 1009 are separated by nine hundred, yet both carry collision deviation +8. The primes 191 and 991 both carry -9. Their last two decimal digits decide the value.
Every prime greater than one hundred lands in one of the forty unit classes modulo one hundred. An unbounded sequence of inputs therefore produces a fixed finite object. At lag one, the signed collision deviation does not retain the size of an integer. It retains only the integer’s residue class modulo the square of the base.
The lag-one finite table first appears in The Collision Fluctuation Sum [1]. The Centered Collision Sum [2] carries finite determination and reflection through every lag. At lag one, the table can now be opened cell by cell. The task is to determine its sharp range, exact sign balance, and complete decimal form.
Primality is not needed for the finite law. The table belongs to the division structure itself and applies to every integer coprime to the base. Primes sample the same table through their residue classes.
The table is also rigid. Complementary classes have values summing to -1. That reflection fixes the mean and the number of negative cells. A direct floor count fixes the two extreme cells.
Collision fingerprints
Let N\geq2 and b\geq2 with \gcd(N,b)=1. For an integer x not divisible by N, let [x]_N denote its representative in \{1,\ldots,N-1\}. Define the base-b digit function by \delta_{N,b}(r)=\left\lfloor\frac{br}{N}\right\rfloor, \qquad 1\leq r<N. For a unit multiplier g\bmod N, define C_{N,b}(g)= \#\left\{r\in\{1,\ldots,N-1\}\ \middle| \delta_{N,b}(r)=\delta_{N,b}([gr]_N)\right\}. The lag-one collision count and its signed deviation are C_b(N)=C_{N,b}([b]_N), \qquad S_b(N)=C_b(N)-\left\lfloor\frac{N-1}{b}\right\rfloor.
Put q=b^2 and G_b=\{d(b+1):0\leq d\leq b-1\}. For every unit class a\bmod q, represented by 1\leq a<q, define D_n(a)= \left\lfloor\frac{(n+1)a}{q}\right\rfloor -\left\lfloor\frac{na}{q}\right\rfloor and T_b(a)= -1-\left\lfloor\frac{a}{b}\right\rfloor +\sum_{n\in G_b}D_n(a). This formula defines the collision periodic table without choosing a prime in any residue class.
The fingerprint theorem
Theorem 1 (Finite determination at lag one). Let b\geq2 and let N\geq2 satisfy \gcd(N,b)=1. If a is the representative of N\bmod b^2 in \{1,\ldots,b^2-1\}, then S_b(N)=T_b(a). Thus the lag-one collision deviation depends only on N\bmod b^2.
Proof. For 1\leq r<N, set n(r)=\left\lfloor\frac{b^2r}{N}\right\rfloor. The nested-floor identity gives \delta_{N,b}(r) =\left\lfloor\frac{n(r)}{b}\right\rfloor. Write br=kN+[br]_N. Multiplication by b gives \delta_{N,b}([br]_N) =\left\lfloor\frac{b[br]_N}{N}\right\rfloor =n(r)-bk. Here k=\lfloor br/N\rfloor=\lfloor n(r)/b\rfloor, so \delta_{N,b}([br]_N)=n(r)\bmod b. Writing n=bu+v with 0\leq u,v<b, a collision occurs exactly when u=v. The collision slices are therefore the b indices in G_b.
Because \gcd(N,b^2)=1, every interior slice boundary is nonintegral. For 0\leq n\leq b^2-2, the floor difference \left\lfloor\frac{(n+1)N}{b^2}\right\rfloor -\left\lfloor\frac{nN}{b^2}\right\rfloor counts exactly the allowed positive residues in slice n. On the terminal slice it also counts the excluded endpoint N. The terminal index b^2-1 belongs to G_b, and hence C_b(N)= -1+\sum_{n\in G_b} \left( \left\lfloor\frac{(n+1)N}{b^2}\right\rfloor -\left\lfloor\frac{nN}{b^2}\right\rfloor \right).
Write N=b^2t+a with t\geq0. Each of the b floor differences contributes t+D_n(a), so C_b(N)=bt-1+\sum_{n\in G_b}D_n(a). Since a is a unit modulo b, it is not divisible by b. Consequently \left\lfloor\frac{N-1}{b}\right\rfloor =bt+\left\lfloor\frac{a}{b}\right\rfloor. Subtracting this quantity gives (6). ◻
Corollary 2 (Prime fingerprints). Every prime p not dividing b carries the table value T_b(p\bmod b^2). In base ten, every prime greater than one hundred has a collision fingerprint determined by its last two decimal digits.
Reflection
Theorem 3 (Antisymmetry). For every base b\geq2 and every unit class a\bmod b^2, T_b(a)+T_b(b^2-a)=-1.
Proof. Let q=b^2. If 1\leq n\leq q-2, neither na nor (n+1)a is divisible by q. Complementary floors give D_n(a)+D_n(q-a)=1. At n=0 the two increments sum to zero. At n=q-1 they sum to two. The set G_b contains both endpoints and has b-2 interior indices. Therefore \sum_{n\in G_b}\bigl(D_n(a)+D_n(q-a)\bigr)=b.
Write a=bh+s with 1\leq s<b. Then \left\lfloor\frac{a}{b}\right\rfloor +\left\lfloor\frac{q-a}{b}\right\rfloor=b-1. Adding the two instances of (6) and using (9) gives T_b(a)+T_b(q-a)=-2-(b-1)+b=-1. ◻
The same identity can be read directly from the paired floor increments. Every interior diagonal slice contributes one unit across the two complementary classes. The initial slice contributes none, while the terminal slice contributes two. The endpoint imbalance is exactly what leaves the final value -1 after subtracting the bin-scale term.
Extremes and sign balance
Theorem 4 (Exact extrema). For every base b\geq2, \max_a T_b(a)=b-2, \qquad \min_a T_b(a)=-(b-1). The maximum occurs at a=b-1, and the minimum occurs at a=b^2-b+1.
Proof. Since 0<a<b^2, every increment D_n(a) is either zero or one. The initial increment D_0(a) is always zero. Thus T_b(a)\leq -1+(b-1)=b-2.
Take a=b-1. For n=d(b+1) with 1\leq d\leq b-1, na=d(b^2-1)\equiv -d\pmod{b^2}. In general, if na=b^2k+r with 0\leq r<b^2, then D_n(a)=\lfloor(r+a)/b^2\rfloor. Thus D_n(a)=1 exactly when r\geq b^2-a. The least nonnegative residue is b^2-d, which is at least b^2-b+1=b^2-a. Hence D_n(a)=1 for every nonzero n\in G_b. It follows that T_b(b-1)=b-2.
Apply the upper bound to b^2-a and use antisymmetry. This gives T_b(a)\geq-(b-1) for every unit class. The complement of b-1 attains equality. ◻
Corollary 5 (Negative classes). Exactly \varphi(b^2)/2 unit classes satisfy T_b(a)<0.
Proof. Negation has no fixed point on the unit classes modulo b^2. Indeed, a\equiv-a\pmod{b^2} and \gcd(a,b^2)=1 would imply b^2\mid2, which is impossible for b\geq2.
The unit classes therefore split into complementary pairs. Each pair contains two integers whose sum is -1. They cannot both be nonnegative, and two negative integers would have sum at most -2. Exactly one member of every pair is negative. ◻
Corollary 6 (Mean). \frac{1}{\varphi(b^2)} \sum_{a\in(\mathbb Z/b^2\mathbb Z)^*}T_b(a)=-\frac12.
Proof. There are \varphi(b^2)/2 complementary pairs, and every pair has sum -1. ◻
Exact tables
Base ten
The forty unit classes modulo one hundred and their lag-one collision fingerprints are
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 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. The extremes are T_{10}(9)=8 and T_{10}(91)=-9. The primes 109 and 1009 share the first fingerprint. The primes 191 and 991 share the second.
The same calculation gives the following exact summaries in selected bases.
| b | \varphi(b^2) | minimum | maximum | positive | zero | negative |
|---|---|---|---|---|---|---|
| 3 | 6 | -2 | +1 | 1 | 2 | 3 |
| 5 | 20 | -4 | +3 | 4 | 6 | 10 |
| 6 | 12 | -5 | +4 | 2 | 4 | 6 |
| 7 | 42 | -6 | +5 | 10 | 11 | 21 |
| 8 | 32 | -7 | +6 | 7 | 9 | 16 |
| 10 | 40 | -9 | +8 | 8 | 12 | 20 |
| 12 | 48 | -11 | +10 | 11 | 13 | 24 |
nfield [3] evaluates both the finite floor formula and the original collision count in exact integer arithmetic. Its bounded verification exhausts every coprime input N with 2\leq N<4b^2 for bases two through twelve. It also reproduces the complete base-ten table and every row of Table 2.
The finite object
The prime-indexed collision sequence is not a collection of unrelated measurements. At lag one it is the pullback of a finite integer-valued function on the unit classes modulo b^2. The base fixes the table before any prime is sampled.
The individual cells depend on the base, but the frame does not move. Complementation fixes the center at -1/2, places one negative entry in every pair, and links the two extremes. The collision periodic table makes that finite structure visible all at once.
Every finite function on the unit group has a Dirichlet-character expansion. The cells and their reflection pairs are now exact. What happens when the primes sample those character channels through a Dirichlet series?
References
[1]A. S. Petty, The Collision Fluctuation Sum, research note, April 2023, revised August 2026. doi:10.5281/zenodo.21852234.
[2]A. S. Petty, The Centered Collision Sum, research note, October 2023, revised August 2026. doi:10.5281/zenodo.21852556.
[3]A. S. Petty, nfield, software repository. https://github.com/alexspetty/nfield