The Composite Interleaving

The Composite Interleaving
Two lattices interleaved. The composite's bins cut across both. Ghost collisions appear where neither prime collides individually. The composite creates its own structure.

Does the collision spectrum of a composite factor through its primes? It does not.

The tensor test

For $n = pq$, the characters modulo $n$ factor as $\chi = \chi_p \times \chi_q$ via the Chinese Remainder Theorem. The collision transform $\hat{S}_n(\chi)$ is indexed by pairs $(\chi_p, \chi_q)$.

If the composite spectrum were a tensor product of the prime spectra, then $|\hat{S}_n(\chi_p \times \chi_q)|$ would factor as $f(|\hat{S}_p(\chi_p)|) \cdot g(|\hat{S}_q(\chi_q)|)$ for some functions $f$ and $g$.

At $n = 13 \times 17 = 221$: only 22 of 192 character pairs have ratios within 10% of any constant. The coefficient of variation is 0.88. The tensor structure fails.

The composite knows something the primes individually don't.

The interleaving

The digit function $\lfloor br/(pq) \rfloor$ depends on both residues $(r \bmod p, r \bmod q)$ simultaneously. The composite's digit is not determined by the individual prime digits alone. The composite bin is wider than either prime's bin, and it absorbs shifts that the prime bins reject.

The interleaving function assigns a composite digit to each point on the $p \times q$ CRT grid. Its column collisions (different prime-digit pairs mapping to the same composite digit) are the source of the ghost collisions that dominate the composite's collision structure.

Ghost collisions

A ghost collision is a collision at the composite that doesn't exist at either prime. Computation shows ghost collisions account for 98-100% of all collisions at semiprimes tested through $n = 667$.

The composite's collision structure is not inherited from the primes. It is created by the interleaving. The composite sees what the primes cannot.

The gate width failure

The gate width theorem gives exactly $b - 1$ deranging multipliers at primes. At composites, the count can exceed $b - 1$. The extras satisfy $g \equiv 1 \pmod{p}$ for a factor $p$, and $\gcd(g-1, n)$ reveals the factor.

The universal count breaks. The extras encode the factorization.

Try it yourself

./nfield 77              # n = 7 x 11
./nfield 91              # n = 7 x 13

Code: github.com/alexspetty/nfield


Alexander S. Petty
December 2025
.:.