The Collision Spectrum and the L-Function Landscape

Here is the centered collision table in base five. Square every entry and add them. The answer is $48$.

The small numerals label the twenty residues from $1$ to $24$ that are not multiples of five. Each large fraction records what is left of the collision deviation after subtracting the average for its column. A positive entry sits above that average. A negative entry sits below it. The numerals are written in ordinary decimal notation, although the digit comparisons use base five.

These are the same counts that begin with long division. A remainder emits a digit. Multiplication by the base gives the next remainder. Count how often the two remainders emit the same digit, then remove the predictable baseline and the final-digit average. For prime denominators greater than $25$, the resulting value depends only on the denominator’s residue modulo $25$.

The centering is easy to check. The first column contains four copies of $4/5$ and one $-16/5$. They sum to zero. Reflection pairs entries of opposite sign. The value at $4$ is $16/5$, and the value at $21$ is $-16/5$.

Squaring stops those partners from cancelling. Add all twenty squares and the fractions give an integer.

What interests me is what else that integer determines. The same $48$ gives an exact sum built from eight Dirichlet $L$-functions. Their values are defined by infinite series. This table is enough to calculate their fourth moment.

The patterns in the table

The character decomposition gives us a way to read a finite arithmetic table through its multiplicative patterns. A character assigns a point on the unit circle to each allowed residue. Multiplying two residues multiplies their assigned points.

Compare the table with each of these patterns and take an average. The result is a Fourier coefficient. Its magnitude measures how much of that pattern the table contains. Together, the coefficients describe the table completely.

In base five there are twenty possible characters. Twelve coefficients vanish. Some vanish because the table changes sign under reflection. Others vanish because each column has been centered. Eight remain. In the terminology of character theory, they are the primitive odd characters modulo $25$. Odd means that their signs reverse under reflection. Primitive means that their pattern cannot be read from the remainder modulo five alone.

Each surviving coefficient has a direct definition as a sum of twenty weighted fractions. That definition lets us calculate it. It does not immediately tell us what is inside it.

The coefficients factor.

Two finite sums

For every odd prime base $b$, the magnitude of each primitive odd coefficient is

$$|\widehat S^\circ(\chi)| =\frac{|B_{1,\bar\chi}|\,|S_G(\chi)|}{b(b-1)}.$$

Read this as two factors multiplied together, then divided by the number of entries in the table. The symbol $\chi$ names the character pattern being tested. The left side is the magnitude of its collision coefficient.

The first factor is a generalized Bernoulli number. Here it is a finite weighted sum. Take each residue, multiply it by its character value with the complex phase reversed, add, and divide by $b^2$. It uses the whole residue table.

The second factor comes from differences across the digit boundaries. Those differences simplify. Its magnitude is twice the magnitude of a much shorter sum, using only the character values at $1,2,\ldots,b-1$. At base five, only four terms are needed.

The proof gets the product by opening up the floor differences that define a collision. Multiplication permutes the allowed residues. That change of variable brings out the same Bernoulli factor from each interior term. The terms at the two ends and the centering contribution can be accounted for exactly, leaving the diagonal boundary sum as the other factor. The paper keeps the sign and complex phase as well as the magnitudes shown here.

This is where the $L$-value enters. For a nontrivial character, the series

$$L(1,\chi)=\chi(1)+\frac{\chi(2)}2+\frac{\chi(3)}3+\cdots$$

converges through cancellation among the character values. A classical identity gives $|B_{1,\bar\chi}|=(b/\pi)|L(1,\chi)|$ for these primitive odd characters. The finite Bernoulli sum and the value of the infinite series are tied together exactly.

That identity is established mathematics. The collision factorization puts it inside a coefficient built from digit equality. We now know one of the two factors in every surviving coefficient through an $L$-value. The other factor still comes from the digit boundaries.

Base five

At base five, the two factors have an additional relation. The diagonal magnitude is exactly $\sqrt5$ times the Bernoulli magnitude, for all eight characters.

The reason is finite. The residue $2$ generates the twenty allowed residues modulo $25$. Character values can therefore be written as powers of a twentieth root of unity. The four terms from the digit boundaries become a short polynomial in that root. The Bernoulli sum becomes another. An identity between the two polynomials gives the magnitude relation.

Both factors now carry the same $|L(1,\chi)|$. Their product gives

$$|\widehat S^\circ(\chi)| =\frac{5\sqrt5}{4\pi^2}|L(1,\chi)|^2.$$

The digit function squares the $L$-value in this precise sense. Each of the eight coefficient magnitudes is the same fixed multiple of the corresponding squared $L$-value magnitude. There is no fitted constant.

The calculation that starts with 48

Fourier coordinates preserve the sum of squares, with a normalization for the size of the table. This is Parseval’s identity. For our twenty entries it says

$$\sum_\chi |\widehat S^\circ(\chi)|^2 =\frac{48}{20}.$$

Only eight coefficients contribute. Each is already quadratic in the magnitude of an $L$-value. Squaring the coefficient raises that magnitude to the fourth power. Substitution gives

$$\sum_{\substack{\chi\bmod25\\ \chi\ \mathrm{primitive\ odd}}} |L(1,\chi)|^4 =\frac{192\pi^4}{625}.$$

The sum runs over exactly those eight characters. A fourth moment here means this sum of fourth powers. It measures the size of a whole family of values at once.

The integer $48$ came from squaring fractions in a small table. The factors of $\pi$ enter through the classical relation between Bernoulli numbers and $L$-values. The square law and Parseval account for the rest. Every part of the answer has a place in the calculation.

Larger bases

The factorization holds at every odd prime base. The extra proportionality at five invites a separate question. How closely do the two factors track one another elsewhere?

For each prime base from $5$ through $71$, I calculated both finite sums for every primitive odd character and compared their magnitudes. A correlation of $1$ means the points lie on a line with positive slope. Smaller positive correlations indicate a looser tendency to rise together.

The correlation is $0.8440$ at base seven, $0.7203$ at base twenty-nine, and $0.6753$ at base seventy-one. It falls overall in this range, with several reversals along the way. The calculation covers $14{,}372$ character evaluations. Because the Bernoulli identity supplies the $L$-value magnitude from a finite sum, no cutoff in a prime sum enters these measurements.

The graph leaves the behavior at growing bases open. It also measures a different comparison from the overlap between collision weights and prime-sum weights in The Spectral Repulsion. Factoring a coefficient does not, by itself, prove the observed overlap or tell us where an $L$-function’s zeros lie.

The route from long division has kept a particular piece of arithmetic with it. Digit equality supplies a finite table. The table supplies a weight for each character. The factorization identifies what those weights contain. Reaching the $L$-functions gives us an exact way to use the weight, through a moment identity whose second factor comes from the digit boundaries.

At base five that factor has enough additional structure to evaluate the fourth moment outright. At larger bases, controlling it is still a mathematical question. The table gives us something specific to investigate within the classical theory.

I can check that table by hand. Following the same quantity through its Fourier coefficients takes it into a family of infinite series. The finite calculation reaches all the way through, with an exact value at the other end.