The Secondary Term of the Cubic Law

The cubic law established the leading growth of digit-collision energy. In an odd prime base $p$, the leading term is $p^3$. This paper identifies the first correction beneath it.

The omitted terms are smaller than the correction shown as $p$ grows through the odd primes. All logarithms here are natural.

There is no circle in the original count. There are integer remainders, digit bins, and the floor function that governs long division. Yet the first correction carries $1/\pi^2$.

That constant enters through coprimality, the arithmetic of pairs that share no common factor. Finding it in a continuous average is one part of the result. The other is proving that the finite digit table preserves the same coefficient. The averaging helps expose the correction; the sampling theorem establishes that it belongs to the original integer count.

The table beneath the count

The continuous average in the proof reduces to a finite square table of rational weights. Put the integers from 1 through $N$ along both sides. In each cell, divide the greatest common divisor of the two labels by their least common multiple.

	1	2	3	4
1	$1$	$1/2$	$1/3$	$1/4$
2	$1/2$	$1$	$1/6$	$1/2$
3	$1/3$	$1/6$	$1$	$1/12$
4	$1/4$	$1/2$	$1/12$	$1$

The diagonal is all ones. Away from it, the weights remember the ratio between the two integers after their common factor has been removed. Call the sum $A(N)$. At $N = 4$, the sixteen entries add to $23/3$.

There are $N^2$ cells, but their total grows only like $3N$. Hilberdink, Luca, and Tóth proved $A(N) = 3N + O((\log N)^2)$. That leaves room for a correction as large as the square of a logarithm, but does not say what its coefficient is.

The calculation here gives

The exponent $5/3$ is smaller than two, so the remaining error cannot absorb the displayed correction. The deficit below $3N$ grows asymptotically like a squared logarithm, with coefficient $3/\pi^2$.

The sieve inside the rounding

Take the pair $(2, 3)$. It can be scaled to $(4,6)$, $(6,9)$, $(8,12)$, and so on. Each copy carries weight $1/6$. In a table of size $N = 10$, three copies fit. The fourth does not.

Ten divided by three is $3\frac{1}{3}$. Replacing the whole-number count by that quotient adds a third of a copy. At weight $1/6$, the overcount is $1/18$.

Every reduced pair has a corresponding fraction. This paper keeps track of all of them. Together they build the correction.

For a fixed larger coordinate $k$, the smaller coprime coordinates supply a harmonic sum, roughly logarithmic in $k$. There is also a factor $1/k$. Adding those weighted logarithms through $N$ produces a squared logarithm. The fractional parts contribute a weighted average of one half, with smaller indices counting more heavily. Saffari and Vaughan’s theorem controls how these weighted fractions fill the unit interval. It makes the average of one half precise, with an error smaller than the squared logarithm.

Removing common factors by Möbius inversion introduces $1/\zeta(2) = 6/\pi^2$. Among pairs chosen independently from 1 through $N$, the proportion with no common factor tends to $6/\pi^2$ as $N$ grows. In this proof the same factor arises in an exact inclusion-and-exclusion sum. It weights the fractional remainders that build the correction. That is the source of $\pi^2$ here.

Together with the two sides of the square, the coefficients combine as

One half is the average fractional part. The other comes from summing $(\log k)/k$. The $6/\pi^2$ is the coprime filter. The 2 counts both sides of the diagonal. In the first example, rounding $3\frac{1}{3}$ down to 3 removed weight from the approximation. That is where the minus sign enters.

The digit table

I reached the square table through digit collisions. In base 5, the two-digit words with matching digits are

00   11   22   33   44

Among the twenty-five possible words, they sit at positions $0, 6, 12, 18, 24$. Five cells on the collision diagonal. The floor function records integer crossings through these cells. Subtract the uniform bulk contribution from each carry count, square the resulting responses, and sum over the twenty-five residues. This square mass is the collision energy $E_5 = 48$.

The collision energy and the square table are connected by an exact identity. Rewrite the floor counts as centered sawteeth and reorder the residues; the reordering preserves the energy. With $N=p-1$, averaging the squared response over the whole interval gives one third of $A(N)$. Average over the $p^2$ grid points, and you get $E_p/p^2$. Their difference is the sampling defect $S_p$.

At base 5, the continuous average is $23/9$. The grid average is $48/25$. Their difference is $143/225$.

The continuous calculation found the coefficient. The question is whether the finite grid preserves it.

The grid preserves it

Increasing the prime makes the grid finer. It also changes the function being sampled. More sawteeth enter the response, more jumps appear between grid points. A finer mesh on a fixed curve would be a simpler problem. Here both the mesh and the curve are moving. If the sampling defect grew like $(\log p)^2$, it could contribute at the same scale as the correction and change the coefficient. The proof must control that discrepancy before the continuous coefficient can be claimed for the digit table.

The paper proves

The $\log\log$ denominator grows slowly, but it grows without bound. The defect is asymptotically smaller than the correction scale. Two exact descriptions of the defect make the proof work. One follows the jumps of the response curve and their positions relative to the sample points. The other uses Dedekind sums and Rademacher reciprocity to express the same discrepancy through smaller moduli. Opening those smaller sums into centered inverse-residue series, and applying Korolev’s incomplete-sum estimate, supplies the strict saving. The divisor factors introduced by the descent have bounded average. That bounded average preserves the saving when all the smaller moduli are added back together.

The finite collision energy therefore has the expansion

The coefficient $-1/\pi^2$ survives. The continuous average determines it, and the sampling estimate carries it back to the finite digit energy. The two averages differ, but their difference is too small to change this term.

Below the correction

The computed sampling defect is smaller than the theorem requires.

Prime base $p$	Sampling defect $S_p$
31	0.951
101	1.027
401	1.011
4001	1.000
10007	1.009

The values move around one. The exact decomposition contains an endpoint contribution $(p-1)^2/p^2$, which tends to one, together with a signed residual and an explicit smaller correction. Proving that the residual vanishes would settle the limit $S_p \to 1$. That limit remains open. So do the stronger bounds $O(\log p)$ and $O(1)$.

The secondary coefficient is proved. Those questions concern the finer arithmetic beneath it.

The same floor operation that produces the digits also rounds the number of copies of each ratio. Those small losses accumulate. Coprimality fixes their coefficient, and the sampling theorem shows that the finite digit table retains it.

The cubic law gave the scale of the energy. The secondary term resolves the first departure from it. What had been left inside an error bound now has a coefficient and a reason.