The Secondary Term of the Cubic Law

Take the pair 2 and 3. The largest integer that divides both is 1. The smallest integer that both divide is 6. Give the pair a weight of $1/6$.

Now multiply both numbers by the same integer. Their greatest common divisor and least common multiple grow by the same factor, so the weight stays the same.

Pair	Common divisor / common multiple	Weight
(2, 3)	1 / 6	$1/6$
(4, 6)	2 / 12	$1/6$
(6, 9)	3 / 18	$1/6$
(8, 12)	4 / 24	$1/6$

Keep only pairs whose two coordinates are at most ten. Three copies fit. The fourth does not.

The count is three, although ten divided by three is $3\frac13$. Replacing the whole-number count by that quotient adds a third of a copy. At weight $1/6$, the overcount is $1/18$.

One eighteenth is easy to lose inside an approximation. This paper keeps track of it, along with the corresponding fractions from every other pair. Together they produce the first systematic correction to a classical sum. The same correction then has to survive a return to the finite arithmetic of digit collisions.

Inside the square

Put the integers from 1 through $N$ along both sides of a square table. In each cell, divide the greatest common divisor of the two labels by their least common multiple. The first four rows and columns look like this.

	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. The entries at $(1,2)$ and $(2,4)$ agree for the same reason our first family did.

Call the sum of the entire $N$-by-$N$ table $A(N)$. At $N=4$, the sixteen entries add to $23/3$.

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

The new calculation gives

Here and below, logarithms are natural. The exponent $5/3$ is smaller than two, so the remaining error is too small to absorb the displayed correction.

The shortfall grows as the table grows, even while it becomes a smaller fraction of the total. Knowing the leading scale and knowing what it leaves out are different levels of accuracy.

The fractions left out

Group the cells by their reduced pairs. A pair $(a,k)$ with no common factor, and with $a<k$, can be repeated at scales

Exactly $\lfloor N/k\rfloor$ copies fit in the square. Each carries weight $1/(ak)$. The reflected pair $(k,a)$ supplies the same contribution on the other side of the diagonal.

This is where rounding enters. Write the number of copies as $N/k$ minus its fractional part. The first piece leads to the main term. The second records what that approximation counted beyond the last whole copy.

An exact identity separates the shortfall from $3N$ into two nonnegative pieces. One comes from the tail of the infinite sum used for the main approximation. The other comes from these discarded fractional copies. The tail grows at most logarithmically. The larger, logarithm-squared correction comes from rounding.

Its size has a recognizable shape. For a fixed larger coordinate $k$, the smaller 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. Making that average precise uses Saffari and Vaughan’s theorem on fractional parts. Removing common factors by inclusion and exclusion introduces $1/\zeta(2)=6/\pi^2$. Together with the two sides of the square, the coefficients combine as

The two halves have different origins. One is the average fractional part. The other comes from summing $(\log k)/k$. The proof controls the errors in both steps. The minus sign was already visible in the first example, where rounding three and a third down to three removed weight from the approximation.

The digit table underneath

I reached this sum through digit collisions. In base five, the two-digit words with matching digits are

00   11   22   33   44

Among the twenty-five possible words, they occupy positions $0,6,12,18,24$. These five cells form the collision diagonal.

The floor function records integer crossings through the selected 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. At base five, its exact value is $E_5=48$.

The paper rewrites the floor counts in terms of fractional parts. Each fractional part rises and resets, giving a sawtooth. After centering these sawteeth and reordering the residues, the collision response becomes a sum of them. The reordering preserves the energy exactly.

This gives two ways to average the squared response. Average over the whole interval, and the result is one third of the GCD/LCM sum. Average over the $p^2$ equally spaced residue points, where $p$ is the odd prime base, and the result is $E_p/p^2$.

At base five, the continuous average comes from the four-by-four table above. It is $23/9$. The average over the twenty-five residue points is $48/25$. Their difference is

Call this difference $S_p$, the sampling defect. For every odd prime, the relation is exact.

The continuous calculation has supplied the correction we want. The remaining question is whether the finite grid changes it.

Between the sample points

Increasing the prime makes the grid finer. It also changes the function being sampled. More sawteeth enter the response, and more jumps appear between the grid points. Ordinary intuition about taking a finer sample of one fixed curve is insufficient here.

A bound of order $(\log p)^2$ for the sampling defect would be too large. That is the very scale of the correction after dividing the energy by $p^2$. The grid could contribute another term of the same size and change the coefficient.

The paper identifies the discrepancy in two exact ways. One follows the jumps of the curve and their positions relative to the sample points. The other uses Dedekind sums, classical sums built from pairs of sawteeth. Rademacher reciprocity expresses the same discrepancy through smaller moduli.

Those smaller sums contain modular inverses. An inverse of 3 modulo 7, for example, is 5, because $3\times5$ leaves remainder 1 on division by 7. As the input runs through the invertible residues, its inverse rearranges them. A complete centered sum has mean zero, but a floor cutoff can stop partway through that rearrangement.

The incomplete sums are the difficulty. Some can stay on one side of zero. Taking the absolute value of every term too early loses the cancellation needed to distinguish the secondary coefficient.

The proof retains the additive phases, reduces each phase to its actual modulus, and uses inverse-residue estimates, including Korolev’s incomplete-sum theorem. Returning from those smaller moduli introduces divisor factors. Their bounded average keeps them from undoing the saving.

The resulting estimate is

The denominator grows slowly, but it grows without bound. That is enough. The sampling defect is smaller than the logarithm-squared scale, so it cannot alter the coefficient found in the continuous calculation.

Substitute into the exact relation between the two averages. The finite collision energy has the expansion

The coefficient survives. The square table determines it, and the sampling estimate carries it into the finite digit energy unchanged.

Below the correction

The computed sampling defect is smaller than the theorem currently requires. Some values from the manuscript are

Prime base $p$	Sampling defect $S_p$
31	0.951176
101	1.026972
401	1.011478
4001	0.999951
10007	1.008630

They 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 disappears would settle the limit $S_p\to1$.

That limit remains open. So do the stronger sampling bounds $O(\log p)$ and $O(1)$. The secondary coefficient is already proved; those questions concern the finer arithmetic beneath it.

The family that began with $(2,3)$ still fits only three times in a ten-by-ten square. Every such integer cutoff leaves a fractional remainder. Taken together, those remainders account for the first correction below the cubic law, and the sampling theorem proves that the digit table preserves it.