Carry Boundaries and Bernoulli Spectra

Long division begins with a remainder crossing an integer boundary.

Multiply the remainder by the base and divide by the denominator. The integer part gives the digit. The fractional part supplies the next remainder. As a function of its input, that fractional part rises between integers and resets at each crossing. Its graph is a sawtooth.

I came to this through digit collisions. Two digits agreed or they did not. Adding those agreements produced a finite table with an exact reflection law. The carry calculation reveals a common mechanism beneath that table, one that continues to work when the digit rule is changed.

Nine cells

Base 3 uses the digits 0, 1, and 2. Split the interval from zero to one into nine equal cells, one for each possible pair of digits. Equal digits select just three cells: the words $00$, $11$, and $22$, or cell indices $G=\{0,4,8\}$.

For denominator 11, the ten fractions $r/11$ land in those marked cells exactly at $r=1,5,6,10$. The count is four. The finite-prefix identity separates it into

The three is the bulk contribution from the three selected cells. The two is the carry count at the residue $11\bmod9=2$. The last term removes the terminal endpoint. Each term is an integer count.

The same floor calculation works for any rule on a fixed number of digits, provided the denominator and base have no common factor. It also allows different weights for different patterns. Once the bulk and endpoint are accounted for, the remaining count depends only on the remainder when the denominator is divided by the number of cells. Many different denominators therefore lead back to the same small carry table.

The boundary factor

A run of selected cells has an entrance and an exit. Add the differences across the run and every interior term cancels. Only the two endpoints remain. If $w(x)$ records the weight of cell $x$, its signed boundary is

An entrance contributes $-1$ and an exit $+1$. The three selected cells in the example have six formal endpoints. On the cyclic nine-cell table, the exit at 9 meets the entrance at 0 and cancels it. Four signed endpoints remain.

To read the table’s spectrum, compare it with regular arithmetic waves, called Dirichlet characters. Each comparison produces a coefficient measuring the table’s response to that wave. Some waves keep their value under reflection; others reverse sign. These are the even and odd characters.

The centered sawtooth reverses sign under reflection, and so does the carry table. All its even coefficients cancel. For a primitive odd character, belonging to the full modulus rather than inherited from a smaller one, the sawtooth response is the classical generalized Bernoulli number $B_{1,\bar\chi}$. I call it the floor potential. The Bernoulli number enters because it is the exact response of the sawtooth to that arithmetic wave.

Let $F_w$ be the centered carry table and let $S_w(\chi)=\sum_x\mu_w(x)\chi(x)$ be its boundary flux. With an unnormalized character transform, the exact factorization is

The equation says that each spectral coefficient is a product of two things: the floor’s response and the digit rule’s signed boundary. Change from digit equality to a transition, a three-digit agreement, or another fixed block pattern. The selected cells and the resulting table change, while the same Bernoulli factor remains. Different digit rules share this part of their arithmetic structure.

This is the finite Stokes principle in the calculation. Summing differences over the chosen cells leaves their edge. The classical sawtooth transform then reads that edge through a character.

Visibility and conductor

Arithmetic waves can come from different resolutions. A character’s conductor is the smallest modulus needed to describe it. Some characters use the full table; others come from a smaller one.

There is also a limit to what an ordinary boundary flux sees. Residues with no common factor with the modulus are called units; Dirichlet characters are zero on all the others. The complete family of ordinary fluxes $S_w(\chi)$ therefore determines exactly the boundary on units. Two rules have the same ordinary fluxes precisely when their unit boundary derivatives agree.

Carry transforms require an additional distinction. A character inherited from a smaller conductor can retain an endpoint that ordinary flux ignores. Modulo 6, selecting cells 0 and 1 leaves endpoints at 0 and 2. Both are nonunits. Every ordinary flux is zero, yet the odd character inherited from modulus 3 gives carry transform $1/3$.

At prime powers, an explicit conductor-depth weight accounts for this response. Lift the three selected cells from modulus 9 to modulus 27 by replacing each cell with three consecutive cells. All the new endpoints are divisible by 3. Primitive characters modulo 27 see zero, while the induced channels retain the response from conductor 9.

The boundary has not disappeared. Its response survives at the smaller resolution. Keeping track of conductors tells us where to find it.

An exact mean and its composite obstruction

For an odd prime base, select the words whose first and last digits agree. At depth at least two, each relevant lower-conductor class receives one entrance and one exit. Their signed masses cancel. Negation also exchanges entrances with exits.

Squaring the magnitudes of the primitive odd fluxes and averaging gives a measure of their collective size. Those two boundary properties determine that average exactly,

The formal boundary counts the interval endpoints before cyclic cancellation. In the nine-cell example it has size six. There are two primitive odd characters, each with squared flux magnitude twelve. Their mean is twelve.

Base 15 exposes the role of the lower conductors. Its two-digit collision support has fifteen cells and thirty formal endpoints. At modulus 225, the top divisor term contributes sixty to the mean. The proper divisors 75 and 45 contribute minus twenty and minus twelve. The divisor 15 contributes zero. Thus

These are the signed terms of a Möbius inclusion-exclusion formula. They remove the lower-conductor content when the primitive family is isolated. They are not negative energies. The exact total moment is 896 across 32 primitive odd characters, so the mean is $896/32=28$.

The same calculation explains both outcomes. Prime-power terminal boundaries satisfy the cancellation needed for the clean mean law. At base 15, two lower-resolution boundary sums survive. The ledger identifies both their locations and their exact effect on the answer.

From flux to energy

Square the magnitude of every carry value and add. This measures the table’s total energy without letting opposite signs cancel. Parseval says that the same total can be read from the arithmetic waves,

For the nine-cell table, the two primitive odd coefficients each have squared magnitude sixteen. The induced odd coefficient is zero. The energy is therefore $32/6=16/3$. At prime powers, grouping these nonnegative squared coefficients by conductor gives the energy decomposition.

On the geometric side, the same energy expands into sawtooth correlations. At unit indices those correlations are finite Möbius combinations of classical Dedekind sums. On the character side, each primitive odd coefficient contains a Bernoulli factor whose magnitude is $\sqrt m\,|L(1,\chi)|/\pi$. Both descriptions preserve the chosen boundary.

These identities determine the finite energy. Estimating its growth as the base or depth increases requires further control of the correlations and boundary moments. A whole remainder orbit also needs a boundary construction beyond a fixed prefix.

This gives me a way to compare different digit rules. I can mark equal digits, a prescribed transition, or a three-digit agreement, then work from the entrances and exits of the selected cells. Each rule has its own boundary. The floor calculation is shared.