Carry Boundaries and Bernoulli Spectra

For a prefix of length $N$ in base $b$, put $m=b^N$. A weight $w(n)$ records the rule on cell $n$. It can be zero or one, or a more general numerical weight. Assume the denominator $q$ is coprime to the base and write $q=hm+a$, with $1\le a<m$.

The exact identity is

$$\begin{aligned} C_w(q)&=h\sum_{n=0}^{m-1}w(n)\\ &\quad+A_w(a)-w(m-1), \end{aligned}$$

where

$$A_w(a)=\sum_{n=0}^{m-1}w(n)\kappa_n(a),$$

$$\kappa_n(a)=\left\lfloor\frac{(n+1)a}{m}\right\rfloor-\left\lfloor\frac{na}{m}\right\rfloor.$$

Each floor difference counts integers in the corresponding right-closed interval. Coprimality prevents a fraction from landing on an interior cell boundary. The last interval includes $q/q$, which accounts for the subtraction. It is an endpoint correction.

At $q=11$, the selected cells are $G=\{0,4,8\}$, and $A_G(2)=2$. At $q=20$, the same $A_G(2)$ appears. The growing denominator changes the bulk count while the carry calculation stays finite.

With indices read cyclically, define

$$\mu_w(x)=w(x-1)-w(x).$$

An entrance has sign $-1$ and an exit $+1$. For $G=\{0,4,8\}$ the nonzero values are $\mu(1)=\mu(5)=1$ and $\mu(4)=\mu(8)=-1$.

The centered sawtooth is

$$\begin{aligned} s_m(n)&=\frac{n}{m}-\frac12\quad(1\le n<m),\\ s_m(0)&=0, \end{aligned}$$

extended periodically. Summing by parts gives the centered carry function

$$F_w(a)=-\sum_{x\bmod m}\mu_w(x)s_m(xa).$$

For unit residues $a$, meaning $\gcd(a,m)=1$, its relation to the floor count is

$$\begin{aligned} F_w(a)&=A_w(a)-\frac{a}{m}\sum_nw(n)\\ &\quad+\frac{w(0)-w(m-1)}2. \end{aligned}$$

The last correction is zero for our example, because both end cells are selected. The complete six-entry table follows, with no interpolation between entries.

The sawtooth reverses sign under reflection, so $F_w(-a)=-F_w(a)$. This is why its comparisons with even characters vanish.

A primitive character belongs to the full modulus rather than being inherited from a smaller one. An odd character satisfies $\chi(-1)=-1$. For such a character, define the ordinary boundary flux and the unnormalized carry transform by

$$S_w(\chi)=\sum_x\mu_w(x)\chi(x),$$

$$\widehat F_w(\chi)=\sum_{a\in(\mathbb Z/m\mathbb Z)^\times}F_w(a)\overline{\chi(a)}.$$

Then the exact factorization is

$$\widehat F_w(\chi)=-B_{1,\bar\chi}\,S_w(\chi),$$

where

$$B_{1,\bar\chi}=\frac1m\sum_{a=1}^{m}a\overline{\chi(a)}.$$

For a unit endpoint $x$, substituting $u=xa$ in the sawtooth transform gives $\chi(x)B_{1,\bar\chi}$. For a nonunit endpoint, the primitive character sum vanishes. Applying this to the boundary sum proves the formula. Even coefficients vanish by reflection.

In the drawing, $m=9$ and $\chi(2)=e^{\pi i/3}$. Multiplication by two visits $1,2,4,8,7,5$ and then returns to one. The arrow turns sixty degrees at each step.

The four boundary arrows sum to

$$\begin{aligned} S_G(\chi)&=1-\chi(4)+\chi(5)-\chi(8)\\ &=3-i\sqrt3. \end{aligned}$$

The floor factor is

$$-B_{1,\bar\chi}=1-\frac{i}{\sqrt3}.$$

Its angle is minus thirty degrees and its length is $2/\sqrt3$. Multiplying gives

$$\widehat F_G(\chi)=2-2i\sqrt3.$$

The lower path adds $F_G(a)\overline{\chi(a)}$ in the numerical order $a=1,2,4,5,7,8$. It produces the same complex number. The conjugate character supplies the reflected coefficient. These are the only two nonzero coefficients of this table.

At modulus twenty-seven the selected cells are

$$G_{27}=\{0,1,2,12,13,14,24,25,26\}.$$

The surviving endpoints are $3,12,15,24$, with signs $+,-,+,-$. All are divisible by three, so every ordinary character flux modulo twenty-seven is zero. The primitive carry transforms are zero too, but the induced carry transforms need not be.

In fact $F_{27}(a)=F_9(a\bmod9)$. Each unit class modulo nine has three lifts. Thus an inherited unnormalized coefficient is three times its old value. The average coefficient, divided by the number of unit classes, is unchanged. The figure uses those average coefficients. Each of the two active characters has power $4/9$, giving $8/9$ at conductor nine at every displayed resolution.

The distinction is essential. Ordinary fluxes determine exactly the boundary on units, because characters are zero on nonunits. They do not determine every induced carry transform. For a small counterexample, select cells zero and one modulo six. The two endpoints, zero and two, are nonunits. All ordinary fluxes vanish. Yet the odd character inherited from modulus three gives carry transform $1/3$.

At a prime power $m=p^N$, a character induced from a primitive character $\chi^*$ of conductor $p^t$ uses a depth-sensitive endpoint weight. For $x=p^vu$ with $p\nmid u$,

$$\eta_\chi(x)=p^v\chi^*(u)\quad\text{if }v\le N-t,$$

and the weight is zero at greater depth, including $x=0$. The complete induced formula is

$$\widehat F_w(\chi)=-B_{1,\bar\chi^*}\sum_x\mu_w(x)\eta_\chi(x).$$

The additional weight recovers the information that ordinary flux discards.

For an odd prime base $p$ and depth $N\ge2$, the first-equals-last rule has $2p^{N-1}$ formal endpoints. Its primitive odd mean is

$$\frac1{n_{\rm po}}\sum_{\chi\ \mathrm{primitive\ odd}}|S_G(\chi)|^2=4p^{N-1}.$$

Here $n_{\rm po}=p^{N-2}(p-1)^2/2$. The proof uses two properties of the unit boundary. Reflection reverses its signs, and each unit residue class at the next lower resolution receives equal entrance and exit mass.

For a general modulus, restrict $\mu$ to units and collect its signed mass modulo each divisor $d$. Write that collected mass as $M_d(r)$ and set

$$\begin{aligned} D_d&=\sum_r|M_d(r)|^2,\\ A_d&=\sum_rM_d(r)\overline{M_d(-r)}. \end{aligned}$$

The primitive odd total is given by the conductor ledger

$$\frac12\sum_{d\mid m}\mu_{\rm Mob}(m/d)\varphi(d)(D_d-A_d).$$

The symbol $\mu_{\rm Mob}$ denotes the Möbius function, not the signed boundary. Its signs perform inclusion-exclusion over conductors.

At base fifteen, $m=225$ and $G=\{0,16,32,\ldots,224\}$. There are sixteen nonzero unit-boundary entries. Their collected masses vanish modulo fifteen but not modulo forty-five or seventy-five. The only nonzero Möbius weights occur at the four divisors shown.

The total is $896$, across $32$ primitive odd characters. Dividing gives $28$. Negative ledger terms remove overlap; the squared responses themselves are nonnegative.

Adding the squared carry values defines an energy. Parseval gives its equivalent spectral expression,

$$E(F_w)=\sum_{a\in(\mathbb Z/m\mathbb Z)^\times}|F_w(a)|^2,$$

$$E(F_w)=\frac1{\varphi(m)}\sum_{\chi\bmod m}|\widehat F_w(\chi)|^2.$$

The six values in our modulus-nine table give $E=16/3$. On the spectral side, each of the two active coefficients has squared magnitude sixteen, so the same energy is $32/6$.

Expanding the boundary formula gives a second description in terms of sawtooth correlations. At unit endpoint indices, these correlations are finite Möbius combinations of classical Dedekind sums. The Bernoulli factor on the character side satisfies

$$|B_{1,\bar\chi}|=\frac{\sqrt m}{\pi}|L(1,\chi)|$$

for primitive odd characters. Thus the finite carry energy retains both the selected boundary and classical arithmetic factors.

These are finite identities. Growth estimates as the base or depth increases require further bounds. A rule on a whole remainder orbit also needs more than the fixed-prefix boundary constructed here.

Carry Boundaries and Bernoulli Spectra in Long Division supplies the proofs, the general weighted formulation and the conductor-depth energy decomposition. Its archived editions retain the research record. The finite-table background is developed in The Collision Invariant and The Collision Spectrum.