The Weight of a Carry

At modulus $m$, use the centered residue row

$$U_a(A)=\frac{[aA]_m}{m}-\frac{m-1}{2m}.$$

Assume $a$, $b$ and $a+b$ are all units modulo $m$. Each row then has variance $v_m=(m^2-1)/(12m^2)$.

On the full addition square, the two inputs and their reduced sum are pairwise independent and uniform. Their centered signed sum is the centered carry. Its variance is therefore $3v_m$.

On the synchronized line $(aA,bA)$, the carry is still zero or one and has the same mean. Its variance must also be $3v_m$. Expanding that variance gives

$$\begin{gathered} \langle U_a,U_b\rangle\ =\\ \langle U_a,U_{a+b}\rangle +\langle U_{a+b},U_b\rangle. \end{gathered}$$

Independence belongs to the full square, not to the synchronized line. The line’s rows are dependent. Their covariance is what the identity preserves.

For a continuous even density $\Phi$ with $\Phi(0)=0$, define

$$E_m^\Phi(f)=\frac1m\sum_A\Phi(f(A))$$

and

$$P_m^\Phi(f,g)=\frac{E_m^\Phi(f+g)-E_m^\Phi(f-g)}4.$$

Carry refinement is required for all real amplitudes and all unit triples at every odd modulus. The root triples at growing moduli give $\Phi(2t)=4\Phi(t)$. The root triple at five gives $\Phi(3t)=9\Phi(t)$. The quotient $\Phi(t)/t^2$, viewed on a logarithmic scale, has the two incommensurable periods $\log2$ and $\log3$. Continuity makes it constant. Nonnegative, nonzero cost gives a positive constant, and one variance calibration makes it one.

The nonlocal rule $E(f)=\|f\|_2^4$ also passes the unit-row carry tests. Thus carry refinement alone does not select squared size among every conceivable measurement. Nor does one five-entry table prove the all-amplitude theorem.

Let $B$ be the base, $\ell$ the lag, $Q=B^\ell$, $m=BQ$ and $N=Q-1$. The equal-digit cells are

$$\begin{gathered} G=\{d(Q+1)+By\},\\ 0\le d<B,\quad 0\le y<Q/B. \end{gathered}$$

Their centered table is

$$\begin{gathered} F(a)=\sum_{n\in G}\left( \left\lfloor\frac{(n+1)a}{m}\right\rfloor -\left\lfloor\frac{na}{m}\right\rfloor\right)\\ -\frac aB. \end{gathered}$$

Write $\mathfrak b(t)=\{t\}-1/2$ off the integers and zero at an integer. The exact identity is

$$F([(1-Q)A]_m)=2\sum_{r=1}^{N}\mathfrak b(rA/m).$$

At base five and lag one, the phase multiplier is $-4\bmod25$. All twenty-five phases, including zero, are retained in the figure. The plotted ramps are the prescribed affine functions, not an arbitrary interpolation between those samples. Their continuous capacity is

$$J(N)=\int_0^1\left(2\sum_{r=1}^N\mathfrak b(rx)\right)^2dx.$$

The finite sampled energy and this integral are different quantities. Their comparison requires an additional sampling estimate.

There is also an endpoint correction between ordinary centered residue rows and sampled midpoint sawteeth. For unit frequencies,

$$U_r(A)=\mathfrak b(rA/m)+e_m(A),$$

where $e_m(A)=1/(2m)-\mathbf1_{A=0}/2$. Omitting it leaves a refinement defect $(m-1)/(4m^2)$, equal to $1/25$ at five. The paper retains the correction before passing to the affine pairing. Higher-lag nonunit rows are covered by straightening, not by this unit-row endpoint formula.

For coprime positive $a,b$, the affine covariance is $1/(12ab)$. The source normalization makes its boundary mass

$$C(a,b)=\frac1{2ab}.$$

It obeys $C(a,b)=C(a,a+b)+C(a+b,b)$. This is the conserved half-length under classical Farey subdivision. The clock contribution is a different quantity,

$$K(a,b)=\frac{C(a,b)}{a+b}=\frac1{2ab(a+b)}.$$

Its period sum is

$$w_k=\sum_{\substack{a+b=k\\(a,b)=1}}K(a,b).$$

The source first admits a reduced pair at cutoff $k=a+b$. Larger cutoffs contain $\lfloor N/k\rfloor$ common-scale copies. Those labeled appearances force the address; conservation by itself does not.

The reciprocal kernel follows directly from the source ledger. It also has an independent uniqueness characterization. Among symmetric degree-minus-three kernels with a continuous normalized angular profile, conservation of the period-weighted boundary under this same subdivision forces the reciprocal form. Unit total mass fixes its constant. The unweighted clock contribution $K$ is not what adds under an individual Farey split.

Expanding the square gives $N/3$ from the individual waves. The cross-pair terms have a common factor $2/3$. Remove the former and divide by the latter. This gives $S(N)=3(J(N)-N/3)/2$, with $S(0)=0$. Its differences obey

$$\Delta S(n)=\sum_{k\mid n}kw_k.$$

This is a triangular recovery. Every proper divisor is smaller than $n$. Subtract its known contribution and divide by $n$.

In particular,

$$w_4=\frac{5/6-1/2}{4}=\frac1{12}.$$

For the kernel inverse, write

$$K_P(a,b)=\frac{P(a/(a+b))}{2ab(a+b)}.$$

The theorem assumes a symmetric $C^3$ profile, positive inside the interval, with unit total primitive mass. Divisor inversion restores full grid sums from primitive sums. After removing the endpoint singularity, grid averaging and Fourier inversion recover the profile. This step needs the complete infinite sequence and the stated regularity.

Symmetry is essential. An antisymmetric perturbation cancels inside every period sum. Fixed phase coordinates are essential too. Translating all waves leaves their squared integrals unchanged but moves their boundaries. The theorem recovers the regular symmetric primitive ledger, not unrestricted source data.

The displayed control uses primes $r=7$, $s=11$ and $\varepsilon=1/(8s^3)$. Set

$$\begin{aligned} h(u)={}&72\cos(2\pi u)\\ &+49\cos(14\pi u)\\ &-121\cos(22\pi u), \end{aligned}$$

and $P(u)=1+2\varepsilon u(1-u)h(u)$. The profile is symmetric, smooth and positive. Its primitive mass remains one. Grid averaging makes its unprimitive rows differ only at seven and eleven. Finite divisor inversion gives the exact changes of period mass and capacity. The first capacity difference is $7/15972$ at seven.

For the signed readout, first form the angular row

$$H_v(n)=\frac1n\sum_{k\mid n}k^3v_k.$$

Normalize its adjacent differences by $c_v=4v_2$, so $a_v(n)=(H_v(n+1)-H_v(n))/c_v$. Logarithmic marking and Dirichlet convolution inversion define

$$\mathfrak P_v(n)=n[(a_v\log)*a_v^{-1}](n).$$

For the collision weights, $a_w(n)=1/n$. The classical identity $\mu*\log=\Lambda$ then gives $\mathfrak P_w=\Lambda$. The whole current recognizes the collision mass. This supplies no new prime-distribution estimate.

The control first changes the normalized increment at six, one step before the first changed capacity. Its current there is $(147/5324)\log6$, while the collision current is zero. The sharp $n^2\log n$ sensitivity concerns a growing first changed address, with preceding increments and the scale held fixed. It does not say that a fixed coefficient is discontinuous or that signed sums cannot cancel.

Every occupied square is a coprime pair. Its diagonal address is the period $a+b$. Colors show the actual kernel contribution on one logarithmic scale across all four panels. The gaps are pairs with a common divisor. All cells are retained. These are the individual contributions whose period sums give the clock weights.