The Double Transversality

Write the sixteen fractions with denominator seventeen, first in base two and then in base three. In each row, compare the first two digits.

In binary, eight rows begin with a repeated digit. In ternary, four do. After subtracting the baseline and family mean, the binary count gives a weight of $+1/2$. The ternary count gives $-2/3$.

Add them.

$$\frac12-\frac23=-\frac16.$$

Most of the two weights disappear in the addition. Is that what happens to the two patterns as well?

Check the two counts at seventeen

The binary matches come from numerators $1,2,3,4$, which begin $0.00$, and $13,14,15,16$, which begin $0.11$. The ternary matches come from $1,8,9,16$, beginning $0.00$, $0.11$, $0.11$, $0.22$ respectively. These are digit prefixes, not terminating expansions.

For lag one, the baseline is $\lfloor(p-1)/b\rfloor$. At seventeen it is eight in binary and five in ternary. The relevant family means of the reduced table are $-1/2$ and $-1/3$.

$$f_2(17)=8-8-\left(-\frac12\right)=\frac12,$$

$$f_3(17)=4-5-\left(-\frac13\right)=-\frac23.$$

Here $f_2$ and $f_3$ denote the centered lag-one tables. The count uses all sixteen nonzero remainders, whether or not a single reciprocal visits them all.

Two rows, six columns

The binary weight depends only on the remainder modulo four. There are two admissible remainders, one and three, with weights $+1/2$ and $-1/2$.

The ternary weight depends on the remainder modulo nine. Its six admissible remainders are $1,2,4,5,7,8$. Put those across the columns and the two binary choices down the rows. Each cell holds the sum of its row’s binary weight and its column’s ternary weight.

Seventeen leaves remainder one modulo four and eight modulo nine. It occupies the upper-right cell.

Average across the top row. The binary part stays at $+1/2$ in every cell. The six ternary weights sum to zero. The row average is therefore $+1/2$.

The bottom row returns $-1/2$. The binary pattern is back.

Average down a column instead. Now the binary weights cancel and the ternary weight remains. All six ternary values come back, including the $-2/3$ at seventeen.

The Chinese remainder theorem supplies the rectangle. Every pair of admissible remainders occurs exactly once among the twelve unit classes modulo thirty-six. These are residue classes, not twelve consecutive primes. No prime distribution enters the recovery.

One sum, such as $-1/6$, cannot tell us which two numbers made it. The twelve sums together retain both patterns.

Read the twelve sums and their residue classes

mod 4 / mod 9	1	2	4	5	7	8	Row mean
1	7/6	11/6	7/6	−1/6	−5/6	−1/6	1/2
3	1/6	5/6	1/6	−7/6	−11/6	−7/6	−1/2
Column mean	2/3	4/3	2/3	−2/3	−4/3	−2/3	0

The row means are $+1/2$ and $-1/2$. The column means are the six ternary weights printed in the last row. The corresponding classes modulo thirty-six are

mod 4 / mod 9	1	2	4	5	7	8
1	1	29	13	5	25	17
3	19	11	31	23	7	35

For example, the two entries $-1/6$ and $-7/6$ in the last column average to $-2/3$. Averaging the six entries in the top row gives $+1/2$.

Three squares

A zero mean can conceal a substantial pattern. Square the weights before averaging and the signs no longer cancel. Call the result the table’s energy.

The binary table has energy $1/4$. For the ternary table, four entries have magnitude $2/3$ and two have magnitude $4/3$.

$$\frac{4(2/3)^2+2(4/3)^2}{6}=\frac89.$$

The twelve-cell sum has energy

$$\frac14+\frac89=\frac{41}{36}.$$

Expand the square of each cell. Besides the two squared weights, there is a cross term. Its average over the rectangle factors into the product of the two separate means. Both means are zero, so the cross term contributes nothing.

This is orthogonality, the same algebra behind the Pythagorean theorem. The vectors here are lists of collision weights. At seventeen, the two entries nearly cancel. Across the whole table, neither loses any energy to the other.

The lag is the distance between the digit positions being compared. The opening example uses lag one. The result holds for any finite collection of pairwise coprime bases, with a separately chosen positive lag in each. Give their tables complex weights if desired. The combined energy is the sum of the component energies multiplied by the squared magnitudes of those weights. Averaging over all the other coordinates recovers each weighted component.

The proof does not change when the tables get large.

Open the joint tables, from twelve cells to 746,496

Each panel adds a binary table down the rows to a ternary table across the columns. The lag pairs are $(1,1)$, $(3,2)$, $(5,3)$, $(7,5)$ and $(9,6)$. Row and column means recover the two component tables in every panel. Their product sizes grow, but no prime sampling is involved.

The cross term when bases share a factor

Let the two table moduli be $m$ and $n$. Their common remainder information lives modulo $d=\gcd(m,n)$. Average each table over the entries with the same remainder $u$ modulo $d$, calling those averages $A(u)$ and $B(u)$.

The cross term on the combined modulus $L=\operatorname{lcm}(m,n)$ is exactly

$$\frac1{\varphi(d)}\sum_{u\in U_d}A(u)\overline{B(u)}.$$

The sum runs over the units of the common modulus. For a fixed common remainder, the remaining choices form a complete rectangle. Their average product is $A(u)\overline{B(u)}$. Averaging these products over $U_d$ gives the formula.

Coprime moduli leave only the one-element common quotient, so the separate zero means suffice. Shared factors can retain nonconstant common information and produce a cross term. They do not have to. If either conditional average vanishes for every $u$, the cross term is still zero.

Two numbers attached to one character

Within a single base, the pairing is different. A Dirichlet character assigns consistent signs or complex phases to the residue classes. Read the centered collision table with it and obtain a coefficient $c_\chi$. Read the primes with the same character and obtain a finite sum.

$$P_X(\chi)=\sum_{m<p<X}\frac{\chi(p)}{\sqrt p}, \qquad m=b^{\ell+1}.$$

The collision sum is the sum of their products.

$$F_{b,\ell,X}(1/2)=\sum_{\chi\ \mathrm{active}}c_\chi P_X(\chi).$$

This finite identity lets us compare the two ingredients separately. For each active character, put the coefficient magnitude on the horizontal axis and the prime-sum magnitude on the vertical axis. A point far to the right and close to the bottom pairs a large collision coefficient with a small prime sum.

At ternary lag two, the four points give a positive Pearson correlation of about $+0.416$. At lag seven, the cloud gives about $-0.202$. The negative value means that larger coefficients tend to meet smaller prime sums in this window. It does not say that every point follows the tendency.

Conjugate characters have equal magnitudes in both coordinates. Each plotted point therefore represents two terms. Counting both, as the full character sum does, leaves these correlations unchanged.

There is no coordinate split like the one between coprime bases. The two quantities belong to the same character. The table fixes one; the primes determine the other.

The row that points the other way

Use all primes between the table modulus and two million. Test base three through lag seven, base five through lag four, and base seven through lag three.

Twelve of the thirteen defined correlations are negative. The positive exception is the ternary lag-two row just plotted. It stays in the calculation.

Base three at lag one gives no correlation value at all. Its two active coefficients both have magnitude $2/3$. There is no horizontal spread, and Pearson’s denominator is zero. Calling this zero correlation would turn a missing value into a result.

At ternary lag seven, there are $2{,}186$ active characters and one inactive odd character. The active terms supply the $1{,}093$ conjugate points above. Thus the negative association survives a substantial increase in the size of the table. The positive lag-two row prevents us from turning that observation into an all-lag sign rule.

Definitions and all fourteen numerical rows

For units $a$ modulo $m=b^{\ell+1}$, the coefficient is

$$c_\chi=\frac1{\varphi(m)}\sum_{a\in U_m}f_{b,\ell}(a)\overline{\chi(a)}.$$

Fourier inversion expresses $f_{b,\ell}(a)$ as $\sum_\chi c_\chi\chi(a)$. Substitute this into the finite prime sum and interchange the two finite sums. This gives the displayed pairing identity without any convergence assumption.

Even characters and those factoring through the base have zero coefficients. In these fourteen rows, they account for all the inactive odd characters. The numerical retention threshold is $10^{-10}$. The following values use $m<p<2{,}000{,}000$ and exponent $1/2$. Conjugates count separately.

Base / lag	Modulus	Primes	Active / inactive odd	Signed sum	Correlation
3 / 1	9	148,929	2 / 1	+0.618770	undefined
3 / 2	27	148,924	8 / 1	+1.182670	+0.416358
3 / 3	81	148,911	26 / 1	+0.616980	−0.419732
3 / 4	243	148,880	80 / 1	+4.825183	−0.275489
3 / 5	729	148,804	242 / 1	+12.770974	−0.326538
3 / 6	2,187	148,606	728 / 1	+29.145248	−0.326717
3 / 7	6,561	148,086	2,186 / 1	+10.357829	−0.201592
5 / 1	25	148,924	8 / 2	+1.034798	−0.451096
5 / 2	125	148,903	48 / 2	+4.009357	−0.176705
5 / 3	625	148,819	248 / 2	+2.482806	−0.277620
5 / 4	3,125	148,488	1,248 / 2	+15.256874	−0.190394
7 / 1	49	148,918	18 / 3	+1.812203	−0.276190
7 / 2	343	148,865	144 / 3	+7.564420	−0.216332
7 / 3	2,401	148,576	1,026 / 3	+4.583791	−0.156945

The signed sum adds $\operatorname{Re}(c_\chi P_X(\chi))$. The correlation instead compares $|c_\chi|$ with $|P_X(\chi)|$. It discards their phases.

In the ternary scatter plots, $2$ generates the units. The label $k$ means $\chi_k(2)=e^{2\pi i k/\varphi(m)}$. One member of each conjugate pair is plotted. At lag two the four representatives are $k=1,3,5,7$.

Five stopping places

Keep the deepest tested lag fixed in each base. Move the prime cutoff through a quarter-million, half a million, one million, two million and five million.

All three correlations become more negative through these five cutoffs. In base three, the value moves from about $-0.052$ to $-0.247$.

The signed sum has other ideas. In that same ternary calculation it goes from $+7.17$ to $-13.77$, then $-1.32$, $+10.36$ and $+4.62$.

The upper plot compares magnitudes. The lower plot adds contributions with their signs and phases. A steady trend in the first need not make the second steady. Here the correlation falls at every step while the sum changes sign twice.

The direct finite prime sums and their character decompositions agree in nfield. That checks the calculation at these cutoffs. It does not supply an estimate beyond them. Nor do prime-sum magnitudes measure distances to zeros of $L$-functions. No zero locations enter these plots.

Three companion views of the calculation

The open question is whether the negative association persists under a fixed normalization as lag and cutoff vary. The complete-table proof supplies no estimate for that pairing within a base.

Return to seventeen. The half and the two thirds leave $-1/6$, but neither disappears from the rectangle. Sweep across the row, then down the column, and both come back.

The table lets us choose the direction. The primes choose the order.