Phase-Filtered Ramanujan Sums and the Spectral Gate

Write the twelve fractions from $1/13$ through $12/13$. Read their first decimal digits.

Ten digits, twelve fractions. Most digits appear once. Three appears twice, from $4/13$ and $5/13$. Six appears twice, from $8/13$ and $9/13$.

Now use the fifty-two fractions over 53. Every digit appears five times, except three and six, which appear six times.

The extra places stay in the same columns.

Both primes end in three. That ending fixes where the extra places go. I want to know whether it also fixes which digit comparisons come out empty.

Keep the two tables. We can test that question by multiplying their remainders.

The extra places

A digit bin collects remainders that emit the same first digit. At thirteen, the digit-three bin is $\{4,5\}$ and the digit-six bin is $\{8,9\}$.

For a prime $p$ not dividing the base $b\ge2$, write

Every bin has $q$ or $q+1$ members. Exactly $r-1$ bins get the extra member. We lose one of the $r$ extra places because remainder zero is excluded. Digit zero is not excluded. At fifty-three, five fractions begin with it.

The remainder $r$ determines which bins are larger. In decimal, primes other than two and five give four possibilities.

Last digit of the prime	Larger bins
1	None
3	Two
7	Six
9	Eight

The quotient grows. The pattern of extra places does not.

Multiply by two. Multiply by six.

Return to thirteen. Double each nonzero remainder, reduce modulo thirteen, and compare its digit before and after the move.

The bin $\{4,5\}$ goes to $\{8,10\}$. Neither result remains in the digit-three bin. The bin $\{8,9\}$ goes to $\{3,5\}$. Neither remains in the digit-six bin.

The other eight bins are singletons. A remainder could keep its digit there only by returning to itself. Doubling fixes no nonzero remainder modulo thirteen.

No matches anywhere.

Now multiply by six. Five goes to four, since $6\cdot5=30$ leaves remainder four. Both emit three. Eight goes to nine. Both emit six.

Call the number of matches $C(a)$, where $a$ is the multiplier. We have counted $C(2)=0$ and $C(6)=2$.

These counts use every nonzero remainder. A multiplier need not be a power of the base. When it is, the count adds the corresponding shifted matches across all remainder cycles. Only in the full-period case does one repetend supply the whole count.

Twelve, or minus one

The Autocorrelation Formula puts digit equality in a square table $K$. A cell is one when its two remainders emit the same digit. The zero row and column stay empty. Its Fourier transform $G(k,k')$ gives the same information in frequency coordinates.

A line through $G$ recovers $C(a)$. Ramanujan’s sum supplies another way to select that line, with two weights and a balance.

For a prime $p$, the sum $c_p(t)$ adds the unit arrows associated with all nonzero residues. When $t=0$ modulo $p$, all $p-1$ arrows point together. Otherwise they form every $p$th root of unity except the arrow at one. The complete set sums to zero. Leaving out that arrow gives minus one.

At thirteen,

This is the classical sum studied in Ramanujan’s 1918 paper. Here its argument is the combined frequency $k+ak'$. The collision identity is

Both frequency labels run from zero to $p-1$. The line $k+ak'=0$ receives weight $p-1$. Every other cell receives minus one.

The off-line cells have not been erased.

The transform and the collision identity

Let $B_d$ be the bin for digit $d$, and let $f_d$ be its indicator, with $f_d(0)=0$. Its Fourier coefficient is the sum of the bin’s rotating arrows,

$$\widehat f_d(k)=\sum_{x\in B_d}e^{-2\pi i kx/p}.$$

The equality table and its transform are

$$K(x,y)=\sum_{d=0}^{b-1}f_d(x)f_d(y),$$ $$G(k,k')=\sum_{d=0}^{b-1}\widehat f_d(k)\widehat f_d(k').$$

There is no conjugate in the second product. Fourier inversion gives

$$K(x,y)=\frac1{p^2}\sum_{k,k'}G(k,k')e^{2\pi i(kx+k'y)/p}.$$

Set $y=ax$ and sum over nonzero $x$. Each exponential becomes part of

$$\sum_{x=1}^{p-1}e^{2\pi i(k+ak')x/p}=c_p(k+ak').$$

The left side counts remainders that retain their digit. The right side is the Ramanujan-weighted collision identity. It holds for every nonzero multiplier, without a primitive-root assumption.

The full table $G$ sums to zero, because Fourier inversion gives $\sum_{k,k'}G(k,k')=p^2K(0,0)=0$. If the selected line sums to $S$, the rest of the table sums to $-S$. Apply the two weights,

Divide by $p^2$. The answer is $S/p$. The double sum has become a single line.

At thirteen with multiplier six, the line sums to 26. The remaining cells sum to $-26$. Their weighted total is $12\cdot26-(-26)=338$. Divide by $13^2=169$ and the answer is two.

There is no rounding in that answer. It counts $5\to4$ and $8\to9$.

Twenty-six survives

The selected line adds contributions from all the digit bins. Each contribution has a magnitude and a phase. Adding them can lose much of the magnitude to cancellation.

To measure that loss, take the magnitude of each bin contribution before adding across bins or frequencies. Call the unsigned total $M(a)$. With $k'=-a^{-1}k$ along the selected line,

The sum after cancellation is $S(a)=pC(a)$. Define the phase-gate coefficient by

The order of operations is essential. Taking $|G(k,k')|$ after its bins have already been added would miss cancellation between bins and give a different denominator.

At thirteen, multiplier six selects $k'=2k$. Its unsigned total is $144.5895015\ldots$. The signed sum is exactly 26.

Thus $\Gamma(6)=0.1798194\ldots$. About eighteen percent of the unsigned mass survives. That is not the fraction of remainders that match, which is $2/12$.

Work out the unsigned total at thirteen

Each singleton bin supplies a Fourier coefficient of magnitude one at every frequency. Eight such bins and thirteen frequencies contribute $8\cdot13=104$ to $M(6)$.

Each two-member bin contains neighbors. Their unit arrows differ in angle by $2\pi k/13$, so the sum has magnitude $2|\cos(\pi k/13)|$. Along $k'=2k$, the product of the two magnitudes is $4|\cos(\pi k/13)\cos(2\pi k/13)|$. Both two-member bins contribute the same amount.

Put

$$t_k=\left|\cos\frac{\pi k}{13}\cos\frac{2\pi k}{13}\right|.$$ $$M(6)=104+8\sum_{k=0}^{12}t_k.$$

The terms satisfy $t_{13-k}=t_k$. These seven values give the entire sum. The displayed decimals are rounded.

$k$	$t_k$
0	1.000000000
1	0.859726283
2	0.502996353
3	0.090223001
4	0.201438535
5	0.265425569
6	0.117034103

Count $t_0$ once and each of the other six twice. Their sum is $5.073687689\ldots$, giving

$$M(6)=144.589501512\ldots,$$ $$\Gamma(6)=\frac{26}{144.589501512\ldots}.$$

Only the decimal evaluation is approximate. The finite cosine sum specifies the denominator exactly.

Zero means no returns

The gate lies between two bounds,

Its denominator is always positive. The zero-frequency terms alone contribute the sum of the squared bin sizes. The gate therefore vanishes exactly when $C(a)$ does.

At the identity multiplier, nothing moves and the gate is one. At thirteen, doubling makes it zero. Multiplication by six puts it just above the lower bound $1/6$.

A zero has a literal meaning in the bins. Every bin moves completely away from itself. If even one remainder stays inside its bin, the gate is positive.

Why the bounds hold

The triangle inequality gives $|S(a)|\le M(a)$. This is the upper bound.

For the lower bound, multiplication by $-a^{-1}$ permutes the frequency labels. Cauchy–Schwarz and Parseval give, bin by bin,

$$\sum_k|\widehat f_d(k)|\,|\widehat f_d(-a^{-1}k)|\le p|B_d|.$$

Adding over bins yields $M(a)\le p(p-1)$. Since $S(a)=pC(a)$, division gives $\Gamma(a)\ge C(a)/(p-1)$.

When $a=1$, opposite-frequency coefficients are conjugates. Every product is a nonnegative squared magnitude. No cancellation occurs, so $\Gamma(1)=1$.

A positive collision count is an integer at least one. Hence a positive gate is at least $1/(p-1)$ at a fixed prime. That bound shrinks as $p$ grows. It supplies no uniform positive gap over all primes.

The companion paper, Phase-Filtered Ramanujan Sums and the Spectral Gate, gives the complete proofs and the Euclidean bin formula.

Four returns at fifty-three

Doubling leaves no return at thirteen. Try the same multiplier at fifty-three.

The digit-zero bin now contains $\{1,2,3,4,5\}$. Doubling keeps $1\to2$ and $2\to4$ inside it. The digit-nine bin contains $\{48,49,50,51,52\}$. Doubling modulo fifty-three keeps $51\to49$ and $52\to51$ inside that bin.

Those are all four matches. The same Ramanujan calculation gives a line sum of $53\cdot4=212$ and recovers $C(2)=4$. The gate is positive.

Six primes with the same two extra places

Every prime below ends in three. Every one has its extra places at digits three and six. The pictures show actual returns, one cell per remainder and multiplier.

Read a row to see which remainders keep their digit under one multiplication. Its number of colored cells is $C(a)$. An empty row means a zero gate. The bottom row is the identity, so it is completely filled.

These are integer comparisons in the remainder table, not frequency plots. Color names the retained digit, with one key shared by all six panels. The largest table has $502^2=252{,}004$ cells. All are retained in the full-size plate. Each row count is also checked against its Fourier slice and phase-gate bounds.

The bins, the moves, and the selector

These wider plates put the two bin populations alongside their multiplication tests and the Ramanujan weights. The selector’s gold cells carry twelve, not the spectral values being weighted.

The extra places are still at three and six. None of the four returns uses either bin.

They occur at zero and nine, where the smaller table had no room for them.