The Spectral Power of the Digit Function

Divide each of the numbers from one through twelve by thirteen. Read just the first decimal digit.

Two fractions begin with a three. Two begin with a six. Every other digit occurs once.

First digit	Numerators that produce it
0	1
1	2
2	3
3	4, 5
4	6
5	7
6	8, 9
7	10
8	11
9	12

These groups are the digit bins. Their sizes are

$$(1,1,1,2,1,1,2,1,1,1).$$

Ten integers, one per digit. At seven, six bins contain one remainder and four are empty. At twenty-nine, two bins contain two remainders and eight contain three.

The pair in the three-bin is 4 and 5. The pair in the six-bin is 8 and 9. Both pairs are neighbors.

I want to keep that small observation in view. The frequency curve we are about to draw can be accounted for by those four numbers.

Slide the bins

Start with the bin $\{4,5\}$. Add one to each remainder and count how many stay in the bin.

Four moves to five and stays. Five moves to six and leaves. There is one match.

Add two instead and both leave. With no shift, both stay. Shifting backward gives the same counts in reverse.

A bin of length three behaves in the same way.

Shift	−3	−2	−1	0	1	2	3
Remainders still in the bin	0	1	2	3	2	1	0

Plot those counts and a triangle appears. Each step loses one more match, until none remain.

Now add the counts from all the bins. Shifts wrap around modulo the denominator. At thirteen, passing twelve takes us to zero and then one. Zero belongs to none of these bins, so it never contributes a match.

At shift zero, all twelve remainders match themselves. At shift one, 4 moves to 5 and 8 moves to 9. Two matches. Shift backward and the same pairs match in reverse. Every other shift gives zero.

That is the whole additive collision profile at thirteen.

Two arrows

A Fourier transform gives us another way to look at this count. It needs a little construction.

Give each remainder an arrow of length one. At frequency one, the arrow turns by one thirteenth of a full turn as we move from one remainder to the next. At frequency six, it turns by six thirteenths, almost half a turn.

Within each bin, add the arrows. Two arrows pointing nearly the same way reinforce one another. Two pointing nearly opposite ways almost cancel.

The figure follows a bin containing two neighbors. The arrows have been turned together on the page so their sum points to the right. That changes neither their separation nor the length of their sum.

Square the length of each bin’s sum, then add those squared lengths across the bins. This is the spectral power, written $\Phi(k)$ at frequency $k$.

A bin containing one remainder contributes one at every frequency. There is nothing for its arrow to cancel against.

At thirteen, eight bins are singletons. They supply a constant eight. The two neighboring pairs supply the changing part. The calculation gives

$$\Phi(k)=8+8\cos^2\left(\frac{\pi k}{13}\right).$$

At frequency zero, all the arrows point the same way. Each pair has sum length two and contributes four. The total is $8+4+4=16$.

Near the middle frequencies, each pair almost cancels, and the total falls close to eight. The eight singleton arrows are still there.

The triangles and the curve

We can write the same formula as

$$\Phi(k)=12+4\cos\left(\frac{2\pi k}{13}\right).$$

Now the shift count is visible inside it. Twelve self-matches supply the constant term. Two forward matches and two backward matches supply the cosine.

At twenty-nine, the same counting gives 28 matches with no shift, 18 after one step, and 8 after two. Beyond two steps, no bin is long enough to retain a match. The counts in the other direction are the same.

The Fourier transform takes these counts to the frequency curve. Its inverse recovers every count. The operation changes how we read the information, while preserving the number of matches at every shift.

Two lengths are enough

The bins are consecutive because the first digit never decreases as the numerator increases. There is a second restriction. Their lengths differ by at most one.

In base $b$, divide the number of nonzero remainders, $p-1$, by the number of possible digits. Call the quotient $q$ and the remainder $\rho$.

$$p-1=bq+\rho,\qquad 0\le\rho<b.$$

Exactly $\rho$ bins have length $q+1$. The other $b-\rho$ have length $q$. At thirteen in decimal, this says $12=10\cdot1+2$. Eight singletons and two pairs.

The arrows in a consecutive bin form a geometric sum. Its squared length has a standard trigonometric expression. Since there are only two lengths, the general power formula has only two terms,

$$\Phi(k)=\tfrac{(b-\rho)\sin^2(\pi kq/p)+\rho\sin^2(\pi k(q+1)/p)}{\sin^2(\pi k/p)},$$

for $k=1,\ldots,p-1$, with $p$ prime and not dividing the base. The first term adds the contributions from the shorter bins. The second adds those from the longer bins. At frequency zero we count the aligned arrows directly, as we did at thirteen.

Monotonicity produces consecutive blocks. Consecutive blocks produce Dirichlet kernels. Squared magnitudes sum to $\Phi$. That is the whole construction.

These are classical Fourier tools, treated by Terras and Zygmund. Comparing equal symbols under shifts also belongs to the sequence theory of Lempel and Greenberger. The digit partition makes the calculation particularly short. Every bin contributes one of just two possible shapes.

Two totals

There are two sums to keep separate.

At frequency zero, each bin contributes its size squared. A bin with two members contributes four ordered pairs. They include each member paired with itself, as well as the two ways of ordering the distinct pair.

At thirteen, the zero-frequency value is 16. It is also the sum of the shift counts, $12+2+2$.

This squared-bin count appears in The Alignment Limit for All Primes when long division has a single remainder cycle. There it enters an average. Here we are keeping the raw count.

Adding the power across all thirteen frequencies gives a different total, 156. More generally, Parseval’s identity gives

$$\sum_{k=0}^{p-1}\Phi(k)=p(p-1).$$

For a fixed prime, that total stays the same when we change the base. The shape can change considerably.

The zero-frequency values are 72, 16 and 12. Each complete spectrum still sums to 156. Changing the bins redistributes the power.

The fixed total comes from something we already counted. With no shift, each of the twelve nonzero remainders matches itself, in every base. The sum across frequencies is thirteen times that self-match count.

A flat spectrum

At seven and eleven in decimal, every occupied bin is a singleton. Each contributes one at every frequency. The spectrum is flat.

The converse holds too. A flat spectrum with total $p(p-1)$ must have value $p-1$ everywhere, including at zero. But the zero-frequency count already contains the $p-1$ self-pairs. Any bin with two members would add distinct pairs and raise it above that value. So every occupied bin must be a singleton.

Digit separation, no matches at a nonzero shift, and flat spectral power describe the same boundary,

$$p\le b+1.$$

We have encountered that boundary through digits and through agreement. Now we can recognize it in the frequency curve as well.

The same bins, a different walk

The power records the length of each arrow sum and discards its direction. That discarded direction is the phase. Moving a bin rotates its sum without changing the power it contributes.

Multiplication also moves through the remainders differently from addition. At seven, compare these two walks beginning at one.

Base	Repeated multiplication, keeping the remainder modulo seven
8	$1\to1\to1$
10	$1\to3\to2\to6\to4\to5\to1$

Both bases give the same flat additive power, six at every frequency. One multiplication fixes the remainder. The other visits all six. The orbit comparisons in The Spectral Structure of Fractional Fields follow this multiplication walk.

The result here describes addition. It tells us exactly how many digit matches survive every shift, and it does so from the two bin lengths.

At thirteen, I can still account for the curve by pointing to the original table. Four and five share a digit. Eight and nine share another. Their forward and backward matches produce all the variation around twelve. The curve gives those few matches a frequency description. Reading it backward brings back the same pairs.