The p-mark dial
Rotation k0
Reading0
Marks reached so far1
Bonus lesson 11 · Polygonal numbers · 60–75 minutes
Beyond Triangular Numbers
The triangular numbers 0, 1, 3, 6, 10, … are the totals of the simplest clock, growth 1. The squares 0, 1, 4, 9, 16, … are the totals of the clock with growth 2, the pentagonal numbers of growth 3, and so on. Every polygonal family is a growing clock with starting circumference 1. In this bonus lesson you find which dials such a clock covers completely, and count the marks on any dial.
Where this lesson is headed
Ps(k) = k + (s − 2)·T(k): the clock with N = 1 and g = s − 2
In words: on an odd prime dial p, an s-gonal clock reaches every mark when p divides the growth s − 2, and (p + 1)/2 marks otherwise. Prime powers and the power-of-two piece each have their own count, and CRT (Lesson 07) multiplies the pieces for any dial. Setup: the clock with starting circumference 1 and growth g = s − 2, read on a dial with m marks.
s: number of sides of the polygon (3 = triangle, 4 = square, …)
g = s − 2: the growth per rotation
p, m: the number of dial marks
Covering prime: an odd p dividing g, where every mark is reached
Square-type prime: an odd p not dividing g, where the count is the number of squares
Q(q): squares on a q-mark dial (Lesson 07)
01. Change the polygon, change the clock
Start where the course started: the triangular clock (s = 3, growth 1) on a 5-mark dial, which reads 0, 1, 3, 1, 0 and reaches 3 marks. Now change the polygon, which changes the growth, and change the dial. Look for the case where suddenly every mark is reached. (Shortcut: Space turns one rotation, ← undoes one.)
Guided experiments
Do these in order. "Set this up" picks the polygon and the dial and turns the clock for you.
Compare polygons and dials
Click a row to load it. What does the "p divides g?" column have to do with the last one?
| Polygon | g = s − 2 | p | p divides g? | Marks reached | All p? |
|---|
02. Why primes come in two kinds
Each step has the general idea on the left and numbers on the right. Most of the work was done in earlier lessons; here it is applied to a clock whose growth is s − 2.
1
Polygonal numbers are clock totals
The s-gonal numbers are usually written
Ps(k) = [(s − 2)k² − (s − 4)k] / 2
Regroup: (s − 2)(k² − k)/2 + k. That is
Ps(k) = k·1 + (s − 2)·T(k)
which is Lesson 01's S(k) with N = 1 and g = s − 2. Each new layer of the polygon adds s − 2 more dots than the last, just as each rotation adds g more centimetres.
2
Covering primes: when p divides the growth
If p divides g, the growth part g·T(k) is whole trips on the p-mark dial, and only the straight part k × 1 shows:
Ps(k) ≡ k (mod p)
So the readings are 0, 1, 2, …, p − 1: every mark, once each. This is the "g a multiple of p" case from Lesson 04's bigger picture.
3
Square-type primes: when p does not divide the growth
If p does not divide g, Ps is a true quadratic on the dial. Completing the square (with A = s − 2, B = s − 4):
Ps(k) = (A/2)(k − B/(2A))² − B²/(8A)
Shifting k, multiplying by the nonzero A/2 and adding a constant never change how many marks are hit, so the count equals the number of squares on the dial: (p + 1)/2 on a prime dial (Lesson 04), and Q(pe) on a prime-power dial (Lesson 07).
| Layer | Squares in it |
|---|
4
Covering primes on prime-power dials
On a pe-mark dial with p dividing g, Ps(k) is no longer just k, but it still hits every mark. Build a solution one digit (base p) at a time: if k gives the right reading on pj marks, try k + pjt. Then
Ps(k + pjt) ≡ Ps(k) + pjt · P′s(k) (mod pj+1)
where P′s(k) = Ak − B/2 reads 1 on the p-mark dial (since p divides A and B = A − 2). So exactly one t fixes the next digit. Every mark is reached exactly once: the count is pe.
5
The power-of-two piece
The odd-prime argument divides by 2, so dials with 2a marks need their own rule, read off from the growth g = s − 2:
g odd (s odd): every mark, over two blocks. This is Lesson 05's pattern: each block turns the dial half way.
g ≡ 0 (mod 4) (s ≡ 2 mod 4): every mark, by the same digit-by-digit lifting as step 4.
g ≡ 2 (mod 4) (s ≡ 0 mod 4, like the squares): the count is Q(2a), the number of squares on the dial.
Indexing matters. T(k) = k(k − 1)/2 skips one mark in its 1st block on 2a marks (Lesson 05), but the triangular clock P₃(k) = U(k) = T(k + 1) hits every mark in the same block. For pentagonal and heptagonal clocks the 1st block misses more and more marks as a grows, so there is no general "+1 bonus" for polygons.
Dial8
1st block
All marks ever
| s | g | 1st block | All ever | Missed by 1st block |
|---|
6
Any dial: split, count, multiply
Exactly as in Lesson 07: split m = 2a × p1e1 × ⋯, count each piece with steps 2 to 5, and multiply. CRT turns any choice of piece readings into one rotation k. For odd s use both blocks (2m rotations), since the power-of-two piece needs them.
Fs(m) = (2-part) × ∏ (pe or Q(pe))
| Piece | Kind | Count |
|---|
Product
Listed directly
Share of marks
03. Prove the count, and its limits
The count Fs(m) is about all marks ever reached. Keep it apart from what a single block of m rotations reaches, and do not assume a polynomial hits every mark just because a small test looks evenly spread.
Proof workshop
Complete proof.
1. Covering primes. Write A = s − 2 and B = s − 4 = A − 2. If an odd p divides A, then Ps(k) ≡ k and P′s(k) ≡ 1 (mod p). Each solution of Ps(k) ≡ r (mod pj) lifts to exactly one modulo pj+1 by replacing k with k + pjt. So all pe marks are reached.
2. Square-type primes. If p does not divide A, completing the square gives Ps(k) = (A/2)(k − B/(2A))² − B²/(8A) on pe marks. Shifting and multiplying by a unit do not change the count, so it is Q(pe).
3. The prime 2. If s ≡ 2 (mod 4) the polynomial lifts one-to-one, as in step 1; if s ≡ 0 (mod 4), completing the square with an odd inverse gives Q(2a). For odd s, each block of 2a rotations turns the readings by half the dial, so two blocks reach every mark.
4. CRT. For pieces that share no factor, choose a rotation class for each and combine them with CRT (odd s may need the full 2m rotations). Multiply the piece counts.
5. Why triangles are special. For odd s, the covering primes are exactly the odd primes dividing s − 2, and s = 3 has none. So among odd polygons the triangular clock reaches the fewest marks on every dial. This says nothing about 1st-block counts, or about even s. ∎
Computer check
Compares the formula with direct listing for polygons s = 3 to 18 on every dial from 1 to 160, checks prime powers separately, and checks the 1st-block gaps for s = 5 and 7 on powers of two. Many matches are strong evidence, but only the proof covers every case.
One misleading pattern
Compare s = 3, 5, 11 and s = 4, 6 on an 8-mark dial.
| s | g | 1st block | All ever |
|---|
Exit problem · use the classification
Predict each count first; then check with the experiments. All counts are numbers of different marks.
What you can now show
Every polygonal family is a growing clock with N = 1 and g = s − 2. On an odd prime dial it either reaches every mark (p divides g) or behaves like the squares. On any dial, the prime-power pieces, the power-of-two rule and CRT give the full count, and you can explain why a single block of rotations can hide marks.