The previous page on Polygonal Numbers introduced flat (plane) shapes made of dots that have fascinated mathematicians
since the times of the Ancient Greek mathematicians such as Pythagoras (around 500BC)
and Diophantus (around 250AD).
I recommend that you look at it first.
Here we extend the idea of counting the dots in dot patterns to
other ways of arranging the dots on a plane: the Central Polygonal Numbers
flat shapes made with sticks (matchsticks) instead of dots
shapes in 3D and even higher dimensions
The last two are called figurate numbers because they are not regular flat polygons in shape.
If the shape is not planar or is not a simple regular polygon, the number of dots (spheres etc)
is called a figurate number.
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Contents of this page
The icon means there is a
You Do The Maths... section of questions to start your own investigations.
The calculator icon
indicates that there is a live interactive calculator in that section.
Matchstick Numbers
Instead of using dots let's use matchsticks to make patterns.
Matchstick Squares
Here, for instance, are the Matchstick square numbers:
size
1
2
3
4
Square
number of matchsticks
4
12
24
40
Can you find the pattern in this sequence of numbers?
What is the next number?
What is a formula for the number of matchsticks in a square of side n?
There is one more way that we can visualise the Matchstick Numbers: as Aztec Diamonds:
size
1
2
3
4
Aztec
Diamond
A(1)
A(2)
A(3)
A(4)
number of squares
4
12
24
40
From top to bottom, each row (and each column) of A(n) has an even number of squares from 2 squares at the top to 2n in the middle
then 2n again in the middle shrinking down to 2 at the bottom and the rows are centred as shown. So each quadrant contains T(n) squares
where T(n) is the n-th Triangle number.
If we add in a diagonal in each square in the Aztec Diamond A(n) as shown here, then we can see that there is one diagonal per square and
the number of diagonals is the number of matchsticks (edges) in a Matchstick Square:
size
1
2
3
4
Square
number of matchsticks
4
12
24
40
If the centre of an Aztec Diamond of 2n rows is the origin (0,0) then
coordinates of the vertices
satisfy the condition that |x|+|y| ≤ n+1 if we exclude the
4 points (±n,0) and (0,±n).
Another way to express this is that the coordinates of A(n) are precisely those
(x,y) with x and y integers and
x −
1
+
y −
1
≤ n
2
2
|r| is the absolute value of r or the positive value of r ignoring the minus sign if r is negative.
You Do The Maths...
How many single edges of squares are there in A(n)?
How many ways are there to tile an Aztec diamond A(1) with dominoes (2x1 tiles)?
How many ways are there to tile an Aztec diamond A(2) with dominoes (2x1 tiles)?
What about the 6-rowed version?
There are 2 ways to tile the A(1) Aztec diamond with dominoes;
There are 8=2^{3} ways to tile A(2);
There are 64=2^{6} ways to tile A(3).
In general there are 2^{n(n+1)/2} ways to tile A(n).
(See the Elkies et al reference below)
The number of squares in A(n) is the longest leg in such triangles.
Right-angled triangles inside regular polygons
But there is another connection too! If we look at right-angled triangles made by joining vertices of a regular polygon, we notice
the same sequence appearing again:
In a square, the only right-angled triangles we can form from the vertices are those with one diagonal like the one on the left here.
Each pair of connected edges makes the two legs of a right-angled triangles, so there are 4 in a square.
Turning to hexagons, we can choose any of the two vertices opposite any edge to obtain a right-angled triangle. Since there are 6
edges and two choices of point opposite, there are 12 right-angled triangles on a hexagon.
This continues for 8, 10, 12, ... sided regular polygons and the number of right-angled triangles is 4, 12, 24, 40, 60, ...
Sums of Squares and Aztec diamonds
If we sum squares of consecutive integers, we have the following patern involving the same sequence, the number of squares in
an Aztec diamond:
3^{2} + 4^{2}
=
5^{2}
10^{2} + 11^{2} + 12^{2}
=
13^{2} + 14^{2}
21^{2} + 22^{2} + 23^{2} + 24^{2}
=
25^{2} + 26^{2} + 27^{2}
The pattern continues with the sum of the n squares ending at A(n)^{2} being the same as the sum of the squares of the next n−1 numbers.
The "other" Aztec Diamonds
All the rows of the Aztec diamonds contain an even number of squares.
What about those formed from an odd number of squares on each row?
size
1
2
3
4
Shape
number of squares
1
5
13
25
What is the next number in this series?
We will meet these numbers in the section below on Centred Polygons. These are the Centred Squares.
References
Aztec diamonds were named in Alternating-Sign Matrices and Domino Tilings (Part 1), N Elkies, G Kuperberg, M Larsen, J Propp
J Algebraic Combinatorics (vol 1 1992) (Part I) pages 111-132,
(Part II) pages 219-234.
Aztec Diamond Generator of random tilings
A Doeraene. This webpage can show either a random domino tiling such as is shown above on the right for A(100)
When we cover the Aztec diamond with a chessboard pattern, each domino covers one black and one white square.
The black square can be on the left or right of a horizontal domino (coloured red or blue)
and at the top or bottom of a vertical domino (in yellow or green).
The webpage can also give an approximation to the number of domino tilings very quickly.
For example,
for A(16) there are 87112285931760246646623899502532662132736 = 8.711×10^{40} = 2^{16×17/2} = 2^{136} tilings,
a number with 41 digits.
A(100) has 200 rows and the number of ways of tiling it with dominoes has 1521 digits.
The Arctic Circle Theorem is Mathologer's Christmas 2020 video
on YouTube (51 minutes) and is a nice presentation on how these domino tiling patterns work.
Matchstick Squares with Diagonals
If we squash our squares a little, we can make a rhombus (a 4-sided shape with all sides the same length)
and have a matchstick as a diagonal.
=
These give rise to many other patterns and sequences.
size
1
2
3
4
Square with Diagonal
Rhombic
number of matchsticks
5
16
33
56
Can you find the pattern in this sequence of numbers?
What is the next number?
What is a formula for the number of matchsticks in a square of
side n?
Surprisingly, these numbers are the octagonal numbers with negative size! The number of
matchsticks in a square of side r is p_{8}(–r).
You can also investigate what happens to p_{n}(–r) for the other
polygons with negative rank. Can we find an interpretation for these numbers in terms of diagrams?
Matchstick Triangles
Let's see what we get if we form equilateral triangles with the matchsticks:
size
1
2
3
4
Triangle
number
3
9
18
30
Can you find the pattern in this sequence of numbers?
What is the next number?
What is a formula for the number of matchsticks in a triangle of
side n?
3, 9, 18, 30, ... A045943
3 n (n+1) / 2 which we see are also just 3 T(n).
These numbers also have a special overlapping runsum property:
the n^{th} is the smallest number
that is a sum of both n-1 and of n consecutive numbers:
9
= 2 + 3 + 4
= 4 + 5
18
= 3 + 4 + 5 + 6
= 5 + 6 + 7
30
= 4 + 5 + 6 + 7 + 8
= 6 + 7 + 8 + 9
...
House of Cards
Once we have the formula for Matchstick Triangles, it is easy to find the formula for
a House of Cards:
tiers
1
2
3
4
House of Cards
number of cards
2
7
15
26
What is the next number?
What is a formula for the number of cards in a House of Cards of
n tiers (n levels high)?
2, 7, 15, 26, 40, 57, 77, ... A045943
are the Matchstick Triangle numbers
n (3n + 1) / 2
These numbers also have a runsum property: The n^{th} is the sum of the next n numbers after n:
1^{st}
2
= 2
2^{nd}
7
= 3 + 4
3^{rd}
15
= 4 + 5 + 6
4^{th}
26
= 5 + 6 + 7 + 8
...
Can you also write each of these House of Cards numbers as a pronic number + a triangle number?
Take n from the n^{th} House of Cards number. You should find a series that we have met before - but which is it?
n(3n+1)/2 = n(n+1) + n(n–1)/2
2−1=1, 7−2=5, 15−3=12, 26−4=22, ... and 1,5,12,22,...
are the Pentagonal Numbers
Matchstick Hexagons
We can extend our matchstick triangle patterns to hexagons:
size
1
2
3
4
Hexagon
number of matchsticks
12
42
90
156
Can you find the pattern in this sequence of numbers?
What is the next number?
What is a formula for the number of matchsticks in a hexagon of
side n?
Up to now, our dot shapes have grown outward from one corner. We have found many useful series and patterns that are used extensively
throughout mathematics. But we can also make polygonal shapes that grow
from the centre by adding a new layer all around the outside.
Here is a table:
Centred Triangles
Rank
1
2
3
4
5
Shapes
Counts
1
4
10
19
31
Centred Squares
Rank
1
2
3
4
5
Shapes
Counts
1
5
13
25
41
Centred Pentagons
Rank
1
2
3
4
5
Shapes
Counts
1
6
16
31
51
Centred Hexagons
Rank
1
2
3
4
5
Shapes
Counts
1
7
19
37
61
You Do The Maths...
Find the next number in each of the series in the tables above by looking at what we add to one
number to get the next. Can you see why this is so from the diagrams?
Find a formula for the Centred Triangle Numbers c_{3}(r),
Centred Square Numbers c_{4}(r)
and Centred Pentagonal Numbers c_{5}(r).
Find a formula for the centred n-gonal numbers c_{n}(r)
Can you find the formula for c_{n}(r) in terms of
Triangle numbers T(n-1) from this pattern which uses shapes of side 6?
Which polygonal shape corresponds to the following series of sums of odd numbers:
1
1+3+1
1+3+5+3+1
1+3+5+7+5+3+1
Express this pattern using polygonal numbers:
2^{2} + 1^{2}
3^{2} + 2^{2}
4^{2} + 3^{2}
5^{2} + 4^{2}
Here is an alternative pattern for the even-shaped centred polygonal numbers. Can you use it
express C_{2n}(r) algebraically?
* Note that the formulas in this table are not the same as those in the OEIS;
here we use r as the number of dots on an outside edge of the polygon
in order to be consistent with the planar non-centred p_{n}(r) formula.
Centred Polygonal Numbers c_{n}(r)
_{↓n}^{r→}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
3
1
4
10
19
31
46
64
85
109
136
166
199
235
274
316
4
1
5
13
25
41
61
85
113
145
181
221
265
313
365
421
5
1
6
16
31
51
76
106
141
181
226
276
331
391
456
526
6
1
7
19
37
61
91
127
169
217
271
331
397
469
547
631
7
1
8
22
43
71
106
148
197
253
316
386
463
547
638
736
8
1
9
25
49
81
121
169
225
289
361
441
529
625
729
841
9
1
10
28
55
91
136
190
253
325
406
496
595
703
820
946
10
1
11
31
61
101
151
211
281
361
451
551
661
781
911
1051
11
1
12
34
67
111
166
232
309
397
496
606
727
859
1002
1156
12
1
13
37
73
121
181
253
337
433
541
661
793
937
1093
1261
13
1
14
40
79
131
196
274
365
469
586
716
859
1015
1184
1366
14
1
15
43
85
141
211
295
393
505
631
771
925
1093
1275
1471
15
1
16
46
91
151
226
316
421
541
676
826
991
1171
1366
1576
16
1
17
49
97
161
241
337
449
577
721
881
1057
1249
1457
1681
17
1
18
52
103
171
256
358
477
613
766
936
1123
1327
1548
1786
18
1
19
55
109
181
271
379
505
649
811
991
1189
1405
1639
1891
19
1
20
58
115
191
286
400
533
685
856
1046
1255
1483
1730
1996
20
1
21
61
121
201
301
421
561
721
901
1101
1321
1561
1821
2101
The Centred Polygonal Number Calculator
C A L C U L A T O R : C e n t r e d P o l y g o n a l N u m b e r s
The game of Chinese Checkers in played on a star-shaped board
and gives us
another nice series of shapes variously called
the hexagonal star
the USA Sheriff's badge but some have 5 points and others 7
the Star of David
1
13
37
73
121
1, 13, 37, 73, 121, 181, 253 ... A003154
These numbers are also the Centred 12-gonal numbers.
Each 12-gon
is made up of 12 triangles plus an extra dot and can therefore be
can be transformed into the corresponding star shape: for example...
Centred Polygonal Numbers versus Corner Polygonal Numbers
There are several important differences between the ordinary or corner polygonal numbers
p_{b}(r) and the centred polygonal numbers
p_{b}(r).
Integer Representations
On the first Polygonal Numbers page we saw that Fermat's theorem states that
we can represent any integer as a sum of those polygonal numbers in many ways,
as a sum of 3 triangular numbers or a sum of up to 4 square numbers, or 5 pentagonal etc.
This is not true for
centred polygonal numbers as the numbers are too far apart.
For instance, with centred triangular numbers:
1, 4, 10, 19, 31, 46, ...
7 needs at least 4 centred triangle numbers: 7 = 4 + 1 + 1 + 1, and
17 needs at least 5 centred triangle numbers: 17 = 10 + 4 + 1 + 1 + 1.
For centred square numbers:
1, 5, 13, 25, 41, 61
9 needs at least 5 centred square numbers: 9 = 5 + 1 + 1 + 1 + 1, and
22 needs at least 6 centred square numbers: 22 = 13 + 5 + 1 + 1 + 1 + 1.
As sums of triangular numbers
The cornered n-gonal polygonal numbers are made up of n-2 (cornered) triangular numbers of rank
r-1 plus (a line of) r:
p_{n}(r) = r + (n – 2)
r(r–1)
2
p_{4}(6) = 2 T(5) + 6
p_{5}(6) = 3 T(5) + 6
p_{6}(6) = 4 T(5) + 6
p_{7}(6) = 5 T(5) + 6
p_{8}(6) = 6 T(5) + 6
whereas a centred n-gonal polygon is made up of n (cornered) triangular numbers of rank
r-1 plus an extra 1:
c_{n}(r) = 1 + n
r(r–1)
2
which is an easy way to remember their formulas.
The centred polygonal numbers formula is illustrated by these dot-diagrams from the
You Do The Maths... question above:
c_{3}(6) = 3 T(5) + 1
c_{4}(6) = 4 T(5) + 1
c_{5}(6) = 5 T(5) + 1
c_{6}(6) = 6 T(5) + 1
c_{7}(6) = 7 T(5) + 1
The Plain and Centred Polygonal Number Calculator
C A L C U L A T O R : P l a i n & C e n t r e d P o l y g o n a l N u m b e r s
Let's move into three dimensions and start off with at Pyramid Numbers since these are related to the
Polygonal Numbers where our 2-dimensional dots now become 3-dimensional balls!
As these 3D solid shapes are not planar, the dots become balls
and the numbers are called figurate numbers.
Pyramid Numbers with various base shapes
Here are three views of a stack of 5 squares making a Square-based Pyramid:
Square based Pyramid of 5 layers
Side view
Expanded view of layers
Top view
and here is a Pentagonal-based Pyramid of height 5:
The layers are just the plane centred pentagonal numbers we saw earlier:
side
1
2
3
4
5
shape
number of dots
1
5
12
22
35
You can see that each layer is one of our polygonal images.
A Pyramid of r layers has each of
the polygonal numbers with ranks from 1 to r.
We will denote these as P_{n}(r)
using a capital P for our 3D shapes and the small p
is used p_{n}(r) for the 2D polygonal numbers
we looked at earlier.
Since we stack r layers to make a pyramid of height (or rank)
r, we have the definition:
P_{n}(r)
=
p_{n}(1) + p_{n}(2) + ... + p_{n}(r-1) + p_{n}(r)
=
r(r + 1)( (n−2) r − n + 5)
6
=
(n − 2)r^{3} + 3 r^{2} − (n − 5) r
6
This can be simplified to give a formula known in the 5th century:
P_{n}(r) =
r + 1
(2 p_{n}(r) + r)
6
For the Polygonal numbers, we summed arithmetic series, accumulating sums from the beginning,
1, 2, 3, 4, 5, ... to get the series 1, 3, 6, 10, 15, ... The Triangular numbers 1, 3, 4, 7, 9, ... to get the series 1, 4, 9, 16, 25, ... The Square numbers 1, 4, 7, 10, 13, ... to get the series 1, 5, 12, 22, 35, ... The Pentagonal numbers
and so on
If we now do the same thing but sum the polygonal number series, we get the Pyramid Numbers.
If we stack Triangles on top of each other, we make a pyramid which has a Triangle on each layer, and so on.
1, 3, 6, 10, 15, ... to get the series 1, 4, 10, 20, 35, ... The Triangular Pyramid numbers 1, 4, 9, 16, 25, ... to get the series 1, 5, 14, 30, 55, ... The Square Pyramid numbers 1, 5, 12, 22, 35, ... to get the series 1, 6, 18, 40, 75, ... The Pentagonal Pyramid numbers
and so on
Here are some simple examples of Pyramid Numbers. How many objects in each picture?
Write each in P_{n}(r) form.
4=P_{3}(2)
5=P_{4}(2)
6=P_{5}(2)
14=P_{4}(3)
20=P_{3}(4)
3D Pyramid Numbers Calculator
C A L C U L A T O R : 3D P y r a m i d N u m b e r s
You are working in a greengrocers shop and have just arranged the new batch of oranges
into a polygonal pyramid of 3 layers. The shopowner says he now wants them
rearranging into a square-based pyramid.
You manage to arrange them all as he requires with none left over. How many oranges could you have had?
What was the polygon shape that you used in your original pile?
Why is it much easier to arrange round fruit in layers which are Hexagonal polygon numbers?
Make a list of all the Hexagonal Pyramid Numbers up to 100.
Check your answer with
A002412
Find a formula for the Pyramid Number P_{n}(r).
Cubes
Generalising the square into 3 dimensions is easy - the cubes!
1×1×1 = 1
2×2×2 = 8
3×3×3 = 27
4×4×4 = 64
The pattern for higher dimensions is shown n, n^{2}, n^{3}
so the next is n^{4} and so on with n^{D} for D dimensions.
These are called hyper-cubes.
The Centered Squares can be arranged in layers to make an octahedron (two square based pyramids joined at their bases).
The first 4 centered squares are arranged in a pyramid to make the octahedron with sides of 4 dots:
1
5
13
25
1+5+13+25+13+5+1=63
Here are the other smaller octahedral numbers
1
1+5+1 = 7
1+5+13+5+1 = 25
1+5+13+25+13+5+1 = 63
1, 7, 25, 63, 129, 231, 377, ...,
(2n−1)(2n^{2} − 2n + 3)
3
,
... A001845 where n is the number of dots on an outside edge of the octahedron.
Icosahedral Numbers
An icosahedron has 20 identical triangular faces, 30 identical edges and 12 vertices.
Each outer face contains a triangular number of r dots with r dots along each edge.
Each edge and all the vertex dots are shared with other faces.
The outer faces surround the next smaller icosahedron.
Here we classify dots as being
vertex: on a vertex of which there are always 12 except for the rank 1 icosahedron of a single dot
mid-edge: on an edge but not a vertex and so all these are on two faces
mid-face: wholly contained in a unique face, not on any edge and not vertices
rank #edge dots
1
2
3
4
5
Vertices
1
12
12
12
12
surface mid-Edge
0
0
30
60
90
surface mid-Face
0
0
0
20
60
total on surface
1
12
42
92
162
total
1
13
55
147
309
The rank r≥1 icosahedron has r balls along every edge.
Apart from the initial single ball, there are always 12 vertices.
From rank 3 there are 30 (r−2) blue balls that are on the edges but not vertices.
From rank 4 there are 20 T(r−3) red balls that are on the faces but not on an edge or vertex where T(r) is the triangle number of rank r.
Each icosahedron is made up on (new) surface dots with the previous icosahedron as internal points.
1, 13, 55, 147, 309, 561 ...,
(2r−1)(5r^{2}−5r+3)
3
, ... A005902
where r is the number of dots on each edge. (Note that the OEIS formula is for the length of the edges if the dots are a unit distance apart, so their n is our r-1.)
Rhombic Dodecahedron
If we put a cube and stick an identical cube on each of its 6 faces, we can joint the centres of the outer 6 cubes to the vertices of the central
cube to make a new solid shape with 12 faces:
Other views of the edges:
Each face is 4-sided with all sides the same length but its angles are not right-angles. This shape is called a rhombus.
It is formed from
the 8 vertices of the cube plus 6 extra vertices above the centre of each face of the cube giving a total of 14 vertices;
each face of the cube has 4 edges 'attached' making 24 edges of the rhombic dodecahedron
each edge of the cube is a diagonal on one face of the rhombic dodecahedron making 12 faces
Because it has 12 faces it is a dodeca-hedron but its faces are all rhombuses so it is called
a rhombic dodecahedron.
This shape can be found in nature as a crystal of garnet.
We can consider the shape to be formed from 6 cubes of equal size placed on the sides on another cube.
Those 6 cubes then have their centres joined to the vertices of the central cube on the face on which they sit.
If the 8 coordinates of the cube are (±1, ±1, ±1) and then the extra coordinates above the centre of each face
are (±2, 0, 0), (0, ±2, 0) and (0,0,±2). This gives
the edge length of the rhombic dodecahedron as √3.
The Rhombic Faces using the vertex coordinates above:
Each edge is of length √3.
Each face is a rhombus because all its edges are the same length.
Each rhombic face has one edge of the cube as
a diagonal (length 2) and the other is of length 2√2 = √8.
The faces' angles are therefore arccos( 1/3 ) and arccos( −1/3 ) or roughly
70.5288° and its complement 109.4712°.
The rhombic faces have area √8.
By making the cube a 3-by-3 array of points, we have a rank 2 rhombic dodecahedron (2 dots along each edge)
consisting of 27 points in the cube plus
6 points above the cube's faces, making 33 in all.
The number of dots in a rhombic dodecahedron with r dots alongs each edge is
1, 15, 65, 175, ... A005917 and the formula is r^{4}−(r−1)^{4}.
Vertices
1
14
14
14
14
14
surface mid-Edge
0
0
24
48
72
96
surface mid-Face
0
0
12
48
108
192
total on surface
1
14
50
110
194
302
total
1
15
65
175
369
671
Depending of which angle is used to view these, we can see both a square grid of dots and a hexagonal grid:
This shape is used by honey bees when constructing their honeycombs.
If we look at the rhombic dodecahedron with any vertex where 3 rhombuses meet at the centre of our view, we see an exact hexagon.
In 3D this is exactly what we see when looking down into an empty honeycomb cell.
Odd runsums
The rhombic dodecahedral numbers also have a nice runsum property:
We can form another solid by taking a cube and joining the mid-points of the 12 sides to make
triangle faces around the 8 vertices, leaving smaller squares on each of the cube's original faces as shown here.
We can use these 12 coordinates for the vertices: {0, ±1, ±1}, {±1, 0, ±1}, {±1, ±1, 0}.
The cuboctahedron has
12 vertices, 24 edges and 14 faces of which 6 are square and 8 are equilateral triangles.
The rhombic cuboctahedron has
12 faces, 24 edges and 14 vertices of which 6 are where 4 faces and edges meet and 8 are where 3 faces and edges meet.
The rhombic octahedron and the cuboctahedron are therefore the dual of each other.
Higher Dimensions
If objects with straight edges in 2D are called polygons and as a solid in 3D are called polyhedrons
then we use the term polytope for a
straight-edged shape in a dimension higher than 3.
Triangular numbers in more dimensions - Pascal's Triangle
The Triangular shapes of 2D and 3D are part of a more general pattern:
r:
1
2
3
4
5
6
p_{3}(r):
1
3
6
10
15
21
P_{3}(r):
1
4
10
20
35
56
We could even invent a 2D version since each row is formed by summing the items on the previous row from the beginning up to
that column, and continue with a Fourth Dimension and Fifth too, although it would be difficult to see how we
could represent these in a (2D) diagram:
If we look at the upwards diagonals in this table starting each in the leftmost column you might find that this new
triangle of numbers is one you have met before:
Pascal's Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...
So the formula for the entries in the Trianglular Numbers in D dimensions table above is