More on Polygonal and Figurate Numbers

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 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 You Do The Maths... icon means there is a You Do The Maths... section of questions to start your own investigations.
The calculator 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:
number of
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?

4, 12, 24, 40, ... A046092
2 n (n+1) So these are twice the Oblong numbers

The Aztec Diamonds

There is one more way that we can visualise the Matchstick Numbers: as Aztec Diamonds:
number of
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:
number of
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
|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...

  1. How many single edges of squares are there in A(n)?
  2. How many vertices (corners of single squares) are there in an Aztec Diamond of 2n rows ?
    1. How many ways are there to tile an Aztec diamond A(1) with dominoes (2x1 tiles)?
    2. How many ways are there to tile an Aztec diamond A(2) with dominoes (2x1 tiles)?
    3. What about the 6-rowed version?
    There are 2 ways to tile the A(1) Aztec diamond with dominoes;
    There are 8=23 ways to tile A(2);
    There are 64=26 ways to tile A(3).
    In general there are 2n(n+1)/2 ways to tile A(n). (See the Elkies et al reference below)

Pythagorean Triangles and Aztec Diamonds

The Aztec diamonds have a connection to the Pythagorean triangles with longest side and hypotenuse differing by 1. Examples of these are
(3, 4, 5)
(5, 12, 13)
(7, 24, 25)
(2n+1, 2n(n+1), 2n(n+1)+1)
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:
32 + 42 = 52
102 + 112 + 122 = 132 + 142
212 + 222 + 232 + 242 = 252 + 262 + 272
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?
number of
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.


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.
number of
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?

5, 16, 33, 56, ... A045944
n (3n + 2) = 4 p3(n) + p4(n) = 4 T(n) + n2
Surprisingly, these numbers are the octagonal numbers with negative size! The number of matchsticks in a square of side r is p8(–r).
You can also investigate what happens to pn(–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:
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 nth 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:
HoC1 HoC1 HoC1 HoC1
number of
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 nth is the sum of the next n numbers after n:
1st2 = 2
2nd7 = 3 + 4
3rd15 = 4 + 5 + 6
4th26 = 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 nth 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:
number of
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?

12, 42, 90, 156, ... A045945
3 n (3 n + 1)

Centred Polygonal Shapes

Centred shapes

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
Shapes tri tri tri tri tri

Centred Squares
Shapes tri tri tri tri tri

Centred Pentagons
Shapes tri tri tri tri tri

Centred Hexagons
Shapes tri tri tri tri tri

/ You Do The Maths...

  1. 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?
  2. Find a formula for the Centred Triangle Numbers c3(r), Centred Square Numbers c4(r) and Centred Pentagonal Numbers c5(r).
  3. Find a formula for the centred n-gonal numbers cn(r)
  4. Can you find the formula for cn(r) in terms of Triangle numbers T(n-1) from this pattern which uses shapes of side 6?
  5. Which polygonal shape corresponds to the following series of sums of odd numbers:
  6. Express this pattern using polygonal numbers:
    22 + 12 32 + 22 42 + 32 52 + 42
  7. Here is an alternative pattern for the even-shaped centred polygonal numbers. Can you use it express C2n(r) algebraically?

    2n = 4, r=6
    Square Centred

    2n = 6, r=6
    Hexagonal Centred

    2n = 8, r=6
    Octagonal Centred

Centred polygon formulas

Centred Polygonal Numbers cn(r)
r=1r=2r=3r=4r=5r=6r *OEIS *
Triangularn=31 4 10 19 31 46½ ( 3r2 – 3r + 2) A005448
Squaren=41 5 13 25 41 61½ ( 4r2 + 4r + 2)
= 2r2 – 2r + 1
Pentagonaln=51 6 16 31 51 76½ (5r2 – 5r + 2) A005891
Hexagonaln=61 7 19 37 61 91½ (6r2 – 6r + 2)
=3r2 – 3r + 1
n-gon n11+n1+3n1+6n1+10n1+15n ½(n r2 – n r + 2)
= n p3(r – 1) + 1
* 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 pn(r) formula.
Centred Polygonal Numbers cn(r)

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
Numbers common to several Centred shapes
simultaneously: -shaped
in the range

up to (optional)
Which polygonal shapes?

all Centred-polygon shapes for cn(r) =
exclude rank 2?

for centred -gonal


calculator: Centred Plain&Centred Pyramid

The Star Centred Figurates

The game of Chinese Checkers in played on a star-shaped board and gives us another nice series of shapes variously called

sq+2tri=hex sq+2tri=hex sq+2tri=hex sq+2tri=hex sq+2tri=hex
11337 73121
sq+2tri=hex sq+2tri=hex sq+2tri=hex sq+2tri=hex sq+2tri=hex
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 pb(r) and the centred polygonal numbers pb(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:
pn(r) = r + (n – 2) r(r–1)
= 2 T(5) + 6
= 3 T(5) + 6
= 4 T(5) + 6
= 5 T(5) + 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:
cn(r) = 1 + n r(r–1)
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:
= 3 T(5) + 1
= 4 T(5) + 1
= 5 T(5) + 1
= 6 T(5) + 1
= 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
Numbers common to several shapes

-shaped and
show their ranks too?
in the range 
up to


calculator: Centred Plain&Centred Pyramid

Solids in 3 dimensions

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 viewExpanded view of layersTop view
and here is a Pentagonal-based Pyramid of height 5:
side view top view
The layers are just the plane centred pentagonal numbers we saw earlier:
shape tri
number of
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 Pn(r) using a capital P for our 3D shapes and the small p is used pn(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:
Pn(r) = pn(1) + pn(2) + ... + pn(r-1) + pn(r)
= r(r + 1)( (n−2) r − n + 5)
= (n − 2)r3 + 3 r2 − (n − 5) r
This can be simplified to give a formula known in the 5th century:
Pn(r) = r + 1(2 pn(r) + r)

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 Pn(r) form.
oranges plums apples tins 1 tins 2
4=P3(2) 5=P4(2) 6=P5(2) 14=P4(3) 20=P3(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
pyramid numbers with shape n=
up to
and height r=
up to

calculator: Centred Plain&Centred Pyramid

/ You Do The Maths...

  1. 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?
  2. 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
  3. Find a formula for the Pyramid Number Pn(r).


Generalising the square into 3 dimensions is easy - the cubes!
1x1x1=1 2x2x2=8 3x3x3=27 444=64
1×1×1 = 12×2×2 = 83×3×3 = 274×4×4 = 64
The pattern for higher dimensions is shown n, n2, n3 so the next is n4 and so on with nD for D dimensions.
These are called hyper-cubes.

Cubes and Runsums

The cubes appear in this odds runsum pattern :
13 = 1 = 1
23 = 8 = 3 + 5
33 = 27 = 7 + 9 + 11
43 = 81 = 13 + 15 + 17 + 19
53 = 125 = 21 + 23 + 25 + 27 + 29


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 7 1+5+12+5+1=24 1+5+12+22+12+5+1=58
Here are the other smaller octahedral numbers
1 7 1+5+13+5+1=24 1+5+13+25+13+5+1=58
11+5+1 = 71+5+13+5+1 = 251+5+13+25+13+5+1 = 63
1 7 1+5+12+5+1=24 1+5+12+22+12+5+1=58
1, 7, 25, 63, 129, 231, 377, ...,
(2n−1)(2n2 − 2n + 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
#edge dots
on surface
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 ...,
, ... 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:
rhombic dodeca rhombic dodeca on cube
Other views of the edges:
view1 view2 view3
rhodod net
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
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 r4−(r−1)4.
rho dod rho dod rho dod rho dod rho dod rho dod
Vertices 11414141414
on surface
total 11565175369671
Depending of which angle is used to view these, we can see both a square grid of dots and a hexagonal grid:
sq grid sq grid
view3 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:
1 = 1
15 = 3 + 5 + 7
65 = 9 + 11 + 13 + 15 + 17
175 = 19 + 21 + 23 + 25 + 27 + 29 + 31
369 = 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49
671 = 51 + 53 + 55 + 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71



cubocta net
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:
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:
"Triangular" numbers in D dimensions
1:r123456...A000027 the natural numbers
2: p3(r)136101521...A000217 the plane triangle numbers
3:P3(r)1410203556...A000292 the triangular-based pyramid numbers
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  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
(D + r) = (D + r) = (D+r)(D−1+r)...(r)

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