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### Figurate and Polygonal Numbers

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Date: 11/21/98 at 18:24:55
From: Megan
Subject: Figurate numbers

I need to know everything about figurate numbers - no one here knows
anything.
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```
Date: 11/22/98 at 15:48:54
From: Doctor Floor
Subject: Re: Figurate numbers

Hi Megan,

Figurate numbers are number sequences that are found by creating
consecutive geometrical figures. Most well known are so-called polygonal
numbers, and most well known among these are squares.

Polygonal numbers are found by making consecutive nested polygons.
When we take squares as an example, they can be created this way:

1   1 2    1  2  3  (the numbers should be thought of as numbered dots)
2 2    2  2  3
3  3  3

The first square has 1 dot, so the first square number is 1.
The second square has 4 dots, so the second square number is 4.
And the third is 9.
Of course you know the formula for square numbers is f(n)=n^2.

For bigger polygons the type of nesting must be considered carefully.
I will give pentagonal numbers as an example:

Pentagonal numbers are the numbers of dots you need for consecutive
nested pentagons (figures with five sides).

Of course the first pentagon is 1 dot, so the first pentagonal number
is 1.

The second pentagon has sides of two dots. The first pentagon must be
included:

2  2
1
2  2

So the second pentagonal number is 5.

The third pentagon has sides of three dots. For the third pentagonal
number, the first two pentagons must be nested inside this one (the

3  3  3
2  2
1         3
2  2
3  3  3

So the third pentagonal number is 12.

As a final example I will give the fourth pentagonal number. First
create the figure:

4  4  4  4
3  3  3
2  2         4
1         3
2  2         4
3  3  3
4  4  4  4

We see that the fourth pentagonal number is 22.

So we see that the polygons are nested in one vertex. This is true for
all polygonal numbers. Other types of figurate numbers are acquired by
other types of nesting, but I don't know about them.

We can make a general formula that gives the rth n-gonal number (where
n is the number of vertices in the polygon):

n-2   2   n-4
---  r  - --- r
2         2

or

0.5r[(n-2)r-(n-4)]

To derive the formula, first note that it must be quadratic. Then
consider the 0th, 1st and 2nd n-gonal numbers for some n. They are very
easy:

0   1   n

Now say the formula for the rth n-gonal number is

F_n(r) = ar^2 + br + c.

Since F_n(0) = 0 it follows that c = 0.

F_n(1) = 1 gives a+b = 1, or a = 1-b [1].

F_n(2) = n gives 4a+2b = n [2].

Inserting [1] in [2] gives 4(1-b)+2b = n,        so
-2b = n-4       and
b = -0.5(n-4) [3].

Inserting [3] in [1] gives         a = 1+0.5(n-4)
= 0.5(n-2).

And we find the formula f_n(r) = 0.5r[(n-2)r-(n-4)] as given above.

I hope this helps!

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Sequences, Series